SAT Inequalities on Desmos: Graph the Solution Set

To solve an SAT inequality on Desmos, type it with <, >, ≤ or ≥ and read the shaded region: every shaded point is a solution. A dotted boundary means the line itself is not included. A solid boundary means it is. For a system, the answer is where the shadings overlap, and you can test any answer choice by plugging the point into each inequality.

Open the College Board version of Desmos on our Desmos on the SAT page and type along. Below you get the boundary rules, two worked examples, one guided example, four practice problems with checked answers, and three mistakes to watch for.

The rule: boundary first, then shading

Every inequality has a boundary. It is the graph you get when you swap the inequality sign for an equals sign. Desmos draws that boundary and shades the side where the inequality is true. The Desmos Help Center puts it this way: strict inequalities (< and >) draw dotted lines, and non-strict inequalities (≤ and ≥) draw solid lines.

On a keyboard, type <= and Desmos turns it into ≤. For absolute value, type abs( in front of the expression.

An inequality with only x in it has a vertical boundary, such as x = 3. So Desmos shades a vertical band across the whole screen, not a piece of a number line. Read your answer from the x-values at the edges of the band.

Two algebra rules still decide the answer when you work by hand. First, flip the sign only when you multiply or divide both sides by a negative number: -2x < 8 becomes x > -4. Second, an absolute value "greater than" splits into two pieces joined by or. The inequality |x| > 5 means x < -5 or x > 5. No number is both less than -5 and greater than 5, so writing "and" would leave you with no solutions at all.

Worked example 1 (I do): which point is in the solution region?

SAT-style question. Which point (x, y) satisfies both 5x + 3y > 15 and y < x + 2?
A) (0, 5)   B) (3, 0)   C) (2, 3)   D) (-1, 4)

Step 1. Type each inequality on its own line.

1  5x + 3y > 15
2  y < x + 2

Step 2. Find the overlap. Desmos shades each inequality in its own color. The system's solutions are the points shaded by both, the darker region to the right of where the two dotted lines cross.

Step 3. Put a choice on the graph. Type (2, 3) on line 3. The point lands inside the darker region, so C looks right. Before you pick it, check why the others fail. Choices A and B both give 5x + 3y = 15 exactly. They sit on the dotted boundary, and 15 > 15 is false. Choice D gives 5(-1) + 3(4) = 7, far short of 15.

By-hand check. For (2, 3): 5(2) + 3(3) = 19, and 19 > 15. Also 3 < 2 + 2 = 4. Both are true, so the answer is C.

Worked example 2 (I do): the greatest value in a word problem

SAT-style question. Four consecutive odd integers, in order from least to greatest, have first integer x. The product of 28 and the third integer is at most 50 less than the sum of the first and fourth integers. What is the greatest possible value of x?

Step 1. Translate. The integers are x, x + 2, x + 4 and x + 6. The third is x + 4. The first plus the fourth is x + (x + 6) = 2x + 6. "At most 50 less than" means ≤ (2x + 6) - 50. So the inequality is 28(x + 4) ≤ (2x + 6) - 50.

Step 2. Graph it.

1  28(x + 4) <= (2x + 6) - 50

Desmos shades everything to the left of a solid vertical line at x = -6. The line is solid because ≤ includes -6.

Step 3. Apply the integer rule. Desmos gives you x ≤ -6. The question also says x is odd. -6 is even, so the greatest odd integer that fits is -7.

By-hand check. 28x + 112 ≤ 2x - 44, so 26x ≤ -156 and x ≤ -6. You divided by 26, a positive number, so the sign stays. Test x = -7: 28(-3) = -84, and -84 ≤ (-14 + 6) - 50 = -58. True. Test x = -5: 28(-1) = -28, and -28 ≤ -54 is false. The answer is -7.

Want more questions like these? Work through the Linear Inequalities lesson notes.

Guided example (we do): an absolute value inequality

SAT-style question. What are all the solutions of |x - 3| < 7?

Step 1. Read it as distance: x is less than 7 units away from 3.

Step 2. Write it without the absolute value bars: -7 < x - 3 < 7.

Step 3 (your step). Add 3 to all three parts.

Check your answer: -4 < x < 10, or (-4, 10) in interval notation. To see it, type this line:

1  abs(x - 3) < 7

Desmos shades a vertical band between dotted lines at x = -4 and x = 10. The lines are dotted because < leaves out -4 and 10.

Quadratic inequalities: read where the parabola sits

For x² - 5x + 6 < 0, you have two options on Desmos. Type the inequality as written, and Desmos shades a band between dotted lines at x = 2 and x = 3. Or graph y = x² - 5x + 6 and look for where the parabola dips below the x-axis. Factoring gives the same answer: (x - 2)(x - 3) is negative only between its zeros, so the solution is 2 < x < 3.

Your turn (you do): four practice problems

1. Three consecutive even integers have smallest integer n. Twice the largest is at most 10 more than three times the smallest. What is the least possible value of n?
2. Which point (x, y) satisfies both 3x - y < 9 and y ≤ 2x + 1?
   A) (3, 0)   B) (0, 2)   C) (2, 3)   D) (-1, 0)
3. Solve |2x + 4| ≥ 8. Write the answer in interval notation.
4. How many integers x satisfy x² - 7x + 10 ≤ 0?

Answers

1. -2. The largest integer is n + 4, so 2(n + 4) ≤ 3n + 10. That gives 2n + 8 ≤ 3n + 10, so n ≥ -2. The smallest even integer that is at least -2 is -2 itself. Notice the question asks for the least value: the solutions keep going up forever, so a greatest value does not exist.
2. C. For (2, 3): 3(2) - 3 = 3, which is less than 9, and 3 ≤ 2(2) + 1 = 5. Choice A sits on the dotted line 3x - y = 9, and 9 < 9 is false. Choice B gives 2 ≤ 1 and choice D gives 0 ≤ -1, both false.
3. (-∞, -6] ∪ [2, ∞). Split it into 2x + 4 ≥ 8 or 2x + 4 ≤ -8, so x ≥ 2 or x ≤ -6. The brackets are square because ≥ includes -6 and 2. On Desmos you see two shaded regions with solid edges at x = -6 and x = 2.
4. 4. The expression factors as (x - 2)(x - 5), which is at most 0 when 2 ≤ x ≤ 5. The integers are 2, 3, 4 and 5.

When algebra is faster

College Board's calculator guidance says some Math questions are better done without a calculator, even though one is allowed. A one-step inequality such as 3x + 4 > 19 takes a few seconds by hand: x > 5. We save Desmos for systems, absolute values and quadratics, and for testing an answer choice against more than one inequality.

Mistakes to avoid

• Counting a point on a dotted line. In worked example 1, the point (3, 0) lies exactly on 5x + 3y = 15. With >, that boundary is dotted, so (3, 0) is not a solution. When a choice looks like it touches the line, plug it in.
• Writing "and" for an absolute value "greater than". |x| > 5 is x < -5 or x > 5. When Desmos shows two separate shaded regions, write "or" between them.
• Stopping at the boundary in integer problems. In worked example 2, Desmos shows x ≤ -6, but the question wants an odd integer, so the answer is -7. The graph gives you the boundary, and you apply the integer rule yourself.

Practice next

Start with the Linear Inequalities lesson notes. For more questions, browse the SAT question bank.

Related guides: to find where two lines cross, see how to solve a system of equations on Desmos. For absolute value graphs, read SAT absolute value functions on Desmos. For systems where two lines match, see infinitely many solutions: find k.

Checked on September 26, 2026 against College Board's SAT Suite calculator policy, the College Board version of the Desmos graphing calculator and the Desmos Help Center article on inequalities and restrictions. Every answer on this page was recomputed before publishing. The Test Advantage is independent of College Board and Desmos.