Absolute Value Equations

Solve absolute value equations for the Digital SAT. Understand the split-case method and identify when no solution exists.

Absolute value represents the distance of a number from zero, so x|x| is always non-negative. Absolute value equations, such as 2x3=7|2x - 3| = 7, can have two solutions (one from the positive case and one from the negative case), one solution, or no solution. The Digital SAT tests your ability to solve these and interpret the results.

Core Concepts

Definition of Absolute Value

a={aif a0aif a<0|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}

Key properties:

  • a0|a| \geq 0 always
  • a=0|a| = 0 only if a=0a = 0
  • a=a|-a| = |a|

Solving ax+b=c|ax + b| = c

If c>0c > 0: Split into two cases.

Case 1: ax+b=cax + b = c

Case 2: ax+b=cax + b = -c

Example: 2x3=7|2x - 3| = 7

Case 1: 2x3=72x - 3 = 72x=102x = 10x=5x = 5

Case 2: 2x3=72x - 3 = -72x=42x = -4x=2x = -2

Solutions: x=5x = 5 or x=2x = -2.

If c=0c = 0: Only one solution. ax+b=0ax + b = 0.

3x+6=0|3x + 6| = 03x+6=03x + 6 = 0x=2x = -2

If c<0c < 0: No solution. Absolute value can never be negative.

x+4=3|x + 4| = -3no solution.

Isolate the Absolute Value First

If the equation has extra terms, isolate the absolute value before splitting.

Example: 3x1+2=143|x - 1| + 2 = 14

3x1=123|x - 1| = 12

x1=4|x - 1| = 4

Case 1: x1=4x - 1 = 4x=5x = 5

Case 2: x1=4x - 1 = -4x=3x = -3

Absolute Value with Variables on Both Sides

x+2=3x4|x + 2| = 3x - 4

Case 1: x+2=3x4x + 2 = 3x - 46=2x6 = 2xx=3x = 3. Check: 5=5|5| = 5 and 94=59 - 4 = 5

Case 2: x+2=(3x4)x + 2 = -(3x - 4)x+2=3x+4x + 2 = -3x + 44x=24x = 2x=12x = \frac{1}{2}. Check: 2.5=2.5|2.5| = 2.5 and 1.54=2.51.5 - 4 = -2.5. 2.52.52.5 \neq -2.5 ✗ Extraneous.

Solution: x=3x = 3 only.

Strategy Tips

Tip 1: Check If the Right Side Is Negative

If expression=negative|\text{expression}| = \text{negative}, immediately write "no solution" and move on.

Tip 2: Always Isolate First

Don't split into cases until the absolute value expression is alone on one side.

Tip 3: Check Both Solutions

Especially when the equation has variables outside the absolute value, one solution may be extraneous.

Tip 4: |a| = |b| Means a = b or a = −b

If you see x3=2x+1|x - 3| = |2x + 1|, set up: x3=2x+1x - 3 = 2x + 1 or x3=(2x+1)x - 3 = -(2x + 1).

Tip 5: Distance Interpretation

x5=3|x - 5| = 3 means "xx is 3 units from 5" → x=8x = 8 or x=2x = 2.

Worked Example: Example 1

Problem

Solve 4x+1=9|4x + 1| = 9.

4x+1=94x + 1 = 9x=2x = 2

4x+1=94x + 1 = -9x=104=52x = -\frac{10}{4} = -\frac{5}{2}

Solution

Worked Example: Example 2

Problem

Solve 2x5=0|2x - 5| = 0.

2x5=02x - 5 = 0x=52x = \frac{5}{2}

Solution

Worked Example: Example 3

Problem

Solve 2x+35=72|x + 3| - 5 = 7.

2x+3=122|x + 3| = 12x+3=6|x + 3| = 6

x+3=6x + 3 = 6x=3x = 3

x+3=6x + 3 = -6x=9x = -9

Solution

Worked Example: SAT-Style

Problem

How many solutions does x4=2|x - 4| = -2 have?

Zero. Absolute value cannot equal a negative number.

Solution

Worked Example: Example 5

Problem

Solve 3x1=x+5|3x - 1| = x + 5.

Case 1: 3x1=x+53x - 1 = x + 52x=62x = 6x=3x = 3. Check: 8=8|8| = 8 and 8=88 = 8

Case 2: 3x1=(x+5)3x - 1 = -(x + 5)3x1=x53x - 1 = -x - 54x=44x = -4x=1x = -1. Check: 4=4|-4| = 4 and 1+5=4-1 + 5 = 4

Solutions: x=3x = 3 and x=1x = -1.

Solution

Practice Problems

  1. Problem 1

    Solve x7=3|x - 7| = 3.

    Problem 2

    Solve 5x+2=8|5x + 2| = -8.

    Problem 3

    Solve 32x1=153|2x - 1| = 15.

    Problem 4

    Solve x+4=2x1|x + 4| = 2x - 1.

    Problem 5

    How many solutions does x24=0|x^2 - 4| = 0 have?

    Problem 6

    Solve x3=2x+1|x - 3| = |2x + 1|.

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Common Mistakes

  • Forgetting the negative case. x=5|x| = 5 gives x=5x = 5 AND x=5x = -5.
  • Setting up only one equation. Always split into two cases (unless the right side is 0 or negative).
  • Not isolating first. In 3x+2=83|x| + 2 = 8, don't split until you have x=2|x| = 2.
  • Not checking for extraneous solutions. When the right side contains a variable, check both solutions.
  • Claiming no solution when there is one. x=0|x| = 0 has the solution x=0x = 0.

Frequently Asked Questions

Can absolute value equations have more than 2 solutions?

For linear expressions inside, at most 2. For quadratic expressions inside, potentially more.

How often does this appear on the SAT?

Approximately once per test. It's a reliable topic.

What about absolute value inequalities?

x<3|x| < 3 means 3<x<3-3 < x < 3. x>3|x| > 3 means x<3x < -3 or x>3x > 3. These occasionally appear on the SAT.

Is $|x|$ the same as $\sqrt{x^2}$?

Yes! x=x2|x| = \sqrt{x^2} for all real xx.

Can I graph absolute value on Desmos?

Yes — y=x3y = |x - 3| graphs as a V-shape with vertex at (3,0)(3, 0).

Key Takeaways

  • ax+b=c|ax + b| = c → two cases: ax+b=cax + b = c and ax+b=cax + b = -c.

  • If c<0c < 0: no solution.

  • If c=0c = 0: one solution.

  • Isolate the absolute value before splitting.

  • Check both solutions, especially when the right side contains variables.

  • Distance interpretation: xa=d|x - a| = d means xx is dd units from aa.

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