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Where two lines meet
A pair of lines that are not parallel and not the same line share exactly one point, and that point is the pair’s answer.
A STORY
Where the two tracks meet
Tara and Arjun stand by the railway, a barrier beside them. Two tracks come in
from different directions. They cross once, then run on past each other.
“Only one crossing,” Arjun says, pointing at the tracks. “What if the tracks ran side by side
instead? Would they ever cross?”
“Never,” Tara says. “Two lines like that stay the same distance apart forever.”
“And if the tracks were laid exactly on top of each other?” Arjun asks.
“Then every point is a crossing,” Tara says. “Infinitely many crossings, not one.”
“So two straight lines only ever give one of three answers,” Arjun says. “One crossing, no
crossing, or every point.”
“Yes,” Tara says. “And you can tell which one, before you draw a single line.”
Which of the three would your own two equations give, before you plot either line?
Let us take the pair $x + y = 5$ and $x - y = 1$. Plot both as straight lines on the same graph. They meet at exactly one point.
Now nudge the second line, in your mind, until it runs exactly parallel to the first. It never meets the first line again. That gives no solution.
Slide it once more, so it lands exactly on top of the first line. Every point on that line is now a shared solution. That gives infinitely many solutions.
A pair of linear equations in $x$ and $y$ is nothing but two such lines. It always lands on exactly one of three outcomes: one solution, no solution, or infinitely many.
Try it yourself. Sketch $x + y = 5$ and $x - y = 1$ on paper, and find where they cross. (Answer: the point $(3, 2)$.)
Which of the three holds can be decided before you plot a single point. We compare three ratios instead: $a_1/a_2$, $b_1/b_2$, and $c_1/c_2$, read straight off the two equations’ own coefficients. This chapter builds that comparison first, then two algebraic methods that solve any such pair directly, and closes on situational problems that turn ordinary words into exactly this kind of pair.
Every method in this chapter finds the same point: wherever the two lines actually meet.
Three pairs of lines are shown side by side: one crosses once, one never meets, and one is the same line drawn twice.
The crossing is read straight off the axes, and the line underneath puts those two numbers back into both equations as a check.
Check yourself
A pair of linear equations in $x$ and $y$ is really a pair of straight lines. How many outcomes are possible for such a pair?
always exactly one solution, since two lines must cross somewhere
either no solution or infinitely many, never a single point
as many solutions as there are values of $x$
exactly three: one solution, no solution, or infinitely many
Check your answer
always exactly one solution, since two lines must cross somewhere — Two lines need not intersect at all — they can also run parallel or lie on top of each other.
either no solution or infinitely many, never a single point — Two lines with different slopes intersect at exactly one point, giving a unique solution.
as many solutions as there are values of $x$ — One equation alone has infinitely many solutions, but the PAIR narrows this to one of just three outcomes.
✓ exactly three: one solution, no solution, or infinitely many — (D) A pair of straight lines can only cross once, never, or lie on top of each other, so exactly three outcomes are possible.
Before drawing either line, what tells you whether a pair of linear equations will have one solution, none, or infinitely many?
comparing the signs of $a_1$ and $a_2$ alone
comparing the ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$
comparing which equation has the larger constant term $c$
counting how many terms each equation has
Check your answer
comparing the signs of $a_1$ and $a_2$ alone — The sign of a coefficient carries no information here; only the ratio between the two equations’ coefficients matters.
✓ comparing the ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ — (B) The three ratios decide the outcome before anything is drawn or solved.
comparing which equation has the larger constant term $c$ — It is the ratio $c_1/c_2$ that matters, never the raw size of either constant term.
counting how many terms each equation has — Every linear equation in two variables has the same three terms by its general form; counting them tells you nothing new.
On a graph, two lines cross at exactly one point. Which outcome does this correspond to?
infinitely many solutions, since the lines meet
no solution, since the lines are not parallel
a unique solution, and the pair is consistent
a solution that cannot be found algebraically
Check your answer
infinitely many solutions, since the lines meet — Infinitely many solutions needs the lines to coincide entirely, not merely cross at one point.
no solution, since the lines are not parallel — Lines that are not parallel are exactly the ones that DO cross, giving a solution rather than none.
✓ a unique solution, and the pair is consistent — (C) A single crossing point is exactly the unique-solution, consistent case.
a solution that cannot be found algebraically — Substitution and elimination find precisely the same point a correct graph shows — the methods agree.
You have met every tool you need here, one at a time. Now two lines share one graph. Try each check below. It takes a minute.
Coordinates of a point (Class 9, Ganita Manjari, page 6). Here: every solution of an equation like $x + y = 5$ is a point you can plot. Check: $(2, 3)$ is on the line $x + y = 5$, since $2 + 3 = 5$.
A line from two points (Class 9, Ganita Manjari, page 27). Here: the graphical method draws each equation from two solutions. Check: On $y = 2x + 1$, $x = 0$ gives $y = 1$ and $x = 3$ gives $y = 7$.
The slope of a line (Class 9, Ganita Manjari, page 31). Here: two different lines with the same slope never cross. Check: $y = 2x + 1$ and $y = 2x - 1$ stay $2$ apart at every $x$.
Solving a linear equation (Class 9, Ganita Manjari, page 20). Here: substitution and elimination each end in one such equation. Check: $20 - 3y = 5$ gives $3y = 15$, so $y = 5$.
Words into an equation (Class 9, Ganita Manjari, page 20). Here: a word problem now needs two letters, $x$ and $y$, and two equations. Check: Two numbers add to $64$, and one is $10$ more. They are $27$ and $37$.
Ratios in proportion (Class 8, Ganita Prakash Part 1, page 162). Here: the ratio test compares $a_1/a_2$ with $b_1/b_2$ in just this way. Check: $2 : 4$ and $3 : 6$ are both $1 : 2$, so they are in proportion.
Same step, both sides (Class 8, Ganita Prakash Part 2, page 141). Here: elimination multiplies both sides by a number to match one letter. Check: $x + y = 5$ times $2$ is $2x + 2y = 10$, and $(2, 3)$ fits both.
If any of these felt new, read the page named before going on.
One line carries six dots as sample solutions, and a second line crossing it marks the single pair that fits both.
Let us take the equation $x + y = 5$. Try the pair $(1, 4)$. Substitute it in: $1 + 4 = 5$. It checks out, so $(1, 4)$ is a solution.
Now you try one. Check the pair $(2, 3)$ in the same equation. (Answer: $2 + 3 = 5$, so yes, it is a solution.)
Two more pairs work too: $(0, 5)$ and $(5, 0)$. A linear equation in $x$ and $y$ always has this same shape: $a x + b y + c = 0$, with $a$ and $b$ not both zero.
Every pair $(x, y)$ that satisfies it is a solution, and there are infinitely many such pairs. Plot every one of them, and they all land on the same straight line.
One equation on its own cannot pin down a single point. Finding one exact point takes a second equation, a second line, and that is what turns one equation into a pair.
Check that $a$ and $b$ are not both zero before calling an equation linear in $x$ and $y$: if both are zero, there is no line left to talk about.
The line for x plus y equals 5 passes through three marked points that satisfy it, and one nearby point that does not.
A SCHOLAR INDIA REMEMBERS
How many solutions does $x + y = 5$ have? Let us count a few. $(0, 5)$, $(1, 4)$, $(2, 3)$, $(2.5, 2.5)$. We can keep going, with no end. Each pair is a point on the same line. So one equation alone never fixes $x$ and $y$. We need a second equation.
Aryabhata · the hoopoe
Check yourself
The general form of a linear equation in $x$ and $y$ is $a x + b y + c = 0$. What must be true of $a$ and $b$?
$a$ and $b$ must both be positive
$a$ and $b$ must be equal
$a$ and $b$ are not both zero
$a$ and $b$ must both be whole numbers
Check your answer
$a$ and $b$ must both be positive — Neither coefficient’s sign is restricted — only that they are not BOTH zero at once.
$a$ and $b$ must be equal — $a$ and $b$ may take any values at all, equal or not, as long as they are not both zero.
✓ $a$ and $b$ are not both zero — (C) The only restriction the definition places is that $a$ and $b$ are not both zero.
$a$ and $b$ must both be whole numbers — The definition allows any real-number coefficients; whole numbers are not required.
Why does a single linear equation such as $x + y = 5$ have infinitely many solutions $(x, y)$?
because $x$ and $y$ can be swapped without changing the equation
every point on the line it describes satisfies the equation
because the equation has two variables written in it
because $5$ can be split into infinitely many pairs of factors
Check your answer
because $x$ and $y$ can be swapped without changing the equation — This equation happens to allow swapping $x$ and $y$, but that is a special feature, not the general reason.
✓ every point on the line it describes satisfies the equation — (B) A linear equation’s solution set is exactly the points on its line, and a line has infinitely many points.
because the equation has two variables written in it — Having two letters does not by itself explain the infinite set; it is the line the equation describes that does.
because $5$ can be split into infinitely many pairs of factors — Factor pairs of $5$ have nothing to do with the $(x, y)$ pairs satisfying $x + y = 5$.
A student writes $0x + 0y + 7 = 0$ and calls it a linear equation in $x$ and $y$. What is wrong?
both $a$ and $b$ are zero
the constant $7$ should be zero too
nothing is wrong; writing $x$ and $y$ is enough to qualify
it is fine, but only if $7$ is replaced with a variable
Check your answer
✓ both $a$ and $b$ are zero — (A) The not-both-zero condition on $a$ and $b$ fails here, so the definition is not met.
the constant $7$ should be zero too — The constant term $c$ is never restricted by the definition — only $a$ and $b$ are.
nothing is wrong; writing $x$ and $y$ is enough to qualify — Writing $x$ and $y$ in an expression does not make their coefficients non-zero — here both are $0$.
it is fine, but only if $7$ is replaced with a variable — The definition never asks for the constant term to be a variable; the real gap is $a = b = 0$.
The three words are not three equal boxes: dependent is a room inside the consistent house, and only the inconsistent pairs stand outside it.
Let us go back to $x + y = 5$ and $x - y = 1$, the pair from the frame section. They share the point $(3, 2)$.
Check it in your own working: $3 + 2 = 5$ and $3 - 2 = 1$. Both hold, so $(3, 2)$ really is a shared solution.
A pair of linear equations with at least one shared solution like this is called CONSISTENT. A pair with no shared solution at all is INCONSISTENT.
A consistent pair can go further. Suppose every point on one line also lies on the other. Then there are infinitely many shared solutions, not just one, and the pair is also called DEPENDENT.
Try it yourself. Is a dependent pair also consistent, or is it a separate case? (Answer: dependent is always consistent, since sharing every point on the line means sharing at least one point too.)
Check yourself
A pair of linear equations is called consistent when
both equations look identical when written out
it can be solved graphically
the two equations have the same constant term
it has at least one common solution
Check your answer
both equations look identical when written out — Equations do not need to look identical to be consistent — a unique shared solution is already enough.
it can be solved graphically — Every pair of equations can be graphed, whether or not the lines ever meet — graphing is not the test.
the two equations have the same constant term — Two equations can share no solution even with identical constant terms, and can share one with very different constants.
✓ it has at least one common solution — (D) Consistency needs only one shared solution, however many the pair actually has.
Why is a dependent pair of equations always also consistent?
dependent pairs are graphed differently, so consistency does not apply to them
dependency and consistency are two names for the same thing
sharing infinitely many solutions is more than enough for consistency
a dependent pair has no solution, and no solution counts as consistent
Check your answer
dependent pairs are graphed differently, so consistency does not apply to them — The same consistency definition applies to every pair; dependent pairs are not a special exception.
dependency and consistency are two names for the same thing — A consistent pair may have only one solution; a dependent pair always has infinitely many — the two are not the same claim.
✓ sharing infinitely many solutions is more than enough for consistency — (C) Dependent is a stronger claim than consistent, so every dependent pair automatically satisfies the weaker one.
a dependent pair has no solution, and no solution counts as consistent — No solution is the definition of INCONSISTENT, the opposite of what a dependent pair actually has.
$2x + 3y - 6 = 0$ and $4x + 6y - 5 = 0$ share no common solution. What should this pair be called?
dependent, since the coefficients of $x$ and $y$ are proportional
inconsistent
consistent, since both equations are linear
inconsistent only if graphed, not otherwise
Check your answer
dependent, since the coefficients of $x$ and $y$ are proportional — Proportional $a$ and $b$ coefficients only make the lines parallel — dependency needs the constants proportional too.
✓ inconsistent — (B) No shared solution at all is exactly the definition of an inconsistent pair.
consistent, since both equations are linear — Every equation here is linear, but that alone never guarantees a shared solution exists.
inconsistent only if graphed, not otherwise — Whether a graph is drawn changes nothing about the equations themselves or their consistency.
Let us take $3x + 2y - 12 = 0$ and $5x - 2y - 4 = 0$. Read the coefficients straight off: $a_1 = 3$ and $a_2 = 5$, so $a_1/a_2 = 3/5$.
Now find the other ratio yourself. Read off $b_1$ and $b_2$, and divide. (Answer: $b_1 = 2$ and $b_2 = -2$, so $b_1/b_2 = -1$.)
Put any pair into the form $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ first. Every pair then gives you the same three ratios to compare: $a_1/a_2$, $b_1/b_2$ and $c_1/c_2$.
These three ratios compare the same three things every time: the $x$-coefficients, the $y$-coefficients, and the constant terms. Nothing else about the two equations matters for this comparison.
Write both equations in the form $a x + b y + c = 0$ before you read off a single coefficient. A stray sign or a rearranged term throws every ratio off.
Two equations in standard form line up in three columns of ratios: x-coefficients, y-coefficients and constants.
Writing 3x + 2y = 12 in standard form first turns the third ratio’s sign from wrong to c one over c two equals 3.
Check yourself
For $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$, which three ratios are compared to classify the pair?
$a_1/b_1$, $a_2/b_2$, $c_1/c_2$
$a_1/a_2$, $b_1/b_2$, $c_1/c_2$
$a_1/a_2$ and $b_1/b_2$ only, with no constant ratio
$c_1/a_1$, $c_2/a_2$, $b_1/b_2$
Check your answer
$a_1/b_1$, $a_2/b_2$, $c_1/c_2$ — The test matches a coefficient AGAINST its counterpart in the OTHER equation, not against another coefficient in its own.
✓ $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ — (B) The test matches each coefficient in the first equation against its counterpart in the second.
$a_1/a_2$ and $b_1/b_2$ only, with no constant ratio — Without $c_1/c_2$, the test cannot tell a parallel pair from a coincident one.
$c_1/a_1$, $c_2/a_2$, $b_1/b_2$ — Each ratio compares the SAME kind of coefficient across the two equations, not a constant against a coefficient.
Why does the ratio test need the third ratio $c_1/c_2$, and not just $a_1/a_2$ and $b_1/b_2$?
the third ratio decides which line is drawn first
without it, the equations cannot be written in general form
it is only needed when $x$ and $y$ both have coefficient $1$
it settles coincident versus merely parallel
Check your answer
the third ratio decides which line is drawn first — Drawing order is never decided by any ratio — the ratios classify the pair’s solution count.
without it, the equations cannot be written in general form — The general form only needs the coefficients to exist; comparing their ratios is a separate, later step.
it is only needed when $x$ and $y$ both have coefficient $1$ — The third ratio is needed for every pair, not only ones where a coefficient happens to equal $1$.
✓ it settles coincident versus merely parallel — (D) The third ratio is what splits the parallel case into no solution and infinitely many.
For $3x + 5y - 7 = 0$ and $6x + 10y - 3 = 0$, what is $a_1/a_2$?
$2$
$1/2$
$3/6 + 5/10$
$-7/-3$
Check your answer
$2$ — This computes $a_2/a_1$, the ratio the wrong way round; $a_1/a_2$ gives $1/2$.
✓ $1/2$ — (B) $a_1 = 3$ and $a_2 = 6$, so $a_1/a_2 = 3/6 = 1/2$.
$3/6 + 5/10$ — $a_1/a_2$ is one single ratio of the $x$-coefficients, not a sum involving the $y$-coefficients too.
$-7/-3$ — $a_1/a_2$ is built from the $x$-coefficients, $3$ and $6$, not from the constant terms $-7$ and $-3$.
Let us take $x + y = 5$. Set $x = 0$, and $y = 5$ follows. Set $y = 0$, and $x = 5$ follows. Two points, $(0, 5)$ and $(5, 0)$, and that is already enough to draw the line.
Now try the same trick on $x - y = 1$ yourself. Set each variable to zero in turn. (Answer: $(0, -1)$ and $(1, 0)$.)
This is the whole recipe for solving a pair graphically: find at least two solutions of each equation, plot them, and draw the line through each pair of points.
Two points are enough to draw a straight line, so pick easy ones for your own working. Often the easiest points are where the line crosses each axis, since one coordinate is then simply zero.
Once both lines are drawn, the answer sits in how they meet. Every method for solving such a pair finds the one point that lies on both lines, if there is one; this one just finds it by drawing. A single crossing point is that point, and the pair’s unique solution. Parallel lines that never meet mean no solution. Lines that coincide give every point on the line as a solution.
A solution read off a graph is only as exact as the plotting. Coordinates that are not whole numbers are easy to misread, which is exactly why the two algebraic methods ahead matter.
Two lines are each drawn from two points, and they cross at the pair’s one solution.
Check yourself
To draw the line for one linear equation in $x$ and $y$, how many solutions of that equation must be plotted?
exactly one, since a single point already lies on the line
at least three, to be safe
at least two, since two points fix a straight line
as many as the equation’s coefficients
Check your answer
exactly one, since a single point already lies on the line — Infinitely many different lines pass through any single point — a second point is needed to fix the line.
at least three, to be safe — Two points already fix a straight line exactly; a third point only serves as an optional check.
✓ at least two, since two points fix a straight line — (C) One point never fixes a unique line, but two always do.
as many as the equation’s coefficients — The method always needs two points, regardless of how many coefficients the equation has.
After both lines are drawn, how is the number of solutions read off the graph?
by whether the lines cross once, never, or coincide
by measuring the angle between the two lines in degrees
by comparing which line is drawn higher on the page
by counting how many points were plotted in total
Check your answer
✓ by whether the lines cross once, never, or coincide — (A) The three possible pictures map directly onto the three possible outcomes.
by measuring the angle between the two lines in degrees — No angle is ever measured — only whether the lines meet, and where, matters to the reading.
by comparing which line is drawn higher on the page — Where a line sits on the page depends only on how it was drawn, not on the equations it represents.
by counting how many points were plotted in total — At least two points per line are always plotted regardless of the outcome; the count itself signals nothing.
A student plots two points for each equation, joins them into two lines, and reads off a crossing point without checking it. What has been skipped?
plotting a third point on each line
re-drawing the lines using a different pair of points
substituting the point into both original equations
measuring the crossing point’s distance from the origin
Check your answer
plotting a third point on each line — A third point only checks the drawing itself, not whether the crossing point solves the original equations.
re-drawing the lines using a different pair of points — Redrawing produces the same lines again; it does not check the crossing point against the equations.
✓ substituting the point into both original equations — (C) A graphical reading is only a candidate until substitution confirms it satisfies both equations.
measuring the crossing point’s distance from the origin — Distance from the origin is never part of confirming a solution — substitution back into both equations is.
Let us go back to $3x + 2y - 12 = 0$ and $5x - 2y - 4 = 0$ from the last section. There, $a_1/a_2 = 3/5$ and $b_1/b_2 = -1$.
Check it yourself: is $3/5$ equal to $-1$? (Answer: no, $3/5 \neq -1$.)
Whenever $a_1/a_2 \neq b_1/b_2$ like this, the two lines have different slopes. Different slopes can only cross once: two straight lines with different slopes are never parallel, so they meet at exactly one point.
This is the easiest of the three cases to spot in your own work. Check the first two ratios only, and if they differ, the answer is already settled. You never need to check the third ratio $c_1/c_2$ at all.
A decision path compares a1 over a2 with b1 over b2, checking c1 over c2 when those two agree.
Check yourself
The ratio test says a pair of linear equations has a unique solution when
$a_1/a_2 = b_1/b_2$
$a_1/a_2 \neq c_1/c_2$
$a_1 \neq a_2$ and $b_1 \neq b_2$
$a_1/a_2 \neq b_1/b_2$
Check your answer
$a_1/a_2 = b_1/b_2$ — This condition covers two different outcomes depending on the third ratio — it does not by itself signal a unique solution.
$a_1/a_2 \neq c_1/c_2$ — The unique-solution test compares $a_1/a_2$ against $b_1/b_2$, not against $c_1/c_2$.
$a_1 \neq a_2$ and $b_1 \neq b_2$ — The test compares RATIOS of coefficients, not the coefficients themselves side by side.
✓ $a_1/a_2 \neq b_1/b_2$ — (D) Unequal first two ratios is exactly the unique-solution condition.
Why does $a_1/a_2 \neq b_1/b_2$ guarantee the two lines intersect at exactly one point?
it shows the lines pass through the origin
it shows both equations have the same constant term
it shows the lines have different slopes
it shows one line is shorter than the other
Check your answer
it shows the lines pass through the origin — Passing through the origin is not what this comparison tests; it tests the lines’ slopes.
it shows both equations have the same constant term — The constant term $c$ plays no role in comparing $a_1/a_2$ to $b_1/b_2$.
✓ it shows the lines have different slopes — (C) Different slopes force exactly one crossing point between any two straight lines.
it shows one line is shorter than the other — A straight line has no finite length to compare — this idea does not apply.
For $2x + 3y - 6 = 0$ and $4x + 5y - 8 = 0$, what does the ratio test say?
$1/2 = 3/5$, so the pair has no solution
$1/2 \neq 3/5$, so the pair has a unique solution
$1/2 \neq 3/5$, so the pair has infinitely many solutions
the test cannot be applied since the constants are different
Check your answer
$1/2 = 3/5$, so the pair has no solution — $3/5$ does not equal $1/2$; the two ratios are genuinely different, not equal.
✓ $1/2 \neq 3/5$, so the pair has a unique solution — (B) $a_1/a_2 = 2/4 = 1/2$ and $b_1/b_2 = 3/5$, and since these differ the pair has a unique solution.
$1/2 \neq 3/5$, so the pair has infinitely many solutions — Unequal first two ratios give a UNIQUE solution, never infinitely many.
the test cannot be applied since the constants are different — This test compares only $a_1/a_2$ and $b_1/b_2$; the constants play no part in it at all.
A pair of equations has $a_1/a_2 = 4/7$ and $b_1/b_2 = 4/7$ too. Without knowing $c_1$ or $c_2$, can the pair have a unique solution?
yes, provided $c_1/c_2$ also equals $4/7$
yes, since the two ratios given are equal to each other
it depends on whether $x$ or $y$ has the larger coefficient
no, since $a_1/a_2 = b_1/b_2$ rules it out
Check your answer
yes, provided $c_1/c_2$ also equals $4/7$ — Matching $c_1/c_2$ too would give infinitely many solutions — the opposite of a unique one.
yes, since the two ratios given are equal to each other — Equal $a_1/a_2$ and $b_1/b_2$ is precisely the condition that rules a unique solution OUT.
it depends on whether $x$ or $y$ has the larger coefficient — The ratio test never asks which coefficient is larger, only how the two equations’ ratios compare.
✓ no, since $a_1/a_2 = b_1/b_2$ rules it out — (D) Equal first two ratios rule out uniqueness regardless of what the third ratio turns out to be.
Let us take pair 2 from the same comparison: $6x + 3y - 15 = 0$ and $2x + y - 6 = 0$. Here $a_1/a_2 = b_1/b_2 = 3$.
Now find the third ratio yourself, in your own calculation. (Answer: $c_1/c_2 = 5/2$, and $5/2$ is not $3$.)
Whenever $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ like this, the pair has no solution. The first two ratios being equal means the two lines share the same slope. The third ratio failing to match means they are not the same line.
A pair like this is inconsistent. There is no point that satisfies both equations at once, because the lines never touch.
Check yourself
The ratio test says a pair of linear equations has no solution when
$a_1/a_2 = b_1/b_2 = c_1/c_2$, the dependent case
$a_1/a_2 \neq b_1/b_2$
$a_1/a_2 = b_1/b_2 \neq c_1/c_2$
$a_1/a_2 \neq c_1/c_2$
Check your answer
$a_1/a_2 = b_1/b_2 = c_1/c_2$, the dependent case — All three ratios matching is the infinite-solutions condition, not the no-solution one.
$a_1/a_2 \neq b_1/b_2$ — Unequal $a_1/a_2$ and $b_1/b_2$ gives a UNIQUE solution, not no solution at all.
✓ $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ — (C) Equal first two ratios with an unequal third is exactly the no-solution condition.
$a_1/a_2 \neq c_1/c_2$ — The no-solution condition needs $a_1/a_2 = b_1/b_2$ first, both unequal to $c_1/c_2$ — not this comparison alone.
Why does $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ mean the lines are parallel but not the same line?
same slope from $a$, $b$; different position from $c$
an unequal $c$-ratio means the lines have different slopes
equal $a$- and $b$-ratios already prove the lines are identical
the condition means one equation is wrong
Check your answer
✓ same slope from $a$, $b$; different position from $c$ — (A) Slope is fixed by the first two ratios, and position along that slope is fixed by the third.
an unequal $c$-ratio means the lines have different slopes — Slope depends only on $a_1/a_2$ and $b_1/b_2$; the $c$-ratio decides where the line sits, not its slope.
equal $a$- and $b$-ratios already prove the lines are identical — Two lines can share a slope and still sit apart — position, carried by $c_1/c_2$, still has to match too.
the condition means one equation is wrong — Both equations are perfectly valid; they just happen to describe two separate, parallel lines.
For $2x + 3y - 8 = 0$ and $4x + 6y - 5 = 0$, what does the ratio test say?
$1/2 = 1/2 = 8/5$, so the pair has infinitely many solutions
$1/2 = 1/2 \neq 8/5$, so the pair has no solution
$1/2 \neq 1/2$, so the pair has a unique solution
the pair is consistent, since $1/2$ appears twice
Check your answer
$1/2 = 1/2 = 8/5$, so the pair has infinitely many solutions — $8/5$ does not equal $1/2$; the third ratio genuinely differs from the first two.
✓ $1/2 = 1/2 \neq 8/5$, so the pair has no solution — (B) $a_1/a_2 = b_1/b_2 = 1/2$ but $c_1/c_2 = 8/5$, so the lines are parallel and distinct.
$1/2 \neq 1/2$, so the pair has a unique solution — $a_1/a_2$ and $b_1/b_2$ are both $1/2$ here — they are equal, not different.
the pair is consistent, since $1/2$ appears twice — Two matching ratios with a third that differs signals PARALLEL lines with no shared solution.
Two lines are parallel and do not coincide. What must be true of their ratio comparison?
$a_1/a_2 = b_1/b_2 \neq c_1/c_2$
$a_1/a_2 \neq b_1/b_2$
$a_1/a_2 = b_1/b_2 = c_1/c_2$
no ratio comparison can tell parallel lines from intersecting ones
Check your answer
✓ $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ — (A) Parallel and distinct lines are exactly what this ratio pattern describes.
$a_1/a_2 \neq b_1/b_2$ — Unequal $a_1/a_2$ and $b_1/b_2$ gives different slopes, which forces the lines to cross rather than stay parallel.
$a_1/a_2 = b_1/b_2 = c_1/c_2$ — All three ratios matching makes the lines identical, not two separate parallel lines.
no ratio comparison can tell parallel lines from intersecting ones — Telling parallel from intersecting lines is precisely what comparing $a_1/a_2$ and $b_1/b_2$ does.
Let us take pair 3 from the same comparison: $4x + 6y - 10 = 0$ and $2x + 3y - 5 = 0$.
Check all three ratios yourself. (Answer: $a_1/a_2 = 4/2 = 2$, $b_1/b_2 = 6/3 = 2$, and $c_1/c_2 = 10/5 = 2$. All three are equal.)
Whenever $a_1/a_2 = b_1/b_2 = c_1/c_2$ like this, the pair has infinitely many solutions. All three ratios matching means the two equations describe the exact same line, just scaled by a constant factor.
This pair is dependent, and a dependent pair is always consistent. Every point on the shared line is a valid solution for you to check.
Check yourself
The ratio test says a pair of linear equations has infinitely many solutions when
$a_1/a_2 = b_1/b_2 \neq c_1/c_2$
$a_1/a_2 = b_1/b_2 = c_1/c_2$
$a_1/a_2 \neq b_1/b_2 = c_1/c_2$
$a_1 = a_2$, $b_1 = b_2$, $c_1 = c_2$
Check your answer
$a_1/a_2 = b_1/b_2 \neq c_1/c_2$ — This is the parallel, no-solution pattern; infinitely many solutions needs the third ratio to match too.
✓ $a_1/a_2 = b_1/b_2 = c_1/c_2$ — (B) All three ratios agreeing is exactly the infinite-solutions, dependent condition.
$a_1/a_2 \neq b_1/b_2 = c_1/c_2$ — The infinite-solutions case needs $a_1/a_2$ and $b_1/b_2$ equal as well, not just $b_1/b_2$ and $c_1/c_2$.
$a_1 = a_2$, $b_1 = b_2$, $c_1 = c_2$ — The condition is about equal RATIOS, which allows one equation to be any constant multiple of the other.
Why does $a_1/a_2 = b_1/b_2 = c_1/c_2$ mean the two equations describe the same line?
it means both equations have the same number of terms
it means the two lines are parallel but placed side by side
it means $x$ and $y$ have swapped roles between the equations
one equation is a constant multiple of the other
Check your answer
it means both equations have the same number of terms — Every linear equation has the same three terms by definition; that alone signals nothing about the ratios.
it means the two lines are parallel but placed side by side — Parallel-but-separate is the no-solution picture; three equal ratios instead put the lines on top of each other.
it means $x$ and $y$ have swapped roles between the equations — No swapping of $x$ and $y$ is involved; each equation keeps its own coefficients scaled by the same factor.
✓ one equation is a constant multiple of the other — (D) Three equal ratios mean every term scales by the same factor, leaving the same line.
For $3x + 6y - 9 = 0$ and $x + 2y - 3 = 0$, what does the ratio test say?
$3 = 3 \neq 3$, so the pair has no solution
$3 \neq 3 = 3$, so the pair has a unique solution
$3 = 3 = 3$, so infinitely many solutions
the second equation is simply wrong, since it looks smaller
Check your answer
$3 = 3 \neq 3$, so the pair has no solution — All three ratios equal $3$ here — none of them differs from the others.
$3 \neq 3 = 3$, so the pair has a unique solution — $a_1/a_2$ and $b_1/b_2$ are both $3$; they are equal, not different.
✓ $3 = 3 = 3$, so infinitely many solutions — (C) Every ratio, $a_1/a_2$, $b_1/b_2$, $c_1/c_2$, equals $3$, so the pair is dependent.
the second equation is simply wrong, since it looks smaller — The second equation is exactly the first one divided through by $3$ — a perfectly valid, equivalent equation.
$2x + 3y - 7 = 0$ is multiplied through by a constant $k$ to give $4x + 6y - 14 = 0$. What does this pair have?
no solution, since the coefficients doubled
a unique solution, found by dividing the second equation by $k$
the pair cannot be classified without knowing $k$’s value
infinitely many solutions
Check your answer
no solution, since the coefficients doubled — Doubling every term together, constant included, keeps all three ratios equal — the opposite of the no-solution pattern.
a unique solution, found by dividing the second equation by $k$ — Dividing the second equation by $k$ just returns the first equation again, not a genuinely different one to solve against.
the pair cannot be classified without knowing $k$’s value — Any nonzero $k$ scales every coefficient and the constant together, so the three ratios stay equal regardless of its value.
✓ infinitely many solutions — (D) Multiplying every term by the same $k$ keeps all three ratios equal, so the pair stays dependent.
Let us take $x + y = 10$ and $2x - y = 5$. From the first equation, $x = 10 - y$.
The substitution method always runs in three steps:
- Step one: write one variable in terms of the other, from the equation where that is easiest. Here, $x = 10 - y$.
- Step two: substitute that expression into the other equation. Here, that means putting $10 - y$ in place of $x$ in $2x - y = 5$. Two variables become one, and one equation in one variable is something you already know how to solve.
- Step three: substitute the value you find back into either equation, to get the remaining variable.
Try all three steps yourself, before you read the full working in the next worked example. (Answer: the pair solves to $x = 5$ and $y = 5$.)
Solve for the variable with the simplest coefficient. It keeps the numbers small.
*Watch what the final equation looks like: a statement that is always true, like $0 = 0$, means infinitely many solutions, and a statement that is always false, like $0 = 9$, means no solution.*
A SCHOLAR INDIA REMEMBERS
You solved one pair by substitution. Will the same steps solve the next pair? Write the steps so that someone else could follow them. First, make one unknown the subject of one equation. Then put it into the other equation. Solve that. Then go back and find the other unknown.
Aryabhata · the hoopoe
The two original lines cross the extra line, y equals 5, at the same point substitution’s last step finds.
Check yourself
The substitution method solves a pair of linear equations by
multiplying both equations until one variable’s coefficients match
plotting both equations and reading off the crossing point
solve for one variable, then substitute it in
guessing values of $x$ and $y$ until both equations hold
Check your answer
multiplying both equations until one variable’s coefficients match — Matching coefficients by multiplying is the elimination method’s move, not substitution’s.
plotting both equations and reading off the crossing point — Plotting and reading a crossing point is the graphical method, not substitution.
✓ solve for one variable, then substitute it in — (C) The full method writes one variable in terms of the other, substitutes it in, solves, then substitutes back.
guessing values of $x$ and $y$ until both equations hold — Substitution follows a fixed sequence of algebraic steps — it never relies on guesswork.
Why does substituting one variable’s expression into the other equation make the problem solvable?
it removes the constant term from the equation
it leaves one equation, one unknown
it makes both equations identical
it changes the equation from linear to quadratic
Check your answer
it removes the constant term from the equation — The constant term stays exactly where it was; what changes is that one variable is replaced by an expression.
✓ it leaves one equation, one unknown — (B) A single equation in one unknown can be solved directly, which two equations in two unknowns cannot be, term by term.
it makes both equations identical — The equations become identical only when the pair is dependent — not in general.
it changes the equation from linear to quadratic — The resulting single-variable equation stays linear; substitution never introduces a squared term here.
Solving $x + 2y = 8$ and $3x - y = 3$ by substitution, the first equation gives $x = 8 - 2y$. What is the correct next step?
substitute $8 - 2y$ for $x$ in $3x - y = 3$
substitute $8 - 2y$ for $x$ back into $x + 2y = 8$
solve $3x - y = 3$ for $y$ before doing anything else
add the two original equations together
Check your answer
✓ substitute $8 - 2y$ for $x$ in $3x - y = 3$ — (A) The expression for $x$ belongs in the OTHER equation, the one it was not derived from.
substitute $8 - 2y$ for $x$ back into $x + 2y = 8$ — Putting the expression back into the equation it came from just restates that equation — it teaches nothing new.
solve $3x - y = 3$ for $y$ before doing anything else — The expression $x = 8 - 2y$ was found precisely to be used next, in the other equation.
add the two original equations together — Adding the original equations is an elimination move, not the next step of a substitution already underway.
After substitution gives $x = 4$ and $y = 2$ for a pair of equations, what confirms the answer is correct?
the values are whole numbers
$x$ is larger than $y$
the second equation was used to find $y$
both values satisfy both equations
Check your answer
the values are whole numbers — Many correct solutions are not whole numbers at all; being whole proves nothing about correctness.
$x$ is larger than $y$ — Which of $x$ and $y$ is larger has no connection to whether the pair of equations is satisfied.
the second equation was used to find $y$ — How the answer was found does not confirm it; only substituting back into both equations does.
✓ both values satisfy both equations — (D) The final check is always substitution back into the ORIGINAL pair, not any intermediate step.
Let us take $2x + 3y = 13$ and $3x + 2y = 12$. Multiply the first equation by $2$ and the second by $3$.
Check the $y$-coefficient in each yourself. (Answer: both become $6y$.)
That is the elimination method’s first step: multiply one or both equations by constants, so one variable ends up with equal coefficients in both.
Second, add or subtract the two equations to eliminate that variable, leaving one equation in one variable to solve. Elimination, like the other methods here, finds the one point that lies on both lines; it just gets there by cancelling a variable instead of plotting or substituting.
Same sign on the matched coefficients: subtract. Opposite signs: add. Getting this backwards doubles the variable instead of removing it, so check the signs on your own pair before doing either.
As with substitution, an always-true statement left at the end means infinitely many solutions. An always-false one means no solution.
The two original lines cross the extra line, x equals 2, at the same point elimination’s subtraction step finds.
Multiplying each equation by the other’s y-coefficient is what makes the y-chips match, and matched chips are the ones a subtraction can remove.
A SCHOLAR INDIA REMEMBERS
In 499 CE, Aryabhata gave a step-by-step method for a whole class of equations. It was not a trick for one equation. The same steps worked for every equation in the class. Elimination is a method of that kind. It works the same way on every pair of linear equations in two unknowns. Learn the steps once, and they work for the next pair too.
Aryabhata · the hoopoe
Check yourself
The elimination method solves a pair of linear equations by
writing one variable in terms of the other and substituting it in
matching a coefficient, then adding or subtracting
plotting both lines and reading off where they cross
dividing both equations by their constant terms
Check your answer
writing one variable in terms of the other and substituting it in — Writing one variable in terms of the other and substituting is the substitution method’s move.
✓ matching a coefficient, then adding or subtracting — (B) Multiplying to match coefficients, then adding or subtracting, is the full elimination method.
plotting both lines and reading off where they cross — Reading a crossing point off a graph is the graphical method, not elimination.
dividing both equations by their constant terms — The target of the multiplying step is a variable’s coefficient, never the constant term.
Why multiply the equations by constants before adding or subtracting them?
to make one variable’s coefficient equal in both
to make both equations have the same constant term
to make the equations easier to plot on a graph
to remove any variable that has a negative coefficient
Check your answer
✓ to make one variable’s coefficient equal in both — (A) Matching coefficients first is what makes the later addition or subtraction cancel a variable.
to make both equations have the same constant term — The multiplying step targets a variable’s coefficient, not the constant term on the other side.
to make the equations easier to plot on a graph — Elimination never involves a graph; the multiplying step is purely algebraic preparation.
to remove any variable that has a negative coefficient — A coefficient’s sign is not the issue; multiplying is about matching MAGNITUDE so the terms cancel.
To eliminate $y$ from $2x + 5y = 9$ and $3x + 2y = 4$, which multipliers make the $y$-coefficients equal?
multiply the first equation by $2$ and the second by $5$
multiply the first equation by $5$ and the second by $2$
multiply both equations by $7$
multiply the first equation by $3$ and the second by $2$
Check your answer
✓ multiply the first equation by $2$ and the second by $5$ — (A) This gives $y$-coefficient $10$ in both equations, ready to cancel.
multiply the first equation by $5$ and the second by $2$ — Swapping the multipliers gives $25y$ in the first equation and $4y$ in the second — not equal.
multiply both equations by $7$ — Multiplying both equations by the same number never changes the ratio between their $y$-coefficients.
multiply the first equation by $3$ and the second by $2$ — This choice equalises the $x$-coefficients, which eliminates $x$, not the $y$ this question asks for.
For $x + y = 12$ and $x - y = 2$, which method is cheaper: substitution or elimination?
substitution, since $x$ can be found from the first equation
neither, since both need the same number of steps for this pair
elimination — adding cancels $y$ directly
elimination, but only after multiplying both equations first
Check your answer
substitution, since $x$ can be found from the first equation — Finding $x$ in terms of $y$ is only the first of several substitution steps still to come; adding is faster here.
neither, since both need the same number of steps for this pair — Adding the two equations directly takes a single step, fewer than the several steps substitution needs here.
✓ elimination — adding cancels $y$ directly — (C) The $y$-coefficients are already $+1$ and $-1$, so adding cancels $y$ in one step.
elimination, but only after multiplying both equations first — The $y$-coefficients already match in magnitude with opposite signs, so adding cancels $y$ with no multiplying at all.
Let us take a problem: two numbers add to $45$, and differ by $5$.
Name the two unknown quantities first, in words, before you write a single equation. Here, let $x$ and $y$ be the two numbers.
Now translate each condition yourself, one at a time, into one equation. (Answer: $x + y = 45$ and $x - y = 5$.)
That is the first of the two steps a situational problem always takes. Second, solve the resulting pair by substitution or elimination.
A solved $x$ and $y$ that satisfy both equations can still fail the original story. Check them against the words, not just against the algebra: an answer can solve the equations correctly and still make no sense as an age, a distance, or a cost.
A SCHOLAR INDIA REMEMBERS
A word problem asks for two unknown numbers. Before you write any equation, name them. Let $x$ be the cost of one pen and $y$ the cost of one notebook, in rupees. Then find two facts in the problem that link $x$ and $y$. Each fact gives one equation. Two facts give the pair you need.
Aryabhata · the hoopoe
Check yourself
Turning a word problem into a pair of linear equations involves
guessing two numbers that sound reasonable for the problem
writing one very long equation that contains every condition at once
solving the problem by trial and error before writing any equation
naming the unknowns, then writing each condition as an equation
Check your answer
guessing two numbers that sound reasonable for the problem — The unknowns are named and each condition translated deliberately — nothing here is guessed.
writing one very long equation that contains every condition at once — Each stated condition becomes its OWN equation; conditions are never folded into a single combined one.
solving the problem by trial and error before writing any equation — Equations are written first from the conditions; solving them by substitution or elimination comes only after.
✓ naming the unknowns, then writing each condition as an equation — (D) The method is a systematic, two-step translation from words to equations.
Why check the values of $x$ and $y$ against the original word problem after solving?
to make the handwriting neater before submitting
because substitution and elimination sometimes give the wrong answer
to confirm the answer fits the situation described
to round the answer to the nearest whole number
Check your answer
to make the handwriting neater before submitting — Handwriting has nothing to do with whether an algebraic answer fits the problem’s situation.
because substitution and elimination sometimes give the wrong answer — A correctly applied method solves the equations correctly — the check is about whether that solution fits the STORY, not about method error.
✓ to confirm the answer fits the situation described — (C) Solving the equations correctly is not the same as the answer fitting the real situation, so a check is still needed.
to round the answer to the nearest whole number — Many genuine answers are not whole numbers at all; the check is about fit to the problem, not rounding.
Two numbers differ by $15$, and their sum is $65$. Let the larger number be $x$ and the smaller be $y$. Which pair of equations fits?
$x - y = 15$ and $x + y = 65$
$x + y = 15$ and $x - y = 65$
$x - y = 65$ and $x + y = 15$
$x = 15y$ and $x + y = 65$
Check your answer
✓ $x - y = 15$ and $x + y = 65$ — (A) The difference condition and the sum condition each become their own equation, attached to the right operation.
$x + y = 15$ and $x - y = 65$ — The SUM is $65$ and the DIFFERENCE is $15$ — this option attaches each total to the wrong equation.
$x - y = 65$ and $x + y = 15$ — Both totals are attached to the wrong operation here — sum should equal $65$, difference should equal $15$.
$x = 15y$ and $x + y = 65$ — A difference of $15$ means $x - y = 15$, not that $x$ is $15$ TIMES $y$.
No solution and infinitely many solutions can feel like the same kind of failure. Neither one hands you a single tidy answer.
They are opposite outcomes, and the third ratio is the only thing that tells them apart.
Let us take $2x + 4y - 6 = 0$ and $x + 2y - 5 = 0$. The first two ratios match: $a_1/a_2 = b_1/b_2 = 2$. But the third does not: $c_1/c_2 = 6/5$. Same slope, different line: parallel, and no solution.
Now change only the second equation’s constant term yourself, from $-5$ to $-3$. (Answer: $x + 2y - 3 = 0$, and the third ratio becomes $c_1/c_2 = 6/3 = 2$, which now matches the other two.)
The same pair of lines is coincident instead of parallel: infinitely many solutions, not none.
The same first line is paired with two different lines: one pair stays parallel, and the other pair lands exactly on top of it.
Does a pair whose first two ratios match have no solution, or infinitely many?
Weaker. A student checks the first two ratios: $a_1/a_2 = 3$ and $b_1/b_2 = 3$. Both match, so the student stops there and says the pair has infinitely many solutions.
That is wrong. The third ratio is $c_1/c_2 = 12/3 = 4$, and $4$ is not $3$. Two matching ratios are never enough on their own.
Stronger. We check all three ratios before deciding. $a_1/a_2 = 3$, $b_1/b_2 = 3$, and $c_1/c_2 = 12/3 = 4$. The first two match, the third does not, so the pair has no solution: the lines are parallel.
Change only the constant term, from $12$ to $9$: $3x + 6y - 9 = 0$ paired with the same $x + 2y - 3 = 0$. Now $c_1/c_2 = 9/3 = 3$, matching the other two. The same two lines are now one line, and the pair has infinitely many solutions.
The first two ratios stay equal while only the constant changes, and that one change is what moves the line from parallel to coincident.
Check yourself
A student says: “no solution and infinitely many solutions are basically the same kind of failure, since the equations just do not give one clean answer.” What is wrong with this?
nothing is wrong; both cases mean the equations cannot be solved
they are opposite, told apart by $c_1/c_2$
the difference is only about how many times you try substitution
no solution and infinitely many are the same whenever $a_1/a_2 = b_1/b_2$
Check your answer
nothing is wrong; both cases mean the equations cannot be solved — Both cases ARE classified outcomes, not failures — one gives no shared point, the other gives every point on a line.
✓ they are opposite, told apart by $c_1/c_2$ — (B) No solution and infinitely many solutions sit at opposite ends of the ratio test, not next to each other.
the difference is only about how many times you try substitution — How many times substitution is tried has nothing to do with which of the two outcomes holds.
no solution and infinitely many are the same whenever $a_1/a_2 = b_1/b_2$ — That first condition is shared by both cases — it is exactly $c_1/c_2$ that decides which one actually holds.
For $2x + 4y - 6 = 0$ and $x + 2y - 5 = 0$, a student sees $a_1/a_2 = b_1/b_2 = 2$ and concludes the pair has infinitely many solutions. What went wrong?
nothing went wrong; two matching ratios are already enough
the student should use $c_1 = 6$ and $c_2 = 5$ directly, not their ratio
the mistake is in computing $a_1/a_2$, which should be $1$
$c_1/c_2 = 6/5$, which is not $2$, so the pair actually has no solution
Check your answer
nothing went wrong; two matching ratios are already enough — Two matching ratios narrow the outcome to parallel or coincident — the third ratio still has to decide between them.
the student should use $c_1 = 6$ and $c_2 = 5$ directly, not their ratio — The test always compares the RATIO $c_1/c_2$ to the other two ratios, never the bare constants on their own.
the mistake is in computing $a_1/a_2$, which should be $1$ — $a_1/a_2 = 2$ is computed correctly here; the actual gap is skipping the check on $c_1/c_2$.
✓ $c_1/c_2 = 6/5$, which is not $2$, so the pair actually has no solution — (D) The third ratio was never checked, and it breaks the match the first two ratios seemed to promise.
Starting from $2x + 4y - 6 = 0$ and $x + 2y - 5 = 0$ (no solution, as above), which single change turns the pair into one with infinitely many solutions instead?
changing the second equation to $x + 2y - 3 = 0$
changing the coefficient of $x$ in the second equation
changing the sign of $y$ in the first equation
multiplying the whole first equation by $2$
Check your answer
✓ changing the second equation to $x + 2y - 3 = 0$ — (A) Only the third ratio needs to move; the first two already match at $2$.
changing the coefficient of $x$ in the second equation — Changing $a_2$ risks breaking the very slope match that already holds — the fix needed is in $c_2$, not $a_2$.
changing the sign of $y$ in the first equation — Flipping the sign of $y$’s coefficient changes the slope entirely, undoing the parallel condition rather than fixing it.
multiplying the whole first equation by $2$ — Scaling an equation by a constant leaves every one of its ratios to the OTHER equation exactly as before.
Once $a_1/a_2 = b_1/b_2$ is confirmed, which single further check decides between no solution and infinitely many?
whether $a_1$ and $b_1$ are both positive
whether $c_1/c_2$ equals that same ratio or not
whether $x$ or $y$ was eliminated first while solving
whether the two equations were originally written in the same order
Check your answer
whether $a_1$ and $b_1$ are both positive — Sign has no bearing here; the decision rests entirely on comparing $c_1/c_2$ to the shared ratio.
✓ whether $c_1/c_2$ equals that same ratio or not — (B) The third ratio is the only remaining check once the first two already match.
whether $x$ or $y$ was eliminated first while solving — Which variable is eliminated first is a choice made while SOLVING; it plays no part in classifying the pair.
whether the two equations were originally written in the same order — Listing order has no bearing on any of the three ratios or on the pair’s classification.
A student writes: “since $a_1/a_2 = b_1/b_2$, the lines must be the same line.” What is the error?
matching two ratios shows parallel, not necessarily the same line
there is no error; matching two ratios is already enough to prove the lines coincide
the error is in comparing $a_1/a_2$ to $b_1/b_2$ at all
the error is that ratios cannot be compared to each other, only to numbers
Check your answer
✓ matching two ratios shows parallel, not necessarily the same line — (A) The third ratio still has to be checked before the lines can be called the same.
there is no error; matching two ratios is already enough to prove the lines coincide — Matching only two ratios proves the lines are parallel — coinciding still needs the third ratio to match as well.
the error is in comparing $a_1/a_2$ to $b_1/b_2$ at all — Comparing $a_1/a_2$ to $b_1/b_2$ is exactly the right first step — what is missing is checking $c_1/c_2$ too.
the error is that ratios cannot be compared to each other, only to numbers — Comparing one ratio to another is exactly the mechanism the whole ratio test runs on.
Adding the two equations cancels the y-terms to give 6x equals 18, but subtracting doubles them to 2x plus 2y equals 10.
Matching coefficients does not always mean subtract.
Let us take $4x + y = 14$ and $2x - y = 4$. The $y$-coefficients are already matched in size, at $+1$ and $-1$.
Try subtracting the equations yourself. (Answer: $(4x + y) - (2x - y) = 14 - 4$ gives $2x + 2y = 10$, and $y$ has not been eliminated.)
The two coefficients have opposite signs, so subtracting doubles $y$ instead of cancelling it. Adding is what works here: $(4x + y) + (2x - y) = 14 + 4$ gives $6x = 18$, so $x = 3$. Substituting back, $y = 14 - 4(3) = 2$.
First try
The $y$-terms in $4 x + y = 14$ and $2 x - y = 4$ are matched in size, so I subtracted and got $2 x + 2 y = 10$.
Second look
Subtracting cancels a term only when the matched coefficients share a sign. Here they are $+1$ and $-1$, opposite signs, so adding cancels them. That gives $6 x = 18$, so $x = 3$ and $y = 2$.
Matching size is not enough. Same sign, subtract. Opposite sign, add.
The coefficients of one variable already match in size. Subtract, or add?
Weaker. A student sees $+2y$ and $-2y$, both matched in size, and subtracts the equations because that is what worked last time. $(5x + 2y) - (3x - 2y) = 19 - 5$ gives $2x + 4y = 14$.
The $y$ term has not gone anywhere. Subtracting two terms with opposite signs doubles them instead of cancelling them.
Stronger. We check the signs first. The $y$-coefficients are $+2$ and $-2$: opposite signs, so we add, not subtract. $(5x + 2y) + (3x - 2y) = 19 + 5$ gives $8x = 24$, so $x = 3$.
Substituting back, $3(3) - 2y = 5$, so $2y = 4$ and $y = 2$. Check: $5(3) + 2(2) = 19$ and $3(3) - 2(2) = 5$. Both hold.
$x + y = 5$ gives $(0, 5)$ and $(5, 0)$ Set $x = 0$, and $y = 5$. Set $y = 0$, and $x = 5$. Two points are enough to draw the line.
$x - y = 1$ gives $(0, -1)$ and $(1, 0)$ Now you try the same trick on the second equation, setting each variable to zero in turn.
the two lines cross at $(3, 2)$ Plot both pairs of points, draw each line, and read off where they meet.
$3 + 2 = 5$ and $3 - 2 = 1$ Check the crossing point by substituting it back into both original equations. Both hold, so $(3, 2)$ really is the pair’s solution.
the first line
the second line
where the two lines meet
Nila kneels where two straight ropes laid across the yard cross at a single point.
Solve any pair of linear equations by graphing.
Find two points on each line Set $x = 0$ to get one point, then set $y = 0$ to get another. For $x + y = 5$, that gives $(0, 5)$ and $(5, 0)$.
Plot and join each pair Mark both points for one equation, and draw the line through them. Do the same for the second equation, $x - y = 1$, using $(0, -1)$ and $(1, 0)$.
Read the crossing point Look at where the two lines meet. Here, they cross at $(3, 2)$.
Check by substituting Put $(3, 2)$ back into both equations. $3 + 2 = 5$ and $3 - 2 = 1$, so it checks out.
Name the case One crossing point means one solution. Parallel lines mean no solution. The same line twice means infinitely many.
Check yourself
Solving $x + y = 5$ and $x - y = 1$ graphically, the worked example plots $(0, 5)$ and $(5, 0)$ for the first line. Where do these points come from?
they are the midpoint and endpoint of the eventual crossing point
they come from solving both equations together
they are chosen at random from any pair of numbers
they are the midpoint and endpoint of the eventual crossing point — The crossing point is found only later, once both lines are drawn — these two points come from one equation alone.
they come from solving both equations together — Solving both equations together produces the crossing point, a later step — these two points come from ONE equation.
they are chosen at random from any pair of numbers — Each point is deliberately chosen so its coordinates satisfy $x + y = 5$, not picked at random.
✓ setting $x = 0$ gives $y = 5$, and setting $y = 0$ gives $x = 5$ — (D) Each point is one solution of $x + y = 5$, found by fixing one variable at zero.
The two lines in the worked example cross at $(3, 2)$. What confirms this is the pair’s solution?
$(3, 2)$ lies exactly halfway between the four plotted points
$3+2=5$ and $3-2=1$: it satisfies both
$3$ and $2$ are both smaller than $5$
the point was read off the graph, so it needs no further check
Check your answer
$(3, 2)$ lies exactly halfway between the four plotted points — Where a point sits relative to the plotted points proves nothing about whether it solves the equations.
✓ $3+2=5$ and $3-2=1$: it satisfies both — (B) Substituting the crossing point back into both original equations is what confirms it.
$3$ and $2$ are both smaller than $5$ — Comparing $3$ and $2$ to $5$ checks nothing about whether the equations are actually satisfied.
the point was read off the graph, so it needs no further check — A crossing point read off a graph is only a candidate answer until it is substituted back and confirmed.
Using the same graphical approach as $x + y = 5$ and $x - y = 1$, which two points would you plot first for $x + y = 8$?
$(0, 8)$ and $(8, 0)$
$(0, 5)$ and $(5, 0)$, carried over from the earlier example
$(8, 8)$ and $(0, 0)$
$(4, 4)$ only, since one point is enough once the equation is simple
Check your answer
✓ $(0, 8)$ and $(8, 0)$ — (A) Setting $x = 0$ and then $y = 0$ in $x + y = 8$ gives these two fresh points.
$(0, 5)$ and $(5, 0)$, carried over from the earlier example — Those points satisfy $x + y = 5$, not $x + y = 8$ — new points must be found for the new equation.
$(8, 8)$ and $(0, 0)$ — $(0,0)$ gives a sum of $0$, not $8$, so it does not lie on this line at all.
$(4, 4)$ only, since one point is enough once the equation is simple — However simple the equation, a single point never fixes a unique straight line — a second point is always needed.
Nine ratios were possible but only eight were computed: the ratio test reads left to right and stops the moment the verdict is settled.
Worked example
Classifying three pairs by ratio, with no solving or graphing
$3x + 2y - 12 = 0$ and $5x - 2y - 4 = 0$: $a_1/a_2 = 3/5$, $b_1/b_2 = -1$ Read the three coefficients straight off each equation.
$3/5 \neq -1$, so the pair has a unique solution and is consistent The first two ratios already differ, so there is no need to check the third ratio at all.
$6x + 3y - 15 = 0$ and $2x + y - 6 = 0$: $a_1/a_2 = b_1/b_2 = 3$ The first two ratios match this time.
$c_1/c_2 = 5/2$ Check the third ratio too, since the first two matched.
$3 \neq 5/2$, so the pair has no solution Same slope, different line: the lines are parallel.
$4x + 6y - 10 = 0$ and $2x + 3y - 5 = 0$: $a_1/a_2 = b_1/b_2 = c_1/c_2 = 2$ All three ratios come out equal.
the pair has infinitely many solutions All three ratios matching means the two equations describe the same line.
$4(1) + 6(1) - 10 = 0$ and $2(1) + 3(1) - 5 = 0$ Check pair 3 yourself by picking any point and testing it in both equations. Try $x = 1$, $y = 1$: it satisfies both, confirming the two lines really are the same one.
Classify any pair of linear equations by ratio, without solving or graphing.
Write both equations in standard form Put each equation as $a x + b y + c = 0$ before reading off a single coefficient.
Compare the first two ratios Read off $a_1/a_2$ and $b_1/b_2$. If they differ, the pair has a unique solution, and you can stop here.
Check the third ratio if needed If $a_1/a_2 = b_1/b_2$, also find $c_1/c_2$. A match with the first two means infinitely many solutions; a mismatch means no solution.
Match the case to the example $3x + 2y - 12 = 0$ and $5x - 2y - 4 = 0$ give $3/5 \neq -1$: one solution. $6x + 3y - 15 = 0$ and $2x + y - 6 = 0$ give equal first ratios but $3 \neq 5/2$: no solution. $4x + 6y - 10 = 0$ and $2x + 3y - 5 = 0$ give all three ratios equal to $2$: infinitely many.
Check yourself
In the worked example, Pair 1 is $3x + 2y - 12 = 0$ and $5x - 2y - 4 = 0$, classified as a unique solution from $3/5 \neq -1$. Which ratios were actually compared to reach this?
$a_1/a_2 = 3/5$ against $c_1/c_2 = 3$
$a_1/a_2=3/5$ against $b_1/b_2=-1$
$b_1/b_2 = -1$ against $c_1/c_2 = 3$
$a_1/a_2 = 3/5$ against itself
Check your answer
$a_1/a_2 = 3/5$ against $c_1/c_2 = 3$ — The unique-solution test never needs $c_1/c_2$; it compares only $a_1/a_2$ and $b_1/b_2$.
✓ $a_1/a_2=3/5$ against $b_1/b_2=-1$ — (B) The unique-solution test compares only the first two ratios.
$b_1/b_2 = -1$ against $c_1/c_2 = 3$ — The $a$-ratio, $3/5$, is one of the two ratios this test actually compares — it cannot be skipped.
$a_1/a_2 = 3/5$ against itself — A ratio compared to itself is always equal by definition and could never signal a unique solution.
The worked example’s Pair 2, $6x + 3y - 15 = 0$ and $2x + y - 6 = 0$, is classified as no solution. What makes it different from Pair 3, which has infinitely many solutions?
Pair 2’s constant ratio breaks the match; Pair 3’s does not
Pair 2’s $a$- and $b$-ratios are not equal to each other, unlike Pair 3’s
Pair 2 has larger coefficients than Pair 3
Pair 2 was solved by substitution and Pair 3 by elimination
Check your answer
✓ Pair 2’s constant ratio breaks the match; Pair 3’s does not — (A) Both pairs share equal $a$- and $b$-ratios; only the constant ratio decides which outcome each one gets.
Pair 2’s $a$- and $b$-ratios are not equal to each other, unlike Pair 3’s — Both Pair 2 and Pair 3 have equal $a$- and $b$-ratios — the split between them comes from the third ratio alone.
Pair 2 has larger coefficients than Pair 3 — Coefficient size never enters the ratio test; only the ratios formed from the coefficients decide the outcome.
Pair 2 was solved by substitution and Pair 3 by elimination — Neither pair is solved by substitution or elimination in this worked example — both are classified purely by the ratio test.
A fourth pair has $a_1/a_2 = b_1/b_2 = c_1/c_2 = 5$, matching Pair 3’s pattern of three equal ratios. What does the ratio test say about it?
no solution, since a ratio of $5$ is larger than any seen in Pair 1 or Pair 2
a unique solution, since the ratios are all the same value
it cannot be classified without seeing the actual coefficients
infinitely many solutions, exactly as in Pair 3
Check your answer
no solution, since a ratio of $5$ is larger than any seen in Pair 1 or Pair 2 — How large the shared ratio is plays no role; what matters is that all three ratios agree with each other.
a unique solution, since the ratios are all the same value — All three ratios matching is exactly the condition for infinitely many solutions, never a unique one.
it cannot be classified without seeing the actual coefficients — Once the three ratios are known and agree, the classification follows without needing the actual coefficient values.
✓ infinitely many solutions, exactly as in Pair 3 — (D) All three ratios agreeing gives the same outcome regardless of what the shared value happens to be.
Two bars show 20 minus 3y equals 5, so y equals 5, once x is written as 10 minus y.
Worked example
Solving x + y = 10 and 2x - y = 5 by substitution
$x = 10 - y$ We write $x$ in terms of $y$, from the first equation.
$2(10 - y) - y = 5$ Now substitute that into the second equation yourself.
$20 - 3y = 5$, so $y = 5$ Simplify, then solve the single remaining variable.
$x = 10 - 5 = 5$ Substitute $y = 5$ back into $x = 10 - y$.
$x = 5$, $y = 5$ satisfies both equations Checking the solution against both original equations confirms it.
$5 + 5 = 10$ and $2(5) - 5 = 5$ *Check by substituting $x = 5$, $y = 5$ back into both original equations.* Both hold, confirming the solution.
Five boxes carry the substitution method from writing x equals 10 minus y through to the checked answer, 5 and 5.
Solve any pair of linear equations by substitution.
Write one variable in terms of the other Pick the equation and the variable that gives the simplest expression. From $x + y = 10$, that is $x = 10 - y$.
Substitute into the other equation Put $10 - y$ in place of $x$ in $2x - y = 5$. That leaves one equation in one variable, $y$.
Solve for that variable Simplify $2(10 - y) - y = 5$ to $20 - 3y = 5$, so $y = 5$.
Substitute back Put $y = 5$ into $x = 10 - y$ to get $x = 5$.
Check both equations $5 + 5 = 10$ and $2(5) - 5 = 5$. Both hold, so $x = 5$, $y = 5$ is the solution.
Check yourself
Solving $x + y = 10$ and $2x - y = 5$ by substitution, the worked example starts from $x = 10 - y$. Which equation does this come from?
$2x - y = 5$, rearranged to isolate $x$
$x + y = 10$, rearranged to isolate $x$
both equations added together
neither equation; it is assumed as a starting guess
Check your answer
$2x - y = 5$, rearranged to isolate $x$ — Isolating $x$ in $2x - y = 5$ gives $x = (5+y)/2$, not $10 - y$.
✓ $x + y = 10$, rearranged to isolate $x$ — (B) Isolating $x$ in the first equation is exactly where this expression comes from.
both equations added together — Adding the two equations produces a new combined equation, not the simple isolated form $x = 10 - y$.
neither equation; it is assumed as a starting guess — This expression is derived directly by rearranging the first equation, not guessed at the start.
Substituting $x = 10 - y$ into $2x - y = 5$, a student writes $2(10 - y) - y = 5$ and simplifies to $20 - y = 5$. What went wrong?
the substitution itself is wrong; $x = 10 - y$ should not be used here
the final answer $y = 5$ is wrong even though the working shown is correct
$-y$ should have been added to $20$, not subtracted
$2(10-y)$ expands to $20 - 2y$, so the equation should read $20 - 3y = 5$
Check your answer
the substitution itself is wrong; $x = 10 - y$ should not be used here — The substitution $x = 10 - y$ is exactly correct; the error is in expanding the bracket that follows it.
the final answer $y = 5$ is wrong even though the working shown is correct — The working shown already drops a term, so its own final answer cannot be trusted as a starting point for checking.
$-y$ should have been added to $20$, not subtracted — The sign of $-y$ is not the issue; the missing step is expanding $2(10-y)$ into $20 - 2y$ before combining terms.
✓ $2(10-y)$ expands to $20 - 2y$, so the equation should read $20 - 3y = 5$ — (D) The expansion of $2(10-y)$ was not carried through fully before combining the $y$-terms.
Using the same substitution approach as $x + y = 10$ and $2x - y = 5$, what is the first useful expression to write for $x + y = 12$ and $3x - y = 8$?
$x = 8 - y$, borrowed from the second equation’s constant
$y = 12 - x$, then substitute into the first equation
$3x = 8 + y$, since the second equation looks easier
$x = 12 - y$, from rearranging the first equation
Check your answer
$x = 8 - y$, borrowed from the second equation’s constant — The constant $8$ belongs to the second equation; isolating $x$ from the FIRST equation uses $12$, not $8$.
$y = 12 - x$, then substitute into the first equation — Putting this expression back into the same equation it came from restates that equation, learning nothing new.
$3x = 8 + y$, since the second equation looks easier — This form still has $x$ multiplied by $3$ rather than isolated on its own, so it is not yet ready to substitute in.
✓ $x = 12 - y$, from rearranging the first equation — (D) The same move applies: isolate $x$ from the equation with the smaller coefficients.
Elimination is a subtraction of bars: match one block in both, and whatever the longer bar overhangs by is the other unknown on its own.
Worked example
Solving 2x + 3y = 13 and 3x + 2y = 12 by elimination
$4x + 6y = 26$ and $9x + 6y = 36$ We multiply the first equation by 2 and the second by 3, so $y$ has equal coefficients in both.
$5x = 10$, so $x = 2$ Subtract the two equations to eliminate $y$, then solve for $x$ yourself.
$2(2) + 3y = 13$, so $y = 3$ Substitute $x = 2$ back into the first original equation.
$x = 2$, $y = 3$ satisfies both equations Checking against both original equations confirms the solution.
$2(2) + 3(3) = 4 + 9 = 13$ and $3(2) + 2(3) = 6 + 6 = 12$ *Check by substituting $x = 2$, $y = 3$ back into both original equations.* Both hold, confirming the solution.
Solve any pair of linear equations by elimination.
Match one variable’s coefficients Multiply one or both equations by constants. For $2x + 3y = 13$ and $3x + 2y = 12$, multiplying by $2$ and $3$ gives $4x + 6y = 26$ and $9x + 6y = 36$.
Check the signs, then add or subtract Matched coefficients with the same sign are removed by subtracting. Here, subtracting gives $5x = 10$, so $x = 2$.
Substitute back Put $x = 2$ into $2x + 3y = 13$ to get $3y = 9$, so $y = 3$.
Check both equations $2(2) + 3(3) = 13$ and $3(2) + 2(3) = 12$. Both hold, so $x = 2$, $y = 3$ is the solution.
Watch for a stray statement If the last step leaves $0 = 0$, the pair has infinitely many solutions. If it leaves something false, like $0 = 9$, the pair has none.
Check yourself
Solving $2x + 3y = 13$ and $3x + 2y = 12$ by elimination, the worked example multiplies the first equation by $2$ and the second by $3$. Why these particular numbers?
they match the constant terms $13$ and $12$ as closely as possible
they are the $x$-coefficients, applied to the wrong equation each time
any two numbers would work equally well
they make the $y$-coefficient equal to $6$ in both equations
Check your answer
they match the constant terms $13$ and $12$ as closely as possible — The constants $13$ and $12$ are never the target here; a variable’s coefficient is.
they are the $x$-coefficients, applied to the wrong equation each time — $2$ and $3$ are chosen specifically to make the $y$-coefficients match, not because of any mix-up with $x$.
any two numbers would work equally well — Other multiplier pairs would generally leave the $y$-coefficients unequal — this specific pair is what makes them match.
✓ they make the $y$-coefficient equal to $6$ in both equations — (D) Matching the $y$-coefficient at $6$ is exactly what lets subtracting remove $y$.
For the same pair, $2x + 3y = 13$ and $3x + 2y = 12$, which alternative multiplier pair would eliminate $x$ instead of $y$?
multiply the first equation by $3$ and the second by $2$
multiply the first equation by $2$ and the second by $3$, as already done
multiply both equations by $6$
multiply the first equation by $12$ and the second by $13$
Check your answer
✓ multiply the first equation by $3$ and the second by $2$ — (A) This makes the $x$-coefficient equal to $6$ in both equations instead.
multiply the first equation by $2$ and the second by $3$, as already done — That multiplier pair is the one already used to eliminate $y$; eliminating $x$ needs the multipliers swapped.
multiply both equations by $6$ — Multiplying both equations by the same number never changes the ratio between their coefficients.
multiply the first equation by $12$ and the second by $13$ — Matching the constants $12$ and $13$ has no bearing on eliminating either variable.
For $4x + y = 9$ and $2x + y = 5$, is elimination cheaper here than for $2x + 3y = 13$ and $3x + 2y = 12$?
no, both pairs need the same amount of multiplying
yes, but only because the constants $9$ and $5$ are smaller
yes, the $y$-coefficients already match at $1$
no, elimination always needs multiplying before subtracting
Check your answer
no, both pairs need the same amount of multiplying — The $y$-coefficients here are already equal at $1$; the earlier pair needed multiplying by $2$ and $3$ first.
yes, but only because the constants $9$ and $5$ are smaller — Constant size never decides how much multiplying is needed — matching coefficients does.
✓ yes, the $y$-coefficients already match at $1$ — (C) Subtracting cancels $y$ in one step here, with no multiplying step first.
no, elimination always needs multiplying before subtracting — Multiplying is only needed to MAKE coefficients match; when they already match, subtracting works directly.
The line 4x + 6y = 12 is drawn again, dashed, over 2x + 3y = 6, showing they are the same line.
Worked example
Elimination finds infinitely many solutions
$2x + 3y = 6$ and $4x + 6y = 12$ These are the two equations we solve by elimination.
$4x + 6y = 12$ Multiply the first equation by 2, and it now matches the second equation exactly.
$0 = 0$ Subtracting one equation from the other leaves a statement true for every $x$ and $y$.
the pair has infinitely many solutions An always-true statement, with no variable left, signals infinitely many solutions rather than a single answer.
$a_1/a_2 = b_1/b_2 = c_1/c_2 = 1/2$ Check the answer against the ratio test, a completely different method. All three ratios match, confirming the same answer.
Check yourself
Solving $2x + 3y = 6$ and $4x + 6y = 12$ by elimination, multiplying the first equation by $2$ gives $4x + 6y = 12$, identical to the second equation. What does this signal?
an arithmetic mistake was made somewhere in the multiplication
the pair has no solution, since nothing new was learned
$y$ has been eliminated, giving a value for $x$ alone
the two equations describe the same line
Check your answer
an arithmetic mistake was made somewhere in the multiplication — The identical result is exactly what a correct multiplication reveals when one equation is a multiple of the other.
the pair has no solution, since nothing new was learned — Learning nothing new because the two equations coincide is exactly the infinite-solutions signal, not no solution.
$y$ has been eliminated, giving a value for $x$ alone — Neither variable was eliminated here — the two equations turned out identical rather than reduced to one unknown.
✓ the two equations describe the same line — (D) An identical result after scaling means one equation was always a multiple of the other.
Subtracting the two identical equations leaves $0 = 0$. A student concludes this means no solution. What is the error?
$0 = 0$ is always true — infinitely many, not none
no error; $0 = 0$ always means the pair cannot be solved
the error is in the subtraction step itself, which should give $0 = 1$
$0 = 0$ means exactly one solution, at $x = 0$ and $y = 0$
Check your answer
✓ $0 = 0$ is always true — infinitely many, not none — (A) A statement true for every $x$ and $y$ signals the equations hold everywhere together, not nowhere.
no error; $0 = 0$ always means the pair cannot be solved — $0 = 0$ being true for every $x$ and $y$ is exactly why the pair has infinitely many solutions, not none.
the error is in the subtraction step itself, which should give $0 = 1$ — The subtraction is carried out correctly; two identical equations always subtract to $0 = 0$.
$0 = 0$ means exactly one solution, at $x = 0$ and $y = 0$ — A statement true for every value of $x$ and $y$ describes EVERY point on the line, not one single point.
The worked example notes $a_1/a_2 = b_1/b_2 = c_1/c_2 = 1/2$ for this pair. How does this match the elimination result?
all three ratios equal, matching elimination’s result
it does not match; the ratio test and elimination can disagree on the same pair
the ratio $1/2$ means the second equation is smaller and therefore less reliable
the match is a coincidence specific to this pair
Check your answer
✓ all three ratios equal, matching elimination’s result — (A) The ratio test and elimination are two routes to the same classification, and here both agree.
it does not match; the ratio test and elimination can disagree on the same pair — The ratio test and elimination always agree on the same pair, since both classify the same equations.
the ratio $1/2$ means the second equation is smaller and therefore less reliable — A ratio of $1/2$ only compares matching coefficients — it never speaks to how reliable an equation is.
the match is a coincidence specific to this pair — The ratio test and elimination agree on EVERY pair of linear equations, not only this particular one.
A situational problem: two numbers, given their sum and difference
let the numbers be $x$ and $y$ We name the two unknown quantities first, in words, before writing an equation.
$x + y = 45$ and $x - y = 5$ Translate the sum and the difference directly into one equation each.
$2x = 50$, so $x = 25$ Add the two equations yourself. This eliminates $y$ directly.
$y = 45 - 25 = 20$ Substitute $x = 25$ back into either original equation.
the two numbers are $25$ and $20$ Name the two numbers the question asked for, in the order it asked for them.
$25 + 20 = 45$ and $25 - 20 = 5$ Check both numbers against the original conditions. Both hold, so 25 and 20 really are the answer.
The two lines from the sum and difference problem cross at 25 and 20, the numbers that satisfy both conditions.
Two bars laid end to end total 45, and laid from a shared start show a difference of 5.
Turn a word problem into a pair of equations, and solve it.
Name the unknowns Write down, in words, what $x$ and $y$ stand for. Here, let $x$ and $y$ be the two numbers.
Translate each condition Turn each sentence into one equation. Sum $45$ gives $x + y = 45$; difference $5$ gives $x - y = 5$.
Solve the pair Add the two equations to eliminate $y$. $2x = 50$, so $x = 25$.
Find the other unknown Substitute back. $y = 45 - 25 = 20$.
Check against the words $25 + 20 = 45$ and $25 - 20 = 5$. Both conditions hold, so the two numbers are $25$ and $20$.
Check yourself
The sum of two numbers is $45$ and their difference is $5$. The worked example lets the numbers be $x$ and $y$ and writes $x + y = 45$ and $x - y = 5$. Why two separate equations?
one equation for $x$ and a different, unrelated one for $y$
because two numbers always need exactly two equations, regardless of the problem
to make the arithmetic simpler, with no real relationship in the problem
each stated condition, the sum and the difference, becomes its own equation
Check your answer
one equation for $x$ and a different, unrelated one for $y$ — Both equations here involve $x$ and $y$ together; neither equation belongs to just one variable alone.
because two numbers always need exactly two equations, regardless of the problem — The equation count matches the number of CONDITIONS given, not a fixed rule about how many unknowns there are.
to make the arithmetic simpler, with no real relationship in the problem — Each equation restates a condition the problem actually states, not an arbitrary simplification.
✓ each stated condition, the sum and the difference, becomes its own equation — (D) One condition maps to one equation, giving exactly two equations for two conditions.
The worked example adds $x + y = 45$ and $x - y = 5$ to get $2x = 50$. What would subtracting the two equations instead give directly?
$2y = 40$, leading to $y = 20$
$2x = 40$, leading to $x = 20$
$2y = 50$, leading to $y = 25$
$0 = 40$, since $x$ and $y$ both cancel
Check your answer
✓ $2y = 40$, leading to $y = 20$ — (A) Subtracting cancels $x$ instead of $y$, leaving $2y = 45 - 5 = 40$.
$2x = 40$, leading to $x = 20$ — Subtracting these two equations cancels $x$, leaving an equation in $y$ alone, not $x$.
$2y = 50$, leading to $y = 25$ — The correct right-hand side after subtracting is $45 - 5 = 40$, not $50$.
$0 = 40$, since $x$ and $y$ both cancel — Only $x$ cancels here — $y$ remains, with coefficient $2$, giving $2y = 40$.
Two numbers have a sum of $60$ and a difference of $12$. Using the same approach as the worked example, what pair of equations fits?
$x + y = 60$ and $x - y = 12$
$x + y = 12$ and $x - y = 60$
$x - y = 60$ and $x + y = 12$
$x = 60y$ and $x - y = 12$
Check your answer
✓ $x + y = 60$ and $x - y = 12$ — (A) The sum condition and the difference condition each attach to their own equation, unchanged in structure from the worked example.
$x + y = 12$ and $x - y = 60$ — The sum is $60$ and the difference is $12$ — this option attaches each total to the wrong equation.
$x - y = 60$ and $x + y = 12$ — This again mismatches the totals — the sum condition should give $60$, not the difference.
$x = 60y$ and $x - y = 12$ — A sum of $60$ means $x + y = 60$, not that $x$ is $60$ TIMES $y$.
RATIOS: $a_1/a_2 \neq b_1/b_2$ means one solution. For $x + y = 5$ and $x - y = 1$, $a_1/a_2 = 1$ and $b_1/b_2 = -1$.
GRAPH: plot two points on each line, and read where they meet. The same pair crosses at $(3, 2)$.
SUBSTITUTION: write one variable in terms of the other, then substitute. Solving $x + y = 10$ and $2x - y = 5$ this way gives $x = 5$, $y = 5$.
ELIMINATION: match one variable’s coefficients, then add or subtract. Solving $2x + 3y = 13$ and $3x + 2y = 12$ this way gives $x = 2$, $y = 3$.
NO SOLUTION vs INFINITE: the third ratio decides. $6x + 3y - 15 = 0$ with $2x + y - 6 = 0$ gives no solution, $c_1/c_2 = 5/2$; $4x + 6y - 10 = 0$ with $2x + 3y - 5 = 0$ gives infinitely many, all three ratios equal to $2$.
Every pair of linear equations in two variables reduces to one question: do the two lines meet at one point, never meet, or lie on top of each other?
We have now seen the same point reached three separate ways. Look back at the five rules above. The pair $x + y = 5$, $x - y = 1$ runs through the first two of them: once as a ratio comparison, once as a picture, both landing on the same point, $(3, 2)$. Every method here finds that same point, no matter which route you take to it.
Try it yourself. Pick any of the four worked pairs above, and solve it by a different method from the one used there. (Answer: you should land on the same $x$ and $y$ either way.)
One pair of lines runs parallel with no solution, and another pair lies on the same line with infinitely many.
Two bars, x and y, total 30 and differ by 8, giving x equals 19 and y equals 11.
Check yourself
The ratio test, the graphical method, and substitution or elimination all answer the same question for a pair of linear equations. What is that question?
do the lines meet once, never, or coincide
which equation was written first
how many terms each equation contains
which variable, $x$ or $y$, is more important
Check your answer
✓ do the lines meet once, never, or coincide — (A) All three routes ultimately classify the pair into one of these three outcomes.
which equation was written first — None of the three methods cares which equation is written first — the classification is symmetric in both.
how many terms each equation contains — Every linear equation has the same three terms by its general form; none of the three methods look at this.
which variable, $x$ or $y$, is more important — All three methods treat $x$ and $y$ symmetrically — neither is more important than the other.
For $x + y = 9$ and $x - y = 3$, which route answers ‘what are $x$ and $y$’ with the least work: ratio test, graphing, or elimination?
the ratio test, since it needs no arithmetic with $x$ and $y$ at all
graphing, since the crossing point can be read off directly
elimination, since adding the equations immediately gives $x$
all three take exactly the same amount of work for this pair
Check your answer
the ratio test, since it needs no arithmetic with $x$ and $y$ at all — The ratio test classifies the pair as consistent or not — it never produces actual values for $x$ and $y$.
graphing, since the crossing point can be read off directly — Plotting points and drawing two lines takes more steps than simply adding these two equations together.
✓ elimination, since adding the equations immediately gives $x$ — (C) The ratio test never finds the actual values, so only graphing and elimination can even answer this question, and adding the equations here is the fastest.
all three take exactly the same amount of work for this pair — The ratio test cannot answer this question at all, so it cannot be tied for least work with the methods that can.
A pair of equations is confirmed, by the ratio test, to have infinitely many solutions. What would graphing that pair show?
two separate lines crossing at one visible point
two lines that never meet anywhere on the page
no line can be drawn at all for either equation
the lines would coincide exactly
Check your answer
two separate lines crossing at one visible point — One visible crossing point is the picture for a UNIQUE solution, not for infinitely many.
two lines that never meet anywhere on the page — Two lines that never meet is the no-solution picture, the opposite outcome from infinitely many solutions.
no line can be drawn at all for either equation — Each equation is still an ordinary linear equation and draws as a straight line — they simply coincide.
✓ the lines would coincide exactly — (D) Every route to the same classification must show the same picture, and infinitely many solutions means coincident lines.
Check yourself: the whole chapter
A pair of linear equations in $x$ and $y$ is really a pair of straight lines. Before drawing or solving anything, what decides whether the pair has one solution, no solution, or infinitely many?
how large the constant terms $c_1$ and $c_2$ are on their own
whether $x$ or $y$ has a larger coefficient
comparing the ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$
how many terms each equation has
Check your answer
how large the constant terms $c_1$ and $c_2$ are on their own — The raw size of $c_1$ or $c_2$ tells you nothing on its own — it is the ratio $c_1/c_2$, compared with $a_1/a_2$ and $b_1/b_2$, that matters.
whether $x$ or $y$ has a larger coefficient — Comparing which coefficient is larger is not part of the test — the ratios between the two equations’ matching coefficients are what decide the outcome.
✓ comparing the ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ — (C) Comparing the three ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ tells you whether the pair has a unique solution, no solution, or infinitely many, before anything is drawn or solved.
how many terms each equation has — Every linear equation in two variables has the same three terms by its general form — counting them tells you nothing new about the pair.
The substitution method solves a pair of linear equations by writing one variable in terms of the other from one equation, substituting it into the OTHER equation, then solving. If the last step gives $0 = 0$, with no variable left, what does this mean?
There is no solution to the pair.
A mistake was made somewhere in the algebra.
The value of $x$ must be $0$.
The pair has infinitely many solutions.
Check your answer
There is no solution to the pair. — An always-false statement, like $0 = 9$, is what signals no solution — $0 = 0$ is always true, which signals the opposite.
A mistake was made somewhere in the algebra. — Reaching $0 = 0$ is not a mistake — it is a valid outcome that shows the two equations describe the same line.
The value of $x$ must be $0$. — “$0 = 0$” has no $x$ in it at all — it is not solving for $x$, it is a statement about the whole pair of equations.
✓ The pair has infinitely many solutions. — (D) A statement that is always true, such as $0 = 0$, with no variable left, means the two equations describe the same line — the pair has infinitely many solutions.
To eliminate $y$ from $3x + 2y = 12$ and $5x + 4y = 22$ using the elimination method, what should be done first?
add the two equations directly, since both contain $y$
multiply the first equation by $2$, so both equations have $4y$
multiply the second equation by $2$, so it has $8y$
subtract the two equations directly, without multiplying either one
Check your answer
add the two equations directly, since both contain $y$ — Adding directly only cancels $y$ when its coefficients are equal and opposite in sign — here they are $2$ and $4$, not matched yet.
✓ multiply the first equation by $2$, so both equations have $4y$ — (B) Multiplying the first equation by $2$ turns its $2y$ into $4y$, matching the second equation’s $4y$ — now the two can be subtracted to eliminate $y$.
multiply the second equation by $2$, so it has $8y$ — Multiplying the second equation by $2$ gives $8y$, which still does not match the first equation’s $2y$ — the first equation needs the multiplying, to reach $4y$.
subtract the two equations directly, without multiplying either one — Subtracting directly only cancels $y$ once its coefficients match — here they are still $2$ and $4$.
For the pair $3x + 4y - 7 = 0$ and $6x + 5y - 9 = 0$, with $a_1/a_2 = 3/6 = 1/2$ and $b_1/b_2 = 4/5$ (unequal), what does this tell you about the pair?
the pair has no solution
the pair has infinitely many solutions
the pair has exactly one solution
the pair cannot be solved without also comparing $c_1/c_2$
Check your answer
the pair has no solution — No solution needs $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ — the first two ratios equal, not unequal as they are here.
the pair has infinitely many solutions — Infinitely many solutions needs all three ratios equal — the first two ratios here are already unequal, which rules this out.
✓ the pair has exactly one solution — (C) $a_1/a_2 \neq b_1/b_2$ means the two lines have different slopes, so they cross at exactly one point — the pair has a unique solution.
the pair cannot be solved without also comparing $c_1/c_2$ — The third ratio is only needed when the first two ratios are equal — once $a_1/a_2 \neq b_1/b_2$, the pair already has a unique solution.
For the pair $2x + 3y - 6 = 0$ and $4x + 6y - 5 = 0$, the ratios are $a_1/a_2 = 1/2$, $b_1/b_2 = 1/2$, and $c_1/c_2 = 6/5$. What does this tell you about the pair?
the pair has exactly one solution
the pair has infinitely many solutions
the pair is dependent, sharing infinitely many common solutions
the lines are parallel and the pair has no solution
Check your answer
the pair has exactly one solution — A unique solution needs $a_1/a_2 \neq b_1/b_2$ — here the first two ratios are equal, which rules this out.
the pair has infinitely many solutions — All three ratios being equal is what gives infinitely many solutions — here $c_1/c_2 = 6/5$ does not match the first two, so this case does not apply.
the pair is dependent, sharing infinitely many common solutions — A dependent pair is always consistent, sharing infinitely many solutions — a pair with no solution is inconsistent, never dependent.
✓ the lines are parallel and the pair has no solution — (D) $a_1/a_2 = b_1/b_2 \neq c_1/c_2$ means the lines share a slope but are not the same line — they are parallel and never meet, so the pair has no solution.
Two cost lines for Class A and Class B cross at 3 months and ₹1400, after which Class B stays cheaper.
You will compare two costs for the rest of your life: two plans, two shops, two prices. Here are seven of those places.
Watching a savings plan grow. You start with $₹500$ saved. You add $₹200$ every week after that. After $x$ weeks you have $y = 200 x + 500$ rupees. The maths: every $(x, y)$ pair that fits this equation lies on one straight line, out of infinitely many such pairs. At $x = 6$ weeks: $y = 200 \cdot 6 + 500 = 1700$ rupees.
Comparing two tuition fee plans. Class A charges $₹500$ joining plus $₹300$ a month. Class B charges $₹800$ joining plus $₹200$ a month. The maths: since the two monthly rates differ, the ratio test says there is exactly one month when both cost the same. At $x = 3$ months: Class A costs $300 \cdot 3 + 500 = 1400$ and Class B costs $200 \cdot 3 + 800 = 1400$.
Comparing two gym memberships. Two gyms both charge $₹800$ a month. Their joining fees are $₹200$ and $₹500$. The maths: the two lines are parallel, same rate and different start, so the pair has no solution. For any $x$ months: $(800x + 500) - (800x + 200) = 300$ rupees, every time.
A shared rate is what makes a pair unsolvable: with the slopes equal the joining-fee gap never closes, so no month answers the question.
Two bills that tell you nothing new. One bill says $2$ pens and $3$ notebooks cost $₹120$. A friend’s bill says $4$ pens and $6$ notebooks cost $₹240$. The maths: all three ratios are equal, so the pair has infinitely many solutions. You cannot find the price of one pen. Here $2/4 = 3/6 = 120/240$. Pens at $₹30$ and notebooks at $₹20$ fit both bills. So do pens at $₹15$ and notebooks at $₹30$.
Counting cash in a till. Your cash box has $25$ notes of $₹10$ and $₹20$ only, worth $₹380$ in all. The maths: write one note count in terms of the other, then substitute, to find how many of each. Let $x$ be $₹10$ notes: $10x + 20(25 - x) = 380$ gives $x = 12$. So $12$ notes of $₹10$ and $13$ of $₹20$.
Blending rice to a target price. A grocer mixes rice at $₹40$ a kg and $₹60$ a kg to make $10$ kg costing $₹520$ in all. The maths: match the coefficients of one variable, then subtract, to find how many kg of each grade. Multiply $x + y = 10$ by $40$: $40x + 40y = 400$. Subtract from $40x + 60y = 520$: $20y = 120$, so $y = 6$ kg.
Printing posters for a fest. A print shop prints $40$ posters for a college fest. A2 size costs $₹120$ each and A3 costs $₹60$ each, for $₹3600$ in all. The maths: name the two unknown counts as $x$ and $y$, turn each fact into one equation, then solve. Let $x$ be the count of A2 posters: $120x + 60(40 - x) = 3600$ gives $x = 20$. So $20$ of each size.
Your turn. A shop has $25$ shirts, some priced $₹150$ and some $₹300$, worth $₹6000$ in all. How many of each price? Answer: Let $x$ be $₹150$ shirts: $150x + 300(25 - x) = 6000$ gives $x = 10$. So $10$ shirts at $₹150$ and $15$ at $₹300$.
practice Solve $x + y = 7$ and $x - y = 3$ graphically. (Worked in full below — read it, then do the next two the same way.)
practice On comparing the ratios $a_1/a_2$, $b_1/b_2$ and $c_1/c_2$, state whether the lines represented by $2x - y - 5 = 0$ and $4x - 2y - 6 = 0$ are intersecting, parallel, or coincident. (Start the same way — write all three ratios down before deciding anything.)
practice On comparing the same three ratios, state whether the lines represented by $3x - 2y - 7 = 0$ and $x + y - 4 = 0$ are intersecting, parallel, or coincident. (Two ratios are enough to settle this one.)
practice The sum of a father’s age and his son’s age is 40 years, and the father’s age is three times the son’s age. Represent this situation as a pair of linear equations and solve it graphically.
practice On comparing the ratios $a_1/a_2$, $b_1/b_2$ and $c_1/c_2$, state whether the lines represented by $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$ are intersecting, parallel, or coincident.
practice On comparing the same three ratios, state whether the lines represented by $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$ are intersecting, parallel, or coincident.
practice On comparing the same three ratios, state whether the lines represented by $3x + y - 8 = 0$ and $x - y = 0$ are intersecting, parallel, or coincident.
Answers
The lines cross at $(5, 2)$, so $x = 5$ and $y = 2$. Substituting back gives $5 + 2 = 7$ and $5 - 2 = 3$, so both equations hold.
$a_1/a_2 = b_1/b_2 = 1/2$ but $c_1/c_2 = 5/6$, so the lines are parallel and the pair has no solution.
$a_1/a_2 = 3$ and $b_1/b_2 = -2$, and $3 \neq -2$, so the lines intersect and the pair has a unique solution.
Son’s age $y$, father’s age $x$: $x + y = 40$, $x = 3y$. Solved graphically, the lines cross at $(30, 10)$ — the father is 30, the son is 10.
$a_1/a_2 = b_1/b_2 = 1/2$, but $c_1/c_2 = 1/3$: parallel, no solution.
$a_1/a_2 = 3$, $b_1/b_2 = -1$: intersecting, a unique solution.
Exercise 3.1 — further practice
practice On comparing the ratios $a_1/a_2$, $b_1/b_2$ and $c_1/c_2$, state whether the lines represented by $x + y - 7 = 0$ and $2 x - y - 2 = 0$ are intersecting, parallel, or coincident.
practice On comparing the same three ratios, state whether the lines represented by $4 x - 5 y + 7 = 0$ and $8 x - 10 y + 3 = 0$ are consistent or inconsistent.
practice Which of these best describes the pair $3 x - y - 5 = 0$ and $6 x - 2 y - 7 = 0$?
Intersecting lines, a unique solution
Parallel lines, no solution
Coincident lines, infinitely many solutions
Cannot be decided without solving
practice A shop sells 3 pens and 2 pencils together for ₹47, and 4 pens and 3 pencils together for ₹64. Form a pair of linear equations for this situation and solve it graphically to find the cost of one pen and one pencil.
practice Half the perimeter of a rectangular field is 28 m, and the length is 6 m more than the breadth. Form a pair of linear equations for this situation and solve it graphically to find the field’s length and breadth.
practice For what value of $k$ do the equations $2 x + k y = 5$ and $4 x + 6 y = 9$ represent a pair of parallel lines?
practice Write one linear equation that, together with $3 x + 2 y - 12 = 0$, forms a pair of coincident lines.
practice The cost of 2 tables and 3 chairs is ₹4100, and the cost of 3 tables and 2 chairs is ₹4400. Using the ratio test, check first whether this pair of equations has a unique solution, then solve it graphically to find the cost of one table and one chair.
Answers
$a_1/a_2 = 1/2$, $b_1/b_2 = -1$: since $1/2 \neq -1$, the lines are intersecting and the pair has a unique solution.
$a_1/a_2 = b_1/b_2 = 1/2$, but $c_1/c_2 = 7/3$: the lines are parallel, so the pair is inconsistent.
B — Parallel lines, no solution.
$3 x + 2 y = 47$, $4 x + 3 y = 64$: the lines cross at $(13, 4)$ — one pen costs ₹13, one pencil costs ₹4.
$x + y = 28$, $x - y = 6$: the lines cross at $(17, 11)$ — the field’s length is 17 m and its breadth is 11 m.
$k = 3$.
Any non-zero multiple works, since the ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ must all be equal — for example, $6 x + 4 y - 24 = 0$, obtained by multiplying every term by $2$.
$a_1/a_2 = 2/3$ and $b_1/b_2 = 3/2$ are unequal, so the pair has a unique solution: the lines cross at $(1000, 700)$ — one table costs ₹1000, one chair costs ₹700.
The father’s age line and the son’s age line cross at a father of 30 and a son of 10.
practice Solve $x + y = 12$ and $3x - y = 8$ by the substitution method. (Worked in full below — read it, then do the next two the same way.)
practice The sum of two numbers is 30, and one of them is four times the other. Find the two numbers. (Start the same way — let the numbers be $x$ and $y$, write the pair $x + y = 30$ and $x = 4y$, then substitute.)
practice A pen and a notebook together cost ₹90, and the notebook costs ₹30 more than the pen. Find the cost of each. (Name the two costs first, then substitute.)
practice The difference between two numbers is 26, and one number is three times the other. Find the two numbers.
practice Two angles are supplementary. The larger angle exceeds the smaller by 18°. Find both angles.
practice 3 bats and 6 balls together cost ₹1800, while 5 bats and 3 balls together cost ₹2650. Find the cost of one bat and one ball.
practice A taxi charges a fixed amount plus a rate per kilometre travelled. A 10-kilometre ride costs ₹150, and a 15-kilometre ride costs ₹200. Find the fixed charge and the rate per kilometre.
practice A fraction becomes $1$ when $1$ is added to its numerator, and becomes $1/2$ when $1$ is added to its denominator instead. Find the fraction.
practice Five years ago, a mother was five times as old as her daughter. Ten years from now, she will be twice as old as her daughter will be then. Find their present ages.
Answers
$x = 5$ and $y = 7$.
The numbers are $24$ and $6$.
The pen costs ₹30 and the notebook ₹60.
$x - y = 26$, $x = 3y$: the numbers are 39 and 13.
$x + y = 180$, $x - y = 18$: the angles are 99° and 81°.
$x + 2y = 600$, $5x + 3y = 2650$: one bat costs ₹500, one ball costs ₹50.
practice By the substitution method, what is the solution of $x + y = 6$ and $x - y = 2$?
$(3, 3)$
$(4, 2)$
$(2, 4)$
$(5, 1)$
practice One number is 4 more than twice another number, and their sum is 25. Find the two numbers.
practice Two angles are complementary. The larger angle exceeds the smaller by 24°. Find both angles.
practice Solve the pair $3 x - 2 y = 11$ and $4 x + 3 y = 26$ using the substitution method.
practice The cost of 4 kg of apples and 3 kg of oranges is ₹440, while the cost of 2 kg of apples and 5 kg of oranges is ₹360. Find the cost per kilogram of each fruit.
practice A mobile plan charges a fixed monthly rent plus a fixed rate per minute of calls. A month with 100 minutes of calls costs ₹250, and a month with 160 minutes costs ₹310. Find the fixed rent and the rate per minute.
practice In a test of 20 questions, each correct answer earns 3 marks and each wrong answer loses 1 mark. Rhea attempted every question and scored a total of 40 marks. Find the number of questions she answered correctly and the number she got wrong.
practice Two numbers are in the ratio 3 : 5. If 5 is added to each number, the ratio becomes 2 : 3. Find the numbers.
Answers
B — $(4, 2)$.
$x = 2 y + 4$, $x + y = 25$: the numbers are 18 and 7.
$x + y = 90$, $x - y = 24$: the angles are 57° and 33°.
$x = 5$, $y = 2$.
$4 x + 3 y = 440$, $2 x + 5 y = 360$: apples cost ₹80 per kg, oranges cost ₹40 per kg.
$x + 100 y = 250$, $x + 160 y = 310$: the fixed rent is ₹150, and the rate is ₹1 per minute.
$x + y = 20$, $3 x - y = 40$: she answered 15 questions correctly and 5 wrongly.
$y = 5 x/3$, $3(x + 5) = 2(y + 5)$: the numbers are 15 and 25.
The two ride conditions become lines that cross at a fixed charge of 50 rupees and a rate of 10 rupees per kilometre.
practice Solve $3x + 2y = 16$ and $2x + 3y = 14$ by elimination. (Worked in full below — read it, then do the next two the same way.)
practice Solve $4x + y = 14$ and $2x - y = 4$ by elimination. (Start the same way — the $y$ terms already match in size, at $+1$ and $-1$. Read those signs before you choose the operation.)
practice The sum of the digits of a two-digit number is 9, and the number is 27 more than the number formed by reversing its digits. Find the number. (Call the tens digit $x$ and the units digit $y$, so the number is $10x + y$.)
practice Solve the pair $2x + 3y = 11$ and $2x - 4y = -24$, once by elimination and once by substitution.
practice A fraction becomes $9/11$ when $2$ is added to both its numerator and denominator, and becomes $5/6$ when $3$ is added to both instead. Find the fraction.
practice The present age of a father is six years more than three times his son’s age. Three years from now, the father’s age will be ten years more than twice the son’s age. Find their present ages.
practice A two-digit number is 4 more than 6 times the sum of its digits. Subtracting 18 from the number reverses its digits. Find the number.
practice A person has only ₹50 and ₹100 notes. She has 25 notes in all, worth ₹1750 together. Find how many notes of each kind she has.
practice A lending library charges a fixed amount for the first three days a book is kept, and a further amount for each day after that. Keeping a book for 4 days costs ₹27; keeping it for 7 days costs ₹45. Find the fixed charge and the daily charge beyond the first three days.
Answers
$x = 4$ and $y = 2$.
$x = 3$ and $y = 2$ — the matched terms have opposite signs, so the equations are added, not subtracted.
The number is $63$.
$x = -2$, $y = 5$, by both methods.
$11x - 9y = -4$, $6x - 5y = -3$: the fraction is $7/9$.
$x = 3y + 6$, $x = 2y + 13$: father 27, son 7.
$4x - 5y = 4$, $x - y = 2$: the number is 64.
$x + y = 25$, $x + 2y = 35$: 15 notes of ₹50 and 10 notes of ₹100.
practice Solve the pair $3 x + 2 y = 12$ and $5 x - 2 y = 4$, once by elimination and once by substitution.
practice A person has ₹20 and ₹10 notes only, 30 notes in all, worth ₹450 together. How many ₹20 notes does she have?
10
15
20
25
practice The sum of the digits of a two-digit number is 12. Adding 18 to the number reverses its digits. Find the number.
practice Five years ago, an uncle was three times as old as his nephew. Ten years from now, he will be twice as old as his nephew will be then. Find their present ages.
practice A boat travels 30 km downstream in 3 hours, and the same 30 km upstream in 5 hours. Form a pair of linear equations in the boat’s speed in still water and the speed of the stream, and solve them.
practice A fraction reduces to $2/3$ when 2 is subtracted from both its numerator and denominator, and reduces to $3/4$ when 1 is added to both instead. Find the fraction.
Answers
$x = 2$, $y = 3$, by both elimination (adding the equations directly) and substitution (writing $x$ in terms of $y$ from either equation).
B — 15.
$x + y = 12$, $y - x = 2$: the number is 57.
$x - 3 y = -10$, $x - 2 y = 10$: the uncle is 50 and the nephew is 20.
$x + y = 10$, $x - y = 6$: the boat’s speed in still water is 8 km/h and the stream’s speed is 2 km/h.
$3 x - 2 y = 2$, $4 x - 3 y = -1$: the fraction is $8/11$.
