SAT Absolute Value Equations and Inequalities on Desmos

To solve an SAT absolute value equation on Desmos, graph each side on its own line, such as y = abs(2x - 6) and y = 4, then click the points where the graphs cross. Each crossing's x-value is a solution, so |2x - 6| = 4 has two: x = 1 and x = 5.

You can type every line on this page into the College Board version of Desmos on our Desmos on the SAT page. The graph of y = |x - h| + k is a V with its vertex at (h, k), and its two sides mirror each other across the line x = h. That symmetry solves the hardest problem below. The page has four SAT-style problems, one guided problem where a case fails, and four practice problems with answers.

The symmetry rule

The function f(x) = |x - h| is symmetric about the line x = h. Two inputs at the same distance from h give the same output.

If |a - h| = |b - h|, then f(a) = f(b).

The rule holds for any function built only from |x - h|. For example, g(x) = a|x - h|² - b|x - h| + c is also symmetric about x = h.

Two more facts about the V. In y = |x - h| + k, the vertex (h, k) is the minimum point. A negative sign in front, as in y = -|x|, flips the V downward, so the vertex becomes a maximum.

Problem 1: the symmetry trap (worked example)

SAT-style problem. The function q is defined by q(x) = a|x - 9|² - 87|x - 9| + b, where a and b are constants and a > b > 87. If q(639) = h and q(-621) = k, what is the value of 9(-639)^{h - k} + 621(9)^{k - h}?

Step 1: spot the symmetry. q(x) depends only on |x - 9|, so q is symmetric about x = 9. Check each input's distance from 9:

• |639 - 9| = 630
• |-621 - 9| = |-630| = 630

Both inputs sit 630 units from the axis, so q(639) = q(-621). That means h = k.

Step 2: simplify. With h = k, both exponents are 0: h - k = 0 and k - h = 0. Any nonzero number to the power 0 is 1, so the expression is 9(1) + 621(1) = 630.

Step 3: check on Desmos. The question never gives a and b, so pick any values that fit a > b > 87, such as a = 89 and b = 88:

1  q(x) = 89(abs(x - 9))^2 - 87abs(x - 9) + 88
2  q(639)
3  q(-621)

Lines 2 and 3 show the same value, so h = k for these constants too. Answer: 630.

Why the direct route fails. With a = 89 and b = 88, q(639) is 35,269,378. You would then need (-639) raised to h - k, and you don't know a or b anyway. Spot the equal distances first, and the exponents drop to 0.

Problem 2: solve |2x - 6| = 4

SAT-style problem. What is the sum of all solutions to |2x - 6| = 4?

On Desmos. Graph each side on its own line:

1  y = abs(2x - 6)
2  y = 4

The V and the horizontal line cross twice. Click each crossing: x = 1 and x = 5.

By hand. Write both cases.

• Case 1: 2x - 6 = 4, so x = 5.
• Case 2: 2x - 6 = -4, so x = 1.

Both check in the original: |2(5) - 6| = 4 and |2(1) - 6| = 4. The sum is 5 + 1 = 6.

Symmetry shortcut. When c > 0, the two solutions of |ax + b| = c sit the same distance from the vertex at x = -b/a, so their sum is 2(-b/a). Here the vertex is at x = 3, and 2(3) = 6. You get the sum without finding either solution. For more on this pattern, see our guide to the sum of solutions formula.

Problem 3: how many solutions does |f(x)| = c have?

SAT-style problem. f(x) = |x² - 4x - 12|. For how many values of x does f(x) = 5?

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1  y = abs(x^2 - 4x - 12)
2  y = 5

Count the crossings: there are 4.

The absolute value flips the part of the parabola that falls below the x-axis. The parabola x² - 4x - 12 is negative between its zeros, x = -2 and x = 6, and reaches -16 at x = 2. After the flip, that section peaks at 16, which is above 5, so it adds two crossings to the two from the outer branches.

By hand. Solve x² - 4x - 12 = 5 and x² - 4x - 12 = -5. The first gives x = 2 ± √21 and the second gives x = 2 ± √11. All four are distinct. Answer: 4 values of x.

Problem 4: an absolute value inequality

SAT-style problem. Which values of x satisfy |x - 3| < 7?

1  abs(x - 3) < 7

Desmos shades a vertical band between the lines x = -4 and x = 10. Every x inside the band works.

By hand. |x - 3| < 7 means -7 < x - 3 < 7. Add 3 to all three parts: -4 < x < 10.

Answer: -4 < x < 10, or (-4, 10) in interval notation. The ends are open because the sign is <, not ≤.

Type the inequality itself when the question asks which values work. Graph two lines, y = abs(...) and y = c, when the question asks how many solutions an equation has. Our guide to SAT inequalities on Desmos covers shaded regions in more depth.

Guided problem: when one case fails (we do)

SAT-style problem. What value of x satisfies |x + 2| = 3x - 6?

Step 1 (shown). Write both cases. Case 1: x + 2 = 3x - 6, so x = 4. Case 2: x + 2 = -(3x - 6) = -3x + 6, so x = 1.

Step 2 (shown). Graph both sides:

1  y = abs(x + 2)
2  y = 3x - 6

Step 3 (you finish). Count the crossings, then substitute each case's answer into the original equation. Which one survives?

Check your answer: the graphs cross once, at (4, 6). x = 4 works: |4 + 2| = 6 and 3(4) - 6 = 6. x = 1 fails: |1 + 2| = 3, but 3(1) - 6 = -3, and an absolute value can't equal a negative number. The answer is x = 4. When the right side contains x, check every case in the original equation.

The symmetry checklist

1. Find the axis of symmetry. For f(x) = |x - h| it is x = h. For f(x) = |ax + b| it is x = -b/a.
2. Measure each input's distance from the axis. Equal distances mean equal outputs.
3. Use the equal outputs. If two outputs are equal, any difference between them is 0. Simplify the exponents or expressions that depend on that difference.
4. Finish with the power rule. Any nonzero number to the power 0 is 1.

Mistakes to avoid

• Writing only the positive case. |ax + b| = c needs two equations: ax + b = c and ax + b = -c. Drop the second and you lose a solution.
• Mixing up the two inequality shapes. |x| > c means x > c or x < -c, two separate regions. |x| < c means -c < x < c, one connected region.
• Computing before you look. In problems like Problem 1, the direct calculation is too large to do by hand. Check the distances from the axis first.
• Forgetting that |x| ≥ 0. |f(x)| = -3 has no solution, whatever f(x) is.

When algebra is faster

College Board's calculator policy says some Math questions are better solved without a calculator, even though you can use one on the whole Math section. |x + 5| = -2 is one: an absolute value can't be negative, so there is no solution, and you don't need a graph. Problem 4 is another, since adding 3 to all three parts takes one line.

Your turn: four practice problems

We wrote these for this page. Solve each by hand, then check it on Desmos. The answers follow the list.

1. Solve |3x + 9| = 15. What is the sum of the solutions?
2. For f(x) = |x² - 9|, how many values of x satisfy f(x) = 7?
3. g(x) = 5|x - 4|² + 3|x - 4| - 2. If g(10) = m and g(-2) = n, what is m - n?
4. Solve |2x - 8| > 6. Write the solution in interval notation.

Answers

1. -6. The solutions are x = 2 and x = -8. Shortcut: the vertex is at x = -3, and 2(-3) = -6.
2. 4. x² - 9 = 7 gives x = ±4, and x² - 9 = -7 gives x = ±√2. On Desmos, y = abs(x^2 - 9) and y = 7 cross four times.
3. 0. |10 - 4| = 6 and |-2 - 4| = 6, so g(10) = g(-2) and m = n. Both equal 196.
4. (-∞, 1) ∪ (7, ∞). 2x - 8 > 6 gives x > 7, and 2x - 8 < -6 gives x < 1. So x < 1 or x > 7.

Practice next

Drill these question types in the free SkillDrill Solution Counts, Matching Terms and Absolute Value Equations. For more SAT-style math sets, open the TTA SAT question bank.

The same count-the-crossings method works on higher-degree equations: see SAT polynomial equations on Desmos. To practice on the calculator Bluebook uses, read our College Board Desmos guide.

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