Inequalities and shaded overlap

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Type each inequality and Desmos shades it. The SAT answer sits where the shading overlaps: read the boundary lines, test a point and pick the whole number.

Difficulty: Beginner
Estimated time: 35 minutes
Domains: Algebra

What you’ll learn

Type inequalities straight into Desmos.
Read a shaded region as the set of solutions.
Tell a solid boundary, which counts, from a dotted one, which doesn't.
Test a point against an inequality.
Find the overlap where every inequality in a system is true.
Turn the limits in a word problem into inequalities, and use the matching edge of the overlap to find the greatest or least whole-number value.
Tell when plugging in one point by hand is faster.

Prerequisites

Why this matters on the SAT

Find the point that satisfies every inequality

Many SAT inequality questions give you two or more conditions and ask which point makes all of them true. Type the inequalities into Desmos, and each one shades the points that make it true. The answer is wherever all the shadings overlap.

It's the same idea as solving systems at intersections, with one difference. Two equations share the points where their graphs cross. Inequalities can share a whole region.

SAT example

Which ordered pair $(x,y)$ satisfies the system of inequalities below?

\begin{aligned} y&>2x-3\\[1.4em] x+y&\le 6 \end{aligned}

Fast Desmos solution

Type both inequalities exactly as written, then plot the four answer choices. Only $(2,2)$ lands in both shaded regions.

$(0,-3)$ and $(3,3)$ sit right on the dotted line of $y>2x-3$ , and points on a dotted line don't count. The answer is B.

The graph shows you every solution at once, but some checks are still your job: whether a line counts, which number is $x$ , and any limits a word problem adds.

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Only

(2,2)

lands where both shadings overlap.

When should you graph inequalities in Desmos?

Desmos earns its place when several conditions on $x$ and $y$ all have to be true, because you can see where they overlap instead of checking each one in your head.

Start with Desmos when…

the inequalities use both $x$ and $y$ , like $y>2x-3$ .
two or more inequalities have to be true at once, like the pair in the SAT example above.
you’d have to test several answer choices against two or more inequalities, or against one awkward one.
the lines are hard to picture in your head, like the boundary of $15x+9y\ge310$ .
a word problem puts several limits on two quantities, especially when it asks for the most or the fewest of something.

Use substitution when…

you have one simple inequality and one point to check. For $(2,1)$ and $3x+y\le8$ , one line of arithmetic does it: $3(2)+1=7$ , and $7\le8$ .

These questions usually sound like “Which ordered pair satisfies the inequality?”, “Which combination could meet all the requirements?”, “What is the greatest possible value?”, or “What is the least possible value?”

Check your understanding:

Which job would you start in Desmos: checking whether $(2,1)$ satisfies $3x+y\le8$ , or finding which point satisfies three inequalities in $x$ and $y$ ? What makes them different?

Read the shading and the boundary

An inequality splits the graph into two parts:

the boundary line, where the two sides are equal
the shaded half-plane, the side full of points that make the inequality true

Take

y\le -\frac{1}{2}x+4.

Desmos shades below the line. The line itself is solid, because "or equal to" lets points sit right on it.

The three plotted points, all at $x=2$ , show every case:

$(2,3)$ is on the line, and it counts.
$(2,1)$ is in the shading, so it works.
$(2,5)$ is above the line, outside the shading, so it doesn't.

The symbol decides whether the line counts. Solid is in, dotted is out:

Symbol	How Desmos draws the line	Do points on the line count?
$<$ or $>$	Dotted	No
$\le$ or $\ge$	Solid	Yes

To get $\le$ or $\ge$ in Desmos, type <= or >=.

Try it yourself:

Change <= to <. Watch what happens to the line, then decide: does $(2,3)$ still work?

Check your understanding:

For $y<2x+1$ , does the point $(1,3)$ satisfy the inequality?

Common mistake:

A point sitting exactly on a line is a classic trap answer, because it looks like it belongs. Before you pick it, check the symbol. It counts only with $\le$ or $\ge$ .

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The line is solid, so points on it count, along with every point below it.

Plot the choices, then verify

You can type an inequality into Desmos exactly as it's written. There's no need to solve for $y$ first.

Say a question asks which point satisfies

4x+3y\le18.

Here's the routine:

Type 4x+3y<=18.
Plot each answer choice on its own line.
Keep a point only if it's in the shading or on a solid line.
Plug your final pick into the inequality to make sure.

Check your understanding:

Which plotted point satisfies $4x+3y\le18$ : $(0,6)$ , $(2,4)$ , $(3,3)$ , or $(4,1)$ ?

Common mistake:

Watch the order. In $(2,4)$ , the first number is $x$ and the second is $y$ , so $x=2$ and $y=4$ . Swap them and you’re testing a different point.

When a point looks like it's right on the edge, don't trust your eyes. Plug in its coordinates and let the numbers decide.

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The inequality is typed as written, with the four answer choices plotted.

Example: Find the overlapping region

Now let's add a third inequality. The idea doesn't change: the answer lives where all the shadings overlap.

Worked example

Which ordered pair $(x,y)$ satisfies the system of inequalities below?

x>0

y\ge x+1

y<-2x+7

A
$(1,1)$
B
$(2,3)$
C
$(1,3)$
D
$(3,0)$

Step 1

List what the point needs

The point has to be to the right of $x=0$ , on or above $y=x+1$ , and below $y=-2x+7$ . Only $y=x+1$ is a solid line here. The other two are dotted.

Step 2

Graph all three

Type each inequality on its own line. The solutions are the points where all three shadings overlap.

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Find the patch where all three shadings overlap.

Step 3

Check each answer choice

Choice C, $(1,3)$ , sits inside the overlap. Choice A is below the solid line $y=x+1$ . Choice B sits right on the dotted line $y=-2x+7$ , so it's out. Choice D is below the region altogether.

Step 4

Confirm it by hand

Plug in $(1,3)$ :

1>0,\qquad 3\ge1+1,\qquad 3<-2(1)+7.

All three are true, so the answer is C.

Common mistake: A point can sit in two shadings and still miss the third. Check every inequality before you answer.

Some questions flip this around and ask which constant gives a system no solutions, or at least one. Those use sliders and solution counts, which come later.

Find the greatest or least value on a feasible edge

In a word problem, each inequality is one limit. The overlap holds every combination that obeys all the limits at once, so it's called the feasible region, the set of choices that are actually possible. When a question asks for the greatest or least value of something, the direction it asks for tells you which edge of that region to look at.

Say a school buys $x$ notebooks at $2 each and $y$ folders at $3 each, where $x$ and $y$ are whole numbers that can't be negative. It needs at least $24$ items, can spend at most $60, and must buy at least $8$ folders. What's the greatest number of notebooks it can buy?

Check your understanding:

Before you read on, try writing the three inequalities yourself.

Add $x\ge0$ , since the school can't buy a negative number of notebooks, and the system is

\begin{aligned} x+y&\ge24\\[1.4em] 2x+3y&\le60\\[1.4em] y&\ge8\\[1.4em] x&\ge0. \end{aligned}

The question wants the greatest $x$ , and $x$ grows as you move right, so look at the rightmost part of the overlap. The other directions work the same way:

For the least $x$ , look at the leftmost part.
For the greatest $y$ , look at the highest part.
For the least $y$ , look at the lowest part.

The rightmost whole-number point in the overlap is $(18,8)$ . Check it exactly:

18+8=26\ge24,

2(18)+3(8)=60\le60,

and $8\ge8$ . So $18$ notebooks is possible. Could the school buy $19$ ? Even with the fewest folders allowed, $y=8$ , the bill would be

2(19)+3(8)=62,

which is over budget.

Check your understanding:

Checking $(18,8)$ shows that $18$ notebooks is possible. Why does the whole argument show that $18$ is the greatest, not only possible?

Common mistake:

It’s tempting to read a decimal coordinate off the edge and call it your count. But Desmos shades every point, including ones with fractions and decimals, and a school can’t buy $18.5$ notebooks. Move to a whole-number point inside the region, plug it into every original inequality, then check that the next whole number in the direction you want fails.

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For the greatest

x

, look at the right edge.

(18,8)

is on the solid budget line, so it counts, and

(19,8)

is just past it.

Finish the solution

Both inequalities are already graphed for you. Plot the four choices, find the one in the overlap, and pick its letter.

Finish the solution

Which ordered pair $(x,y)$ satisfies the system of inequalities below?

\begin{aligned} y&>-x+2\\[1.4em] y&\le2x+5 \end{aligned}

First steps

Both inequalities are in, with their shading showing.
Plot each answer choice on its own line.
Keep the point that's in both shadings, remembering that a point on a dotted line doesn't count.

Finish it

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Plot each choice and see which one lands in both shadings.

Practice problems

Your turn. Before each one, decide whether you'd graph it or check it by hand, and give it a real try before you open the hint.

Practice problem

Which ordered pair $(x,y)$ satisfies the inequality

2x+5y\le16?

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Practice problem

Which ordered pair $(x,y)$ satisfies the system of inequalities below?

\begin{aligned} y&>x-2\\[1.4em] 2x+y&\le8 \end{aligned}

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Practice problem

A theater sells adult tickets for $15 each and student tickets for $9 each. For one show:

at most $24$ tickets can be sold;
ticket revenue must be at least $310; and
at least $8$ student tickets must be sold.

Let $a$ be the number of adult tickets sold and $s$ be the number of student tickets sold. What is the greatest possible value of $a$ ?

Your answer

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Graph every limit, then look at the right edge of the overlap.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Type inequalities as written. Rearrange one only when that saves real time.
The shading shows every point that makes the inequality true.
Solid is in, dotted is out.
A system's solutions are where every shading overlaps.
Plot the answer choices, then plug your pick in to confirm it.
In a word problem, turn every limit into an inequality, and remember that counts are whole numbers that can't be negative.
For the greatest or least $x$ , look at the right or left edge. For $y$ , look at the top or bottom.
Desmos shades decimals too, so pick a whole-number point, check it, and check that the next value fails.
One point and one short inequality? Plug it in by hand.

Next lesson

Sliders for unknown constants

Vary an unknown constant and observe how it changes an equation, function, or graph.

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Practice

Practice this lesson

818 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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