Graph and model two-variable inequalities

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Difficulty: Intermediate
Estimated time: 35 minutes
Domains: Algebra
Techniques: Boundary-linesTest-pointsFeasible-regionsInteger-constraintsBinding-boundsGreatest-least-feasible-values

What you’ll learn

Picture a two-variable inequality as a line plus a shaded side.
Tell a solid line, whose points count, from a dashed one, whose points don't.
Check whether a point works in one inequality, or in every inequality of a system.
Turn the limits in a word problem into inequalities.
Find the limit that stops a count from going further, and prove the greatest or least whole number it can be.

Prerequisites

Why this matters on the SAT

Find the point where every condition works

These SAT questions are about possible combinations: pairs of numbers that fit several rules at once. One rule might set a minimum and another a maximum, and the answer has to satisfy both. On a graph, each rule becomes a shaded region, so the pairs that fit every rule are right there to see, where the shadings overlap.

SAT example

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

\begin{aligned} y&>2x-4\\[1.4em] x+2y&\le8 \end{aligned}

A
$(1,-2)$
B
$(2,1)$
C
$(3,3)$
D
$(4,4)$

Fast Desmos solution

Type both inequalities into Desmos exactly as written, then add the four answer choices as points. Only $(2,1)$ lands where the two shadings overlap.

Choices A and D sit right on the dashed line $y=2x-4$ . The symbol $>$ is strict, so points on that line don't count. Choice C is outside $x+2y\le8$ .

Now confirm $(2,1)$ by hand:

1>2(2)-4

and

2+2(1)\le8.

That's $1>0$ and $4\le8$ . Both are true, so the answer is B.

Calculator loads as you approach

Only one of the four points sits inside both shadings without touching the dashed line.

Spot the question and choose a method

Each point $(x,y)$ stands for one combination of two quantities: $x$ is the first, and $y$ is the second. In a word problem, $(18,6)$ might mean $18$ standard kits and $6$ advanced kits.

These questions ask which pair lies in a shaded region, satisfies several inequalities at once, or could happen under several real-world limits. The combinations that obey every limit make up the feasible region, the spot where all the shadings overlap.

What you see	What to do first	Why
The question asks whether one point fits one short inequality.	Plug the point in by hand.	One quick calculation beats setting up a graph.
The question already shows a graph.	Read the line and the shading, and plug in a point if you're unsure.	The region is already drawn for you.
There are two or more inequalities, or several answer choices to test.	Graph everything in Desmos.	You see the overlap and every choice at the same time.
A word problem gives several limits, like a budget and a minimum.	Use a hybrid: write the inequalities by hand, graph them, then check your final point by hand.	Desmos can't read the story for you, but it's great at showing the region.

Read the boundary before the shading

Every linear inequality splits the coordinate plane into three parts:

The boundary line. You get it by swapping the inequality sign for $=$ .
One side of the line, where every point makes the inequality true.
The other side, where every point makes it false.

For

3x+2y\le18,

the boundary is

3x+2y=18.

The symbol also tells you whether points on the line count:

Symbol	How the line looks	Points on the line
$<$ or $>$	dashed	excluded: they don't count
$\le$ or $\ge$	solid	included: they count

Here's a way to remember it. The bar under $\le$ and $\ge$ means “or equal to,” and the points on the line are exactly the ones where the two sides are equal. Bar underneath, solid line.

In $3x+2y\le18$ , equality is allowed, so the line is solid. Now look at the three points in the calculator. The point $(2,6)$ is on the line, because

3(2)+2(6)=18.

The two sides are equal, and $\le$ allows that, so $(2,6)$ is a solution. The point $(2,4)$ gives $14$ on the left, which is less than $18$ , so it's in the shaded region. The point $(2,7)$ gives $20$ , which is too big, so it's outside.

Try it yourself:

In the calculator, change $\le$ to $<$ . Watch what happens to the line, then decide which of the three points stops counting.

Check your understanding:

Once $3x+2y\le18$ becomes $3x+2y<18$ , what happens to $(2,6)$ , $(2,4)$ and $(2,7)$ ?

Common mistake:

A common slip is assuming that a point on the line always counts. The equation $3x+2y=18$ only gives you the line. Whether the line itself counts depends on the original symbol. With $\le$ , the point $(2,6)$ counts. With $<$ , it doesn’t. If you’re not sure a point is on the line, plug it in: equal sides mean it is.

Calculator loads as you approach

Change the symbol and watch the line. Reset brings back the original inequality and points.

Use a test point to choose the side

When you type an inequality into Desmos, it shades the correct side for you. But when a question shows only the line, or you're sketching the graph by hand, you have to find the side yourself. Pick a test point, any point that isn't on the line, and plug it into the original inequality.

The origin, $(0,0)$ , is usually the easiest choice, because every $x$ and $y$ term becomes $0$ . Take

2x-y>4.

Its boundary is $2x-y=4$ , and the origin isn't on it. Plug in $(0,0)$ :

2(0)-0>4

That says $0>4$ , which is false. So the origin is on the wrong side, and you shade the side that does not contain it. The line is dashed, because $>$ leaves out equality.

Try it yourself:

The calculator shows only the line and the origin. Decide which side should be shaded, then change $=$ to $>$ and see whether you were right.

If the line passes through $(0,0)$ , the origin can't tell you anything, so use another easy point like $(1,0)$ or $(0,1)$ . A test point has one job: it tells you which whole side works. If one point on a side makes the inequality true, every point on that side does.

When the inequality is already solved for $y$ , you can skip the test point. If it says $y>$ or $y\ge$ an expression, shade above the line. If it says $y<$ or $y\le$ , shade below it.

A vertical line works the same way, sideways. The inequality $x>3$ means shade to the right of $x=3$ , and $x\le3$ means shade to the left, with the line included.

Check your understanding:

For $x+2y\ge6$ , is the test point $(0,0)$ in the solution region? Which side of the line should you shade, and should the line be solid or dashed?

Calculator loads as you approach

No shading yet, only the line

2x-y=4

and the test point

(0,0)

Find the overlap where every inequality is true

A system of inequalities is a set of conditions that all have to be true at once. Each inequality shades its own region, and the solutions to the system are where all the shadings overlap.

Take this system:

\begin{aligned} x&\ge0\\[1.4em] y&>-x+2\\[1.4em] 2x+y&\le8. \end{aligned}

In Desmos, each inequality goes on its own line. The spot where all three shadings overlap is where the whole system is true.

Now look at the four plotted points:

$(0,2)$ sits on the dashed line $y=-x+2$ , so it's excluded.
$(1,2)$ is inside all three regions, so it works.
$(2,4)$ sits on the solid line $2x+y=8$ and satisfies the other two inequalities, so it's included.
$(3,3)$ makes $2x+y=9$ , which is more than $8$ , so it's outside that region.

Here's the Desmos routine for any system:

Type each inequality exactly as written. You don't need to solve for $y$ first.
Add each answer choice as a point $(x,y)$ .
Keep only the points inside every shading, counting solid lines but not dashed ones.
Plug the point you keep into every original inequality.

Step 4 matters most when a point looks like it's touching a line. On a screen, “on the line” and “just off the line” can look the same, so trust the numbers, not the picture. Plugging in tells you for sure whether the point is on the line, and then the symbol tells you whether the line counts.

Check your understanding:

Both $(2,4)$ and $(0,2)$ sit on a boundary line. Why does $(2,4)$ work while $(0,2)$ doesn’t?

Common mistake:

It’s easy to find a point in two of the shadings and stop there. Take $(3,3)$ : it satisfies $x\ge0$ and $y>-x+2$ , but it breaks $2x+y\le8$ . A point has to pass every inequality, so one false statement rules it out.

Calculator loads as you approach

Find each of the four points and check which lines it touches and which shadings it’s in.

Example: Check a feasible combination

Worked example

A science club assembles standard equipment kits and advanced equipment kits. Each standard kit costs $12, and each advanced kit costs $20. The club must assemble at least $24$ kits, spend no more than $360, and assemble at least $6$ advanced kits.

Let $x$ be the number of standard kits and $y$ be the number of advanced kits. Which ordered pair $(x,y)$ could represent the numbers of kits assembled?

A
$(18,6)$
B
$(17,6)$
C
$(16,9)$
D
$(20,5)$

Step 1

Turn each limit into an inequality

Start with the total. The club makes $x+y$ kits, and “at least $24$ ” means

x+y\ge24.

Next, the money. The kits cost $12x+20y$ dollars in all, and “no more than $360” means

12x+20y\le360.

The advanced kits get their own inequality:

y\ge6.

There's one more limit the story doesn't say out loud: the club can't make a negative number of kits, so $x\ge0$ . You don't need $y\ge0$ , because $y\ge6$ already covers it.

Step 2

Graph all four inequalities

Type all four inequalities into Desmos. The overlap shows every pair of numbers that satisfies all of them. Add the four answer choices as points so you can compare them.

Step 3

Bring back the story

Desmos shades every point that fits the inequalities, including ones like $(18.5,6)$ . But nobody makes half a kit, so the counts have to be nonnegative whole numbers. Here, all four choices are already whole numbers. The restriction matters when you pick a point from the region yourself.

Now rule out the wrong choices. Choice B makes only $17+6=23$ kits, one short of $24$ . Choice C goes over the budget, because

12(16)+20(9)=372.

Choice D has only $5$ advanced kits, and the club needs at least $6$ .

Step 4

Check the answer

Confirm $(18,6)$ against every inequality:

18+6=24\ge24,

12(18)+20(6)=336\le360,

and

6\ge6.

The total is exactly $24$ kits, and that's allowed, because “at least $24$ ” includes $24$ . Both numbers are nonnegative whole numbers, so every condition holds. The answer is A.

Common mistake:

If you check only the budget, choice D looks fine. It costs $340, under the $360 limit, and it even makes $25$ kits. But it has only $5$ advanced kits, and the club needs $6$ . Turn every sentence of the story into an inequality before you graph, and check your final pair against all of them.

Find the greatest or least whole-number value

Some questions go one step further and ask for the greatest or least possible value of one count, like the most standard kits a club could make. Now one pair that works isn't enough. You also have to show that nothing past your answer works.

Here's a small case. A club has $100. Standard kits cost $12, advanced kits cost $20, and the club needs at least $1$ advanced kit. What's the greatest number of standard kits it can make?

Let $x$ be the number of standard kits and $y$ the number of advanced kits. The limits are $12x+20y\le100$ and $y\ge1$ . Every advanced kit eats into the budget, so to leave the most money for standard kits, make as few advanced kits as you're allowed: $y=1$ . Then

\begin{aligned} 12x+20(1)&\le100\\[1.4em] 12x&\le80\\[1.4em] x&\le\frac{20}{3}. \end{aligned}

That's about $6.67$ , and the club can't make part of a kit. It's tempting to round up to $7$ , but $12(7)+20(1)=104$ , which is over budget. Now try $6$ : $12(6)+20(1)=92$ , which fits. So the answer is $6$ , and you've proved it: $6$ works and $7$ doesn't.

SAT questions usually give you more than one limit, so here's the full routine:

Name the count the question asks for, and turn every limit into an inequality.
If the question gives or fixes a value for the other variable, put it in. If it only gives a minimum, like $y\ge1$ above, use that minimum when it leaves the most room for your count.
Solve each inequality for your count. Each one gives a ceiling (an upper bound) or a floor (a lower bound). For the greatest value, the lowest ceiling is the one that stops you. It's called the binding bound. For the least value, the highest floor stops you.
Counts are whole numbers, so round a fractional bound down for a greatest value and up for a least value. Then test that number and the next whole number past it, the way you tested $6$ and $7$ .
Plug the full pair into every original inequality.

This routine is for the greatest or least value of one of the two counts. If a question asks for the greatest or least value of a separate expression across the region, like a total profit, you need a different method, taught in Optimize a linear objective over a feasible region.

Calculator loads as you approach

The overlap includes fractional pairs too, but the club can only make whole kits. Reset brings back the full system and the four points.

Practice problems

Your turn. Before you start each one, decide: plug in by hand, or graph?

Plug in one point

Practice problem

Which ordered pair $(x,y)$ satisfies

4x+3y\le19?

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Plugging in by hand is quickest here. Graph it only if you want to see the choices.

Watch the solid and dashed lines

Practice problem

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

\begin{aligned} y&<-x+6\\[1.4em] y&\ge2x-3 \end{aligned}

Calculator loads as you approach

Two inequalities and four choices: graph them all here.

Find the most standard kits

Practice problem

An event organizer will assemble $x$ standard supply kits and $y$ deluxe supply kits, where $x$ and $y$ are nonnegative whole numbers. Each standard kit serves $1$ participant and uses $2$ storage units. Each deluxe kit serves $2$ participants and uses $5$ storage units.

The organizer needs kits for at least $18$ participants, has space for no more than $49$ storage units, and must assemble at least $3$ deluxe kits.

What is the greatest possible value of $x$ ?

Your answer

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Finding the bounds by hand is quickest. Graph here if you want to see the region, or to check your answer and the next whole number.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Swap the inequality sign for $=$ to get the boundary line.
With $<$ or $>$ , the line is dashed and its points don't count. With $\le$ or $\ge$ , it's solid and they do.
A test point tells you which side of the line works.
A system's solutions are where every shading overlaps.
Plug your final point into every original inequality. Near a line, trust the numbers, not the picture.
In a word problem, know what each coordinate counts, and keep counts to nonnegative whole numbers.
For a greatest value, the lowest ceiling decides. For a least value, the highest floor does. Then test your whole number and the next one past it.
Plug in by hand for one quick check, and graph in Desmos for several regions or points. For a word problem, use the hybrid: write the inequalities by hand, graph the region, then check your final point by hand.

Related lessons

Not quite this kind of question? Try these:

Only one unknown, with a range of values that work? Review Solve one-variable inequalities.
Two equations, and you need the point where their lines cross? See Solve linear systems strategically.

For more Desmos practice with shaded regions and plotted points, try Inequalities and shaded overlap.

You can also go back to the SAT Algebra course to see the full lesson path.

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