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The one rule that is not like an equation
◆FRAME
Solve $-x < 5$.
Treat it like an equation for a moment. Multiply both sides by $-1$ and you get $x < -5$.
Now test that. If $x < -5$ then $x = -10$ should work. Put it back into the original: is $-(-10) < 5$? That asks whether $10 < 5$, which is false. The answer is wrong.
Try the other direction. If $x > -5$, take $x = 0$: the original becomes $0 < 5$, true. Take $x = -4$: it becomes $4 < 5$, true as well.
So the answer is $x > -5$. Multiplying by $-1$ flipped the symbol. The same thing happens in the other direction: $x < 5$ becomes $-x > -5$, never $-x < -5$.
The reason fits in one line. On the number line $-3$ sits to the RIGHT of $-8$, but 3 sits to the LEFT of 8. Multiplying by a negative reflects every number through zero, and a reflection swaps left and right. Order goes with it.
Adding and subtracting do nothing of the kind — they slide every number the same distance the same way, and sliding keeps the order. Multiplying by a POSITIVE number stretches without reflecting, so that keeps the order too.
One operation in the whole list behaves unlike an equation, and it is silent. Nothing looks wrong on the page; the answer is simply false.
The top two pictures are the same set of numbers written in two ways: mirror one and you get the other, which is all that multiplying by minus one does. The red line keeps the symbol and shades somewhere else entirely. Test it with zero if the rule ever feels arbitrary — zero satisfies x < 5, and it satisfies minus x > minus 5, but it does not satisfy minus x < minus 5.
The same two places, described twice. Nothing about A and B changed between the lines — only the quantity they are being measured in, and pressure runs the opposite way to altitude. That is the flip rule before it is algebra, and it is why multiplying a comparison by a negative number cannot leave the symbol alone.
$<$ and $>$ are STRICT — “less than” and “greater than”, with the boundary value itself excluded. $\leq$ and $\geq$ are NON-STRICT — “less than or equal to” and “greater than or equal to”, with the boundary included. The extra line under the symbol is the “or equal to”.
The bigger change from equations is what an answer looks like.
Solve $2 x = 6$ and you get $x = 3$. One number. Solve $2 x < 6$ and you get $x < 3$, which is not a number at all. It is 2.9, and 0, and $-100$, and every other value below 3. That collection is the SOLUTION SET.
So “solve” means something wider here: describe every value that makes the statement true. Usually there are infinitely many of them, and the description IS the answer.
That is also why these answers get drawn. A single number is easy to state; a set running off to infinity is easier to see, which is what the number line in the next section is for.
An equation asks which value works. An inequality asks which values work, and expects a range for an answer.
Read down a column and exactly one thing changes: the circle at 2. Hollow leaves 2 out, filled takes it in — that is all the extra bar under the symbol does. Read across a row and only the direction changes. In every one of the four, the answer is a shaded stretch rather than a single number.
One symbol is different between the two rows, and the shape of the answer changes with it. Where the equation stops at 3, the inequality treats 3 as the edge it never reaches — every number to the left of it works, so writing the answer means describing the stretch rather than naming a value.
Three rules cover all of it, and two of them you already use on equations.
RULE 1. Adding or subtracting the same quantity on both sides never changes the direction. If $x < 5$ then $x + 2 < 7$. Both sides slid two to the right, and the order held.
RULE 2. Multiplying or dividing both sides by the same POSITIVE number never changes the direction either. If $x < 5$ then $3 x < 15$. Both sides stretched away from zero by the same factor, and stretching preserves order.
RULE 3. Multiplying or dividing both sides by the same NEGATIVE number REVERSES the direction. If $x < 5$ then $-x > -5$.
Only rule 3 is new, and it is the flip rule from the opening, sitting in its proper place — one entry in a list of three, not a separate topic.
Solving is then the routine you already know: get x by itself, one operation at a time. The only extra work is a question at each step. Was that a multiply or divide by a negative? If so, flip the symbol as you write the next line.
Do the flip in the same stroke as the operation. Planning to fix it afterwards is how it gets forgotten.
Worked example
Solve $-2x+5>9$
$-2x > 4$ subtract 5 from both sides — a subtraction, so rule 1 applies and the symbol stays untouched.
$x < -2$ divide both sides by $-2$ — a division by a negative, so rule 3 fires and the symbol reverses; the arithmetic ($4$ divided by $-2$ is $-2$) and the direction flip are independent decisions.
$-2(-3)+5 = 11$ check a value inside the answer, $x=-3$: $11 > 9$ is true, so it satisfies the original inequality.
$-2(0)+5 = 5$ check a value the answer excludes, $x=0$: $5 > 9$ is false, so it is correctly excluded — testing one value inside your answer and one outside is cheap, and if both behave, the direction of the symbol is right.
The next figure, test-point-method, carries this same check one dimension further. There the boundary is a whole line rather than one crossing value, so instead of substituting a single number back into the inequality, you substitute a single point that is not on that line. Either way the check costs one substitution, and it is exactly where a sign flipped backwards would show up.
The answer to a one-variable inequality is a RAY on the number line: a boundary point, and everything past it in one direction.
Drawing it takes two decisions, and they are independent.
First, the CIRCLE at the boundary. An OPEN circle — hollow — means the boundary value is EXCLUDED, which is what a strict $<$ or $>$ asks for. A FILLED circle means the boundary value IS included, which is what $\leq$ or $\geq$ asks for. The circle carries the whole strict-versus-non-strict distinction by itself.
Second, the DIRECTION of the ray. It runs toward the values that satisfy the inequality. For $x < 3$ it runs left, toward smaller numbers. For $x > 3$ it runs right.
Two decisions with two answers each means four possible pictures at any given boundary, and only one of them is right. Reading either decision wrongly produces a wrong answer that still looks tidy on the page.
The circle answers “is the boundary in?” and the ray answers “which side?”. Settle them one at a time.
Only the first of these answers the question that was asked. The other three are not sloppy drawings; each is a correct picture of a different inequality, which is why getting either decision wrong produces a confident wrong answer rather than an obviously broken one.
The same drawing, three strokes in. Notice that the circle is settled at step 2, before any arrow exists — it comes off the symbol and nothing else. Draw the arrow first and the circle turns into a guess, because by then there is a picture on the page to match instead of a symbol to read.
Worked example
Represent $x \geq -3$ on a number line
$-3 \geq -3$ true — the symbol is $\geq$, non-strict, so the boundary value satisfies the inequality: the circle at $-3$ is FILLED.
$x \geq -3$ the values that work are $-3$ and everything larger; larger means further right, so the ray runs RIGHT from $-3$.
$0 \geq -3$, $-5 \geq -3$ true, then false — reading the picture back as a sentence, everything from $-3$ rightwards with $-3$ included, should match the inequality exactly, and checking one included point and one excluded point confirms it does; if that sentence ever stops matching, the picture is wrong.
The filled circle says minus 3 is itself a solution, which is what the bar under the inequality sign buys you. The arrow carries on without stopping because every larger number qualifies too — the drawing ends, the solution set does not.
Two or more inequalities can be imposed at once. The word that decides everything is AND.
A SYSTEM asks for the values that satisfy every inequality in it, simultaneously. So the method comes in two parts: solve each one on its own, then keep only what they have in common.
That common part is the INTERSECTION of the solution sets — the overlap. It is not the combination of everything either one allows, which is a different and almost always larger set.
The number line makes the difference visible. Draw one ray for each inequality, one above the other. The system’s answer is the stretch where both rays are present. Anything covered by only one of them fails the other, so it is out.
Two things can happen. The rays overlap, and the answer is that overlap — often a bounded stretch with a boundary at each end. Or they do not overlap at all, and no value satisfies both. An empty answer is a legitimate result, not a sign that you made a mistake.
Solve them separately, then take the overlap. The overlap step is where systems are lost.
The two rays point away from each other, so no value sits under both, and the second line is what that looks like as an answer. It is a real answer, not a failed drawing. A reader who shades everything either condition allows has answered a different question — a system asks for the overlap.
Worked example
Solve $2x+1>5$ and $3x-4\leq 11$ together
$x > 2$ solve the first inequality alone — subtract 1 to get $2x>4$, then divide by 2, positive, so no flip.
$x \leq 5$ solve the second alone — add 4 to get $3x\leq 15$, then divide by 3, positive again.
$2 < x \leq 5$ picture the two rays: one open at 2 (since $>$ is strict), one filled at 5 (since $\leq$ is not) — they overlap between 2 and 5, and each end keeps the strictness it arrived with.
$x=3$: $7>5$ and $5\leq 11$; $x=2$: $5>5$ check the ends — both hold for $x=3$, so 3 is in; $5>5$ is false, so 2 is correctly excluded. Write the combined answer on one line, with each boundary keeping its own symbol; they do not have to match.
Two rays stacked above one number line, with the band where they overlap marked below it.
The compact form is two statements sharing one x, and each half governs its own end of the bar. That is the whole reason the ends are drawn differently — reading the expression left to right tells you which circle to fill, before you have looked at the picture at all.
Add a second variable and the answer changes shape.
An inequality like $a x + b y < c$ is a condition on a PAIR of numbers, $(x, y)$ — which is to say, on a point in the coordinate plane. Some points satisfy it and some do not, so the answer is the collection of points that do.
That collection is a REGION, with area. Not a line, and certainly not a single point.
The structure is simpler than it sounds. Replace the inequality sign with an equals sign and you have the BOUNDARY LINE, $a x + b y = c$. That line cuts the plane into exactly two pieces, called HALF-PLANES.
And each half behaves uniformly. Every point in one of them satisfies the inequality, and every point in the other fails it. There is no mixing and no third case — the boundary line is the only place the behaviour changes.
That uniformity is what makes the whole method work. One test settles an entire half-plane, which is why the next two sections need only decide the style of the line and which side to shade.
One line, two sides, one of them the answer. That is the whole picture for a two-variable inequality.
Replace the inequality sign with an equals sign and you have the boundary. That one line does not cut the plane into an answer and a blank — it cuts it into three sets, and every point in the plane falls into exactly one of them before any shading is decided. Which of the three you keep is what the symbol says, and it is why the method on the next pages needs a test point rather than more algebra.
●KEY-TERM
The boundary line has to say whether it is itself part of the answer, and it says it by how it is drawn.
DASHED for strict. Take $a x + b y < c$ or $a x + b y > c$. A point sitting exactly on the line gives $a x + b y = c$. That is neither less than nor greater than c, so the point fails. The line is drawn dashed, meaning “this edge is not included”.
SOLID for non-strict. For $a x + b y \leq c$ or $a x + b y \geq c$, a point on the line gives equality, which the “or equal to” accepts. It passes, so the line is solid and belongs to the region.
This is the number line’s open and filled circle again, one dimension up. There the boundary was a single point; here it is a whole line, and the same question is being answered about it.
Getting it wrong is not a small error, either. A dashed line drawn solid claims that an infinite set of points solves an inequality that none of them solves.
Read the symbol before drawing the line. The “or equal to” is what makes it solid.
The same line and shading drawn twice: dashed for the strict inequality, solid for the non-strict one.
▲CONCEPT
The line is drawn and two half-planes sit either side of it. Exactly one is the answer, and deciding which takes a single substitution.
Pick any point that is NOT on the boundary line. Substitute its coordinates into the inequality and see whether the statement comes out true.
If it is true, the half-plane containing that point is the solution — shade that side. If it is false, the solution is the other half-plane — shade the far side.
One test point settles the whole region, because every point on a given side behaves the same way. You are not sampling the region; you are identifying which half is which.
The origin $(0, 0)$ is almost always the point to use. Substituting zeros is the easiest arithmetic available, and the truth of the result is usually clear at a glance.
The one time the origin will not do is when the boundary line passes through it. Then $(0, 0)$ is on the line rather than off it, and it decides nothing. Pick something else clear of the line, such as $(1, 0)$.
One point, one substitution, one whole half-plane decided. Use the origin unless the line runs through it.
The worked example just before this one checked a solved inequality by substituting a single number; a two-variable inequality has no single crossing value to substitute back into, only a whole boundary line, so the check needs a whole point instead, chosen off that line rather than derived by algebra.
Worked example
Graph $x+y\leq 4$
$x+y=4$ through $(0,4)$ and $(4,0)$ replace $\leq$ by $=$ for the boundary line — two intercepts fix it.
$\leq$: non-strict so the boundary line is drawn SOLID — points on the line do satisfy the inequality.
$0+0=0$ the line does not pass through the origin, so test $(0,0)$: $0 \leq 4$ is TRUE, so the origin is in the solution — shade the half-plane containing it, the side below and left of the line.
$5+5=10$ sanity-check against the far side, $(5,5)$: $10 \leq 4$ is false, correctly outside — two intercepts fix the line, the symbol fixes its style, and one test point fixes the side: three decisions, in that order, every time.
The written inequality settles the first two steps and says nothing at all about the third. That silence is why the method needs a test point rather than more algebra, and step 3 is the one readers skip, because the first two look like the hard ones.
The line x + y = 4, shaded below and left, with the origin and (4, 4) tested and marked.
The worked example above begins from a line whose two ends it already knows. Those ends are not read off a graph — they are solved for, one letter at a time, and every boundary line in the rest of the chapter is fixed the same way. The right-hand case is the line Exercise 5.1 asks the intercepts of.
An inequality can name only one variable and still be a two-variable question. It depends on which plane you are working in.
Take $x < 3$. On a number line that is a ray. In the coordinate plane it is something much larger, because a point there needs BOTH coordinates and the condition constrains only one of them.
Work out what it allows. The point $(1, 0)$ qualifies, since $1 < 3$. So does $(1, 50)$, and $(1, -800)$. The letter y is never mentioned, so no value of y is ruled out.
That means the answer contains a whole vertical line of points above and below each qualifying x. Sweep across every x below 3 and the answer is the entire region to the left of the vertical line $x = 3$.
So the picture is built the usual way. The boundary is the vertical line $x = 3$, with y unrestricted. The style is dashed, since $<$ is strict. The shaded side is everything to the left.
A variable that does not appear is not restricted. Unrestricted means every value, and that is what turns a ray into a half-plane.
Each purple ray is the number-line answer you already know, drawn again at a different height. It comes out the same every time because the inequality never mentions y — and doing that at every height, not just four, is what the shading is.
Worked example
Graph $y\geq -2$ in the coordinate plane
$y=-2$ the variable named is y, so the boundary comes from $y=-2$ — a HORIZONTAL line two units below the x-axis, since every point with that y-coordinate lies on it whatever x may be.
$-2 \geq -2$ $\geq$ is non-strict, so the line is drawn SOLID.
$0 \geq -2$ TRUE — the line does not pass through the origin, so test $(0,0)$: its y-coordinate is 0, so shade the half-plane containing the origin, everything ABOVE the line. Compare with the earlier vertical boundary at $x<3$: the named variable and the direction of its boundary are perpendicular — the variable that appears fixes the boundary line, and the variable that does not appear is the direction that line runs in.
The horizontal line y = minus 2, with everything above it shaded across the full width of the grid.
The two inequalities differ by one letter and nothing else — same minus 2, same "or equal to", same solid edge. Which variable is named is the only thing deciding whether the boundary runs across the page or up it, so that is the thing to read first, before deciding which side to shade.
A system in two variables works exactly as it did in one. The word is still AND, and the operation is still INTERSECTION.
Each inequality in the system contributes one half-plane. The system’s solution is the set of points lying in EVERY one of them at once — the overlap of all the shaded regions, not their combination.
That overlap has a name worth knowing: the FEASIBLE REGION. The term comes from the problems this material was built for. There, each inequality is a real limit — a budget, a supply, a number of hours. The region is the set of plans that respect every limit at once.
Drawing it is mechanical. Graph each inequality separately: boundary, style, shaded side. Then find the area covered by every shading.
Three outcomes are possible.
The region can be BOUNDED, a closed polygon with corners where boundary lines cross.
It can be UNBOUNDED, open in some direction and running off the page.
Or the shadings may share no area at all, in which case no point satisfies every condition and the system has no solution.
The corners of the feasible region are where two boundary lines meet, and they are the points those real problems usually turn on.
Three inequalities, three separate shadings — none of them the answer on its own. The answer is the piece all three cover, and here it is a triangle with corners (1, 1), (5, 1) and (1, 5). Shading straight onto one grid gets you there too, but it is the same three decisions in the same order.
The chapter names three outcomes and shows only the first one twice. An unbounded region is not an unfinished drawing — the two constraints simply never close it off. And an empty answer is still an answer: both bands are shaded, neither shares a point with the other, so no pair of numbers satisfies the system.
Worked example
Graph the system $x+y\leq 4$, $x\geq 0$, $y\geq 0$
$x+y\leq 4$: solid through $(0,4)$, $(4,0)$, shaded below-left done earlier — the origin passes the test.
$x\geq 0$: solid boundary at $x=0$, shaded right names only x, the y-axis is the boundary, so every point with x-coordinate 0 or more satisfies it.
$y\geq 0$: solid boundary at $y=0$, shaded above the mirror image of the previous condition — the x-axis is the boundary.
triangle at $(0,0)$, $(4,0)$, $(0,4)$ the last two conditions keep only the first quadrant; the first condition then cuts that quadrant along $x+y=4$ — bounded by the two axes and the line.
$(1,1)$: $1+1=2$; $(3,3)$: $3+3=6$ check one interior point and one outside — $2 \leq 4$ keeps $(1,1)$ inside all three conditions; $6 \leq 4$ is false, so $(3,3)$ fails the first condition and is correctly out.
Three boundary lines, each with an inward arrow, and the triangle where all three overlap shaded.
Every corner on the left is the answer to a small pair of equations on the right. That matters when the picture is not drawn to scale or the numbers are not whole: the corner is still exactly where two boundaries meet, and solving the pair gets it without measuring anything.
The chapter has one genuinely new rule in it, and one word that does the rest.
The RULE: multiplying or dividing both sides by a negative number reverses the direction. Every other step you already used on equations still works untouched.
In ONE VARIABLE, solving is equation-solving with that single check added at each step. The answer is a ray on the number line, and two independent decisions draw it. The circle is open or filled according to strictness. The ray points toward the values that work.
In TWO VARIABLES, the same solving produces a region instead. The boundary line comes from replacing the inequality with an equation; it is dashed if strict and solid if not; and one test point, usually the origin, decides which side to shade.
The WORD is INTERSECTION. Whenever more than one inequality is imposed, in one variable or in two, the answer is what they have in common, never everything either one allows. On a number line that is the shared stretch; in the plane it is the feasible region.
One flip rule, and intersection every time two conditions meet. The rest of the chapter is those two ideas in a different number of dimensions.
THE TRAP. For $x > 3$, mark a filled circle at 3. It is the boundary value, so it ought to be marked plainly.
THE REALITY. The circle at 3 must be OPEN. A filled circle does not mean “boundary” — it means “this value is IN the solution set”, and 3 is not.
Test it directly. Is $3 > 3$? No. Three is not greater than itself, so it fails the very inequality being drawn. Filling the circle puts a value into the answer that does not belong there.
The rule pairs one symbol with one circle, and there are only two cases to remember. $<$ and $>$ take an OPEN circle. $\leq$ and $\geq$ take a FILLED one. The line underneath the symbol and the filling of the circle carry the same meaning.
This one is easy to slip on because a filled dot is easier to draw and reads as more definite. The picture then looks confident and says something false, which is the worst combination available.
Substitute the boundary value into the inequality before drawing. If it fails, the circle is open.
The circle at 3 is hollow, and that is the whole difference. A strict inequality is satisfied by every number past 3 but not by 3 itself, so filling that circle in would quietly add one wrong answer to the set.
Same boundary, same direction, one symbol apart — and the answer to "what is the smallest whole number that works" is not the same number. That is what filling the circle by habit actually costs, and the practice set asks for exactly this kind of answer more than once.
✕MISCONCEPTION
THE TRAP. $x < 3$ graphs as the vertical line $x = 3$. Since y never appears in the inequality, there is nothing to draw in the y direction.
THE REALITY. $x < 3$ graphs as a whole HALF-PLANE. The line $x = 3$ is only its boundary, and it is not even part of the answer.
Two things are wrong with the trap, and they pull in opposite directions.
It includes too much ON the line. The strict $<$ excludes $x = 3$ entirely, so that boundary should be DASHED. Drawing the line as the answer selects exactly the points that fail.
And it includes far too little OFF the line. Test $(1, 7)$: the condition asks only whether $1 < 3$, which is true, so $(1, 7)$ is in the solution. So is $(0, -40)$, and so is $(2.9, 1000)$.
A variable that does not appear has not been fixed at some value. It is UNRESTRICTED, free to be anything at all, and that freedom is what gives the region its height.
Absent from the inequality means unconstrained, not zero. The solution runs the full length of the plane in that direction.
The hollow point on the right is the one to watch: at (4.2, 0.6) its y-value looks perfectly ordinary, but its x sits past the border, so it fails on x alone — the same reason every point right of the border fails.
The line is not a partial answer waiting for shading. Every point on it has x equal to 3, and a strict inequality rejects all of them — which is why the line is drawn dashed and why the two red circles are hollow. Draw the line alone and you have drawn the one set of points guaranteed to be wrong.
Exercise 5.1 — Solving, graphing, and systems of inequalities
practice Solve $3x - 8 \leq 13$ for $x$.
practice Solve $-5x + 8 > 23$ for $x$.
practice Solve the system $2x - 3 < 7$ and $x + 4 \geq 2$ together.
practice Which circle marks the boundary of the solution set of $x \geq -2$ on a number line?
Filled circle at $-2$
Open circle at $-2$
Filled circle at $2$
Open circle at $2$
practice Is the boundary line for $4x - y > 8$ drawn solid or dashed?
Dashed
Solid
Both
Neither
practice Can the origin be used as a test point for $x - y > 0$? If not, name a point that works.
practice State the $x$- and $y$-intercepts of the boundary line for $2x + y = 6$.
practice For $y > -1$ graphed in the coordinate plane, is the boundary line horizontal or vertical, and is it solid or dashed?
practice The system $x + y \leq 5$, $x \geq 0$, $y \geq 0$ describes a feasible region of which shape?
Triangle
Rectangle
Unbounded strip
Empty set
Answers
$x \leq 7$
$x < -3$
$-2 \leq x < 5$
Filled circle at $-2$
Dashed
No — the origin lies on the boundary line; use $(1, 0)$.
$(3, 0)$ and $(0, 6)$
Horizontal, at $y = -1$; dashed.
Triangle
The last item in Exercise 5.1 asks what shape this system makes. Left is the answer, with all three corners named. Right is the same system with x ≥ 0 deleted — the two constraints that are just the axes are the easy ones to forget, and forgetting one is what this looks like. Count the corners when you are not sure which you have.
Miscellaneous — practice set
practice Solve $2(x-1) < 3(x+2) - 7$ for $x$.
practice Solve the system $3x + 4 > 2x - 1$ and $5 - 2x \geq 1$ together.
practice A student scores 70 and 75 in the first two tests. Find the least third score $x$ so that the average of all three is at least 75.
practice The system $x + 2y \leq 8$, $x \geq 0$, $y \geq 0$ describes a feasible region. Is it bounded, and what are the vertices?
Answers
$x > -1$
$-5 < x \leq 2$
$x \geq 80$
Bounded; vertices $(0, 0)$, $(8, 0)$, $(0, 4)$.
This same check works for any "at least" or "at most" word problem: try a score just below your answer and one above it, and confirm one fails and the other passes before trusting which way the inequality points. Skip the check and a flipped sign reads as correct all the way through.
The miscellaneous set opens with the first solve in this chapter where you choose which side the x terms go to. Both choices are legal, and the two workings do not look alike in the middle — so a line that disagrees with a classmate is not evidence that either route went wrong. What has to agree is the last one.
From the item bank
The boundary line ax + by = c splits the coordinate plane into two half-planes. What determines which half-plane is the solution of ax + by < c?
neither half-plane — only the boundary line itself satisfies a strict inequality.
whichever half-plane happens to be visually on the left of the line.
whichever half-plane satisfies the inequality, found by testing a point.
both half-planes together, since a strict inequality always covers the whole plane except the line.
Check your answer
neither half-plane — only the boundary line itself satisfies a strict inequality. — It is the reverse: a strict inequality excludes its own boundary line entirely, and the solution is exactly one of the two half-planes.
whichever half-plane happens to be visually on the left of the line. — Left and right have no fixed meaning here — which side solves the inequality depends on the inequality itself, checked by testing a point, not on a visual default.
✓ whichever half-plane satisfies the inequality, found by testing a point. — (C) Substituting a test point into the inequality shows which side actually satisfies it — that side is the solution half-plane.
both half-planes together, since a strict inequality always covers the whole plane except the line. — A strict inequality holds on exactly ONE side of the boundary line, not both — testing a point shows which single side qualifies.
From the item bank
When graphing ax + by ≥ c, how should the boundary line ax + by = c be drawn?
solid, but only if the shaded region also happens to include the origin.
dashed, since the inequality is non-strict.
dashed, since ≥ never includes its own boundary.
solid — points on the line satisfy a non-strict inequality too.
Check your answer
solid, but only if the shaded region also happens to include the origin. — Whether the boundary is solid or dashed depends only on strict vs non-strict, never on which side happens to contain the origin.
dashed, since the inequality is non-strict. — Non-strict inequalities (≤ or ≥) use a SOLID boundary line, not dashed — dashed is reserved for the strict symbols.
dashed, since ≥ never includes its own boundary. — ≥ specifically DOES include its own boundary — “greater than or equal to” means the boundary value itself satisfies the inequality, which is why the line is solid.
✓ solid — points on the line satisfy a non-strict inequality too. — (D) ≥ is non-strict, so points on the boundary line itself DO satisfy the inequality — the line is drawn solid.
From the item bank
A graph shows the boundary line for x - y < 3 drawn as a SOLID line. What is wrong?
the line should be dashed, since a strict inequality excludes its own boundary.
nothing is wrong — strict inequalities are always drawn with a solid boundary line.
the line should not be drawn at all, since strict inequalities have no boundary.
the line is correct as drawn, as long as the shaded region is on the correct side of it.
Check your answer
✓ the line should be dashed, since a strict inequality excludes its own boundary. — (A) x - y < 3 is strict, so points on the line x - y = 3 do NOT satisfy it — the boundary should be dashed, not solid.
nothing is wrong — strict inequalities are always drawn with a solid boundary line. — It is the opposite: strict inequalities (< or >) use a DASHED boundary, since the line itself is excluded from the solution.
the line should not be drawn at all, since strict inequalities have no boundary. — A strict inequality still has a boundary line, x - y = 3 — it is simply excluded from the solution, shown by drawing it dashed, not by omitting it.
the line is correct as drawn, as long as the shaded region is on the correct side of it. — Shading the correct side does not fix the boundary style — a strict inequality still requires a dashed line regardless of which side is shaded.
From the item bank
To decide which half-plane solves a two-variable inequality, the test-point method —
only works when the boundary line passes through the origin.
tests every point in the plane, one at a time, until the boundary is found.
tests a point off the boundary line against the inequality.
always tests the point (1,1), since it is the simplest possible choice.
Check your answer
only works when the boundary line passes through the origin. — The method works for any boundary line — the origin is simply a common, easy-to-substitute choice when it is not ON the boundary line itself.
tests every point in the plane, one at a time, until the boundary is found. — Only ONE convenient point needs testing — its result tells you the whole half-plane it sits in, without checking every other point.
✓ tests a point off the boundary line against the inequality. — (C) Any point off the boundary line works — substituting it into the inequality reveals which half-plane is the solution.
always tests the point (1,1), since it is the simplest possible choice. — No single point is mandatory — (1,1) is only useful when it is NOT on the boundary line; the origin is usually simpler, but any off-boundary point works.
From the item bank
Graph 2x - y > 2. Testing the origin: 2(0) - 0 = 0, and 0 > 2 is FALSE. Which half-plane is shaded, and how is the boundary drawn?
half-plane not containing the origin, boundary dashed.
the half-plane NOT containing the origin, boundary solid.
the half-plane containing the origin, boundary dashed.
the half-plane containing the origin, boundary solid.
Check your answer
✓ half-plane not containing the origin, boundary dashed. — (A) A false test result rules out the origin’s side, so the OTHER half-plane is shaded; the inequality is strict, so the boundary is dashed.
the half-plane NOT containing the origin, boundary solid. — The half-plane choice is right, but 2x - y > 2 is strict, so the boundary should be dashed, not solid.
the half-plane containing the origin, boundary dashed. — The false test result rules OUT the origin’s side — the shaded half-plane is the other one.
the half-plane containing the origin, boundary solid. — Both are wrong here: the false test result rules out the origin’s side, and a strict inequality needs a dashed boundary, not solid.
From the item bank
Why is the boundary of x < 3, graphed in the coordinate plane, a VERTICAL line rather than a horizontal one?
because every inequality with only one variable defaults to a vertical boundary, regardless of which variable it names.
because the coordinate plane only has horizontal lines available for two-variable inequalities.
because y is always zero in this inequality.
because the inequality restricts x, and a fixed value of x traces a vertical line as y varies freely.
Check your answer
because every inequality with only one variable defaults to a vertical boundary, regardless of which variable it names. — The orientation depends on WHICH variable is restricted — an inequality like y < -2 instead produces a horizontal boundary, not a vertical one by default.
because the coordinate plane only has horizontal lines available for two-variable inequalities. — The coordinate plane represents vertical lines just as naturally as horizontal ones — x = 3 is a perfectly ordinary vertical line on it.
because y is always zero in this inequality. — y is not fixed to zero at all — it is completely unrestricted; that is exactly why the boundary is vertical, tracing every y at x = 3.
✓ because the inequality restricts x, and a fixed value of x traces a vertical line as y varies freely. — (D) x = 3 is vertical because every point with that x-coordinate, for any y at all, lies on it — exactly the boundary x < 3 needs.
From the item bank
An amusement park ride requires a rider’s height code x to satisfy x ≥ 4 on a two-variable safety chart, where the other axis, y, tracks age and plays no role in this particular rule. Which region of the chart is marked SAFE?
only the single vertical line x = 4, since that is the exact restriction stated.
only the points where both x ≥ 4 and y ≥ 4, treating the restriction as applying to both axes.
the entire plane except a single horizontal line at y = 4.
the entire half-plane where x ≥ 4, for every value of y — age is irrelevant here.
Check your answer
only the single vertical line x = 4, since that is the exact restriction stated. — x = 4 is only the boundary of the safe region. Every point with x ≥ 4, not just the boundary line, satisfies the rule.
only the points where both x ≥ 4 and y ≥ 4, treating the restriction as applying to both axes. — The rule names only x — nothing restricts y at all, so requiring y ≥ 4 as well adds a condition the rule never states.
the entire plane except a single horizontal line at y = 4. — The restriction is on x, the height code, not y, the age — a horizontal line at y = 4 has nothing to do with this rule.
✓ the entire half-plane where x ≥ 4, for every value of y — age is irrelevant here. — (D) Since y never appears in x ≥ 4, the rule holds for every age — the safe region is the whole half-plane x ≥ 4, not narrowed by y at all.
From the item bank
Graph x < 5 in the coordinate plane. Testing the origin: 0 < 5 is TRUE. Which region is shaded, and how is the boundary drawn?
the half-plane to the right of x = 5, boundary solid.
the half-plane to the left of x = 5, boundary solid.
the half-plane to the right of x = 5, boundary dashed.
the region left of x = 5, boundary dashed.
Check your answer
the half-plane to the right of x = 5, boundary solid. — Both are wrong: the true test result keeps the LEFT side, and a strict inequality needs a dashed boundary, not solid.
the half-plane to the left of x = 5, boundary solid. — The shaded side is right, but x < 5 is strict, so the boundary should be dashed, not solid.
the half-plane to the right of x = 5, boundary dashed. — The test result was TRUE, which keeps the origin’s side — the region to the LEFT of x = 5, not the right.
✓ the region left of x = 5, boundary dashed. — (D) The true test result keeps the origin’s side; x < 5 is strict, so the boundary at x = 5 is dashed.
From the item bank
Graph the system x + y ≤ 6, x ≥ 0, y ≥ 0, and additionally x ≤ 4. Which shape best describes the feasible region?
a triangle, the same shape any system of linear inequalities always produces.
an unbounded region extending infinitely, since x ≤ 4 does not close off the region on its own.
a quadrilateral — x ≤ 4 cuts a corner off what would otherwise be a triangle.
a single line segment, since four inequalities over-constrain the system down to one dimension.
Check your answer
a triangle, the same shape any system of linear inequalities always produces. — A feasible region’s shape depends on how many boundary lines actually cut into it — adding the fourth constraint x ≤ 4 here changes the shape to four sides, not three.
an unbounded region extending infinitely, since x ≤ 4 does not close off the region on its own. — x ≤ 4 does not need to close the region alone — together with x ≥ 0, y ≥ 0, and x + y ≤ 6, the whole system is fully bounded.
✓ a quadrilateral — x ≤ 4 cuts a corner off what would otherwise be a triangle. — (C) Without x ≤ 4 the region would be the triangle bounded by the axes and x + y = 6; adding x ≤ 4 slices off its top corner, leaving a four-sided region.
a single line segment, since four inequalities over-constrain the system down to one dimension. — Four inequalities in two variables can still leave a full two-dimensional region — here they bound an ordinary four-sided area, not a single line.
From the item bank
Which of the following points satisfies x < 3, under the correct half-plane interpretation (not the vertical-line-only misconception)?
(3, 0)
none of these — x < 3 only contains points exactly on the line x = 3, which none of these are.
(4, 0)
(2, 100)
Check your answer
(3, 0) — x = 3 fails the strict inequality x < 3 — the boundary value itself is excluded.
none of these — x < 3 only contains points exactly on the line x = 3, which none of these are. — x < 3 is not restricted to the boundary line at all — it describes the whole half-plane to its left, which does include points like (2, 100).
(4, 0) — x = 4 is greater than 3, failing x < 3 entirely — this point sits on the wrong side of the boundary.
✓ (2, 100) — (D) x < 3 only restricts the x-coordinate; (2,100) has x = 2 < 3, so it satisfies the inequality regardless of its y-value.
From the item bank
To find which half-plane solves x + y > 4, a student tests the origin (0,0): 0 + 0 = 0, and 0 > 4 is FALSE. What does this tell you?
The origin’s half-plane IS the solution — shade toward the origin.
The origin’s half-plane is NOT the solution — shade the OTHER half-plane, away from the origin.
The inequality has no solution, since (0,0) fails it.
Try a different point until one satisfies the inequality, then shade based on that point instead.
Check your answer
The origin’s half-plane IS the solution — shade toward the origin. — The origin FAILING the inequality rules OUT the origin’s side — the solution is the OTHER half-plane, not the origin’s.
✓ The origin’s half-plane is NOT the solution — shade the OTHER half-plane, away from the origin. — (B) The origin fails the inequality, so its half-plane is excluded — the solution is the half-plane on the OTHER side of the boundary line.
The inequality has no solution, since (0,0) fails it. — A failing test point only tells you which half-plane to shade — it says nothing about whether a solution exists; the other half-plane still satisfies x + y > 4.
Try a different point until one satisfies the inequality, then shade based on that point instead. — One test point is always enough — since (0,0) fails, the entire half-plane containing it is excluded, and the other half-plane is the full solution.
From the item bank
Graph $x - y < 3$. Testing the origin: $0 - 0 = 0$, and $0 < 3$ is TRUE. Which half-plane is shaded, and how is the boundary drawn?
the half-plane NOT containing the origin, boundary solid.
the half-plane NOT containing the origin, boundary dashed.
the half-plane containing the origin, boundary dashed.
the half-plane containing the origin, boundary solid.
Check your answer
the half-plane NOT containing the origin, boundary solid. — Both are wrong here: a true test result keeps the origin’s side, and a strict inequality needs a dashed boundary, not solid.
the half-plane NOT containing the origin, boundary dashed. — A TRUE test result confirms that side, it does not exclude it — the origin’s half-plane is exactly the solution here.
✓ the half-plane containing the origin, boundary dashed. — (C) A TRUE test result keeps the origin’s side; $x - y < 3$ is strict, so the boundary is dashed.
the half-plane containing the origin, boundary solid. — The half-plane choice is right, but $x - y < 3$ is strict, so the boundary should be dashed, not solid.
From the item bank
Graph $y \leq 2x$. The boundary line $y = 2x$ passes through the origin, so the origin cannot be used as a test point. Testing $(1, 0)$ instead: $0 \leq 2(1) = 2$ is TRUE. Which half-plane is shaded, and how is the boundary drawn?
the half-plane containing (1, 0), boundary solid.
the origin can still be used as the test point — it is always the standard convenient choice.
the half-plane NOT containing (1, 0), boundary solid.
the half-plane containing (1, 0), boundary dashed.
Check your answer
✓ the half-plane containing (1, 0), boundary solid. — (A) The test point (1, 0) satisfies the inequality, so its half-plane is shaded; $y \leq 2x$ is non-strict, so the boundary is solid.
the origin can still be used as the test point — it is always the standard convenient choice. — The origin sits ON this boundary line ($0 = 2(0)$), so it cannot tell the two half-planes apart — a different test point is required whenever the boundary passes through the origin.
the half-plane NOT containing (1, 0), boundary solid. — A TRUE test result confirms that side, it does not exclude it — (1, 0)’s half-plane is exactly the solution here.
the half-plane containing (1, 0), boundary dashed. — The half-plane choice is right, but $y \leq 2x$ is non-strict, so the boundary should be solid, not dashed.
From the item bank
Graph $x \leq -2$ in the coordinate plane. Testing the origin: $0 \leq -2$ is FALSE. Which region is shaded, and how is the boundary drawn?
the region left of x = -2, boundary dashed.
the region below the line, boundary solid.
the region right of x = -2, boundary solid.
the region left of x = -2, boundary solid.
Check your answer
the region left of x = -2, boundary dashed. — The region choice is right, but $x \leq -2$ is non-strict, so the boundary should be solid, not dashed.
the region below the line, boundary solid. — x never appears as y — the named variable is x, so the boundary is vertical, not horizontal, and there is no ‘below’ side to it.
the region right of x = -2, boundary solid. — The false test result rules OUT the origin’s side — the shaded region is the other one, left of x = -2.
✓ the region left of x = -2, boundary solid. — (D) A FALSE test result rules out the origin’s side, so the region left of x = -2 (not containing the origin) is shaded; $x \leq -2$ is non-strict, so the boundary is solid.
board-standard For $x < 3$ graphed in the coordinate plane, does the line $x = 3$ alone show the solution?
board-standard Graph $x + y \leq 4$. Testing the origin, which side is shaded?
board-standard Solve $(3x-2)/4 \geq (x+1)/2$ for x.
board-standard Solve the compound inequality $-3 \leq 2x - 1 < 5$ for x.
board-standard For the system $x \geq 0$, $y \geq 0$, $x + y \leq 6$, what shape is the feasible region?
a triangle
a square
an unbounded strip
a single point
board-standard Solve $5 - 2x > 1$ for x.
board-standard Solve $2x - 7 \leq 3x + 1$ for x.
board-standard A student scored 62 and 48 in two tests out of 100. Find the least score needed in a third test to average at least 60.
board-standard A firm’s profit is $200$ more than 3 times its cost C. Find the least cost for a profit of at least 5000.
board-standard Solve $3x - 7 > 2(x-6)$ and $6 - x \geq 11 - 2x$ together.
JEE Solve $(2x-1)/3 \geq (3x-2)/4 - (2-x)/5$ for x.
JEE x litres of a 40% acid solution are mixed with $(10-x)$ litres of a 15% solution to make 10 litres. Find x so the mixture exceeds 25% acid.
JEE Solve $-8 \leq 5x - 3 < 7$ and count the integer values of x satisfying it.
JEE Notebooks cost 15 rupees and pens cost 8 rupees. A buyer wants twice as many pens as notebooks, spending at most 200. Find the greatest number of notebooks.
JEE For the system $x + y \leq 10$, $x \geq 2$, $y \geq 3$, does the point $(5,4)$ lie in the feasible region?
JEE For the system $2x + y > 4$ and $x - y < 1$, does $(3,0)$ satisfy both?
JEE A product costs $500 + 20x$ to make and sells for $45x$. Find the least whole number of units for a profit.
JEE Solve $2 \leq (3x-4)/5 \leq 8$ for x.
$14/3 \leq x \leq 44/3$
$14/3 \leq x \leq 40/3$
$2/3 \leq x \leq 44/3$
$14 \leq x \leq 44$
Answers
$x < 4$
$x \leq -3$
filled
$x > 4$
$\leq$
$x > 4$
dashed
no
$x \geq -6$
intersection
$x \geq 3$
$-3 < x \leq 4$
no — a half-plane
origin’s side
$x \geq 4$
$-1 \leq x < 3$
a triangle
$x < 2$
$x \geq -8$
$x \geq 70$
$C \geq 1600$
$x \geq 5$
$x \leq 2$
$4 < x \leq 10$
3 integers
6 notebooks
yes
no
21 units
$14/3 \leq x \leq 44/3$
Item 10 of this set asks whether a system is solved by union or by intersection. This is what the rejected answer would have cost: the two conditions already overlap, so taking either one leaves nothing ruled out. Intersection is not a convention to remember here; it is the only one of the two that narrows anything.
The chapter-end set asks whether (5, 4) is in this region. It is — but the two hollow markers are the ones worth looking at. Each clears two of the three constraints comfortably and fails only the third, which is what being outside usually looks like: not far out, just out on the one inequality nobody checked.
The chapter-end set asks how many integers satisfy minus 8 at most 5x minus 3 less than 7. Solving gives the band drawn here, and then the counting is done by eye — four integers sit in the range and only three of them are inside, because the right-hand end is open. Nothing but the circle at 2 decides the answer.
Items 11 and 12 of this set are the same substitution with opposite outcomes. On the left the origin satisfies the inequality, so the half-plane it sits in is the answer. On the right it does not, so the answer is the other half-plane and the test point ends up outside the region it just selected. The test point never marks the answer; it only says which side does.
The perimeter condition gives the top rail and nothing else — every width at most 10, the negative ones included. The lower bound appears nowhere in the item’s wording and nowhere in the algebra: it comes from the rectangle, because a width is a length. That is the whole distance between the answer and the distractor "w at most 10, no lower bound", which is the algebra’s own result, correct and incomplete.