ACT Absolute Value Equations: Two, One, or No Solutions

Solve absolute-value equations by isolating the bars, choosing the right cases, and checking each answer. Includes original examples and four practice questions.

KiboNovember 7, 2026

To solve an absolute-value equation, isolate the absolute value first. If it equals a positive constant, solve two equations: the expression inside the bars equals that constant or its negative. If it equals zero, set the inside expression to zero. If it equals a negative constant, there is no real solution. Check every proposed answer in the original equation.

ACT’s Mathematics Standards include interpreting absolute value as distance and solving absolute-value equations. Every problem below is original Kibo practice, not an official exam item or a prediction of topic frequency.

Use distance to understand the two cases

Absolute value measures distance from zero. For example, $|-7|=7$ and $|7|=7$ . In $|x-4|=7$ , the distance between $x$ and $4$ is $7$ , so $x=11$ or $x=-3$ . Check: $|11-4|=7$ and $|-3-4|=7$ . The two answers are equally far from $4$ ; they need not be opposites.

For $|A|=c$ with $c>0$ , either $A=c$ or $A=-c$ . The absolute value itself stays nonnegative in both cases. OpenStax’s absolute-value equations explanation describes this rule and why a negative right side has no solution.

Count solutions after isolating the absolute value

For the linear-inside form $|ax+b|=c$ , where $a\neq 0$ and $c$ is a real constant, use this table. The nonzero-coefficient condition matters: each linear case then has exactly one solution.

Solution counts for an isolated absolute value with a nonconstant linear expression inside
Right side	Equations to solve	Distinct real solutions
$c>0$	$ax+b=c$ or $ax+b=-c$	Two
$c=0$	$ax+b=0$	One
$c<0$	Stop: an absolute value cannot be negative.	None

This count applies to the stated linear form. More complicated expressions inside the bars, or a variable on the other side, require additional reasoning.

Two solutions: isolate before splitting

Solve $2|3x-2|+5=21$ . Subtract $5$ from both sides, then divide both sides by $2$ :

$2|3x-2|=16$ $|3x-2|=8$

Now solve $3x-2=8$ or $3x-2=-8$ . The first gives $3x=10$ , so $x=\frac{10}{3}$ . The second gives $3x=-6$ , so $x=-2$ .

At $x=\frac{10}{3}$ , the original left side is $2|10-2|+5=2(8)+5=21$ .
At $x=-2$ , it is $2|-6-2|+5=2(8)+5=21$ .

Splitting into $3x-2=21$ and $3x-2=-21$ would ignore the outside multiplier and added constant. Isolate the bars before deciding which two values the inside expression can have. For help with the resulting equations, use the linear-equation balance method.

One solution or no solution

For $4|2x+1|-3=-3$ , add $3$ and divide by $4$ to get $|2x+1|=0$ . The inside must be zero, so $2x+1=0$ and $x=-\frac{1}{2}$ . Substitution gives $4|-1+1|-3=-3$ . Writing a positive-zero case and a negative-zero case would repeat the same answer.

For $|x+6|+4=1$ , subtract $4$ to get $|x+6|=-3$ . No real number has a negative absolute value. You can also see this directly: the original left side is at least $4$ , so it cannot equal $1$ . There is no need to split into cases.

When the right side contains a variable

Consider $|x+5|=2x+1$ . A solution must make the right side nonnegative, so $2x+1\geq 0$ , which means $x\geq -\frac{1}{2}$ . Unlike a fixed positive constant, this right side changes sign as $x$ changes.

Solve $x+5=2x+1$ , giving $x=4$ .
Solve $x+5=-(2x+1)$ . Distribute the minus: $x+5=-2x-1$ , so $3x=-6$ and $x=-2$ .
Check both in the original equation. At $x=4$ , both sides equal $9$ . At $x=-2$ , the left side is $3$ and the right side is $-3$ , so that candidate fails.

The only solution is $x=4$ . The rejected value also violates $x\geq -\frac{1}{2}$ . Solving the two linear equations produces candidates; the sign condition and original-equation check determine which candidates actually work.

Try four original practice questions

Find every real solution. First isolate the absolute value, then decide whether to split into cases. Substitute each candidate into the original equation before opening the answers.

$3|2x+5|-4=23$
$|4x-9|+6=6$
$2|x-8|+7=3$
$|x-6|=2x+3$

Show answers and checks

Isolate to get $|2x+5|=9$ . Solve $2x+5=9$ or $2x+5=-9$ , giving $x=2$ or $x=-7$ . The original checks are $3|4+5|-4=27-4=23$ and $3|-14+5|-4=27-4=23$ .
Isolate to get $|4x-9|=0$ , so $x=\frac{9}{4}$ . Check: $|9-9|+6=6$ . There is one distinct solution.
Subtract $7$ and divide by $2$ : $|x-8|=-2$ . There is no real solution. The original left side is at least $7$ , which cannot equal $3$ .
A solution needs $x\geq -\frac{3}{2}$ . The cases $x-6=2x+3$ and $x-6=-(2x+3)$ give $x=-9$ and $x=1$ . At $-9$ , the original sides are $15$ and $-15$ , so reject it. At $1$ , both sides equal $5$ . Only $x=1$ works.

Record the step that needs another attempt

Use your practice error log to identify whether you forgot to isolate the bars, omitted the negative case, treated zero as two distinct answers, or kept a candidate that failed substitution. Write the corrected step and try the problem again later with the solution hidden.

Try a free ACT practice question with Kibo, and use the same habit of checking your answer in the original problem.

Prepared with AI assistance. All worked examples and practice questions are original; their candidate solutions and substitution checks were verified separately with exact arithmetic. No human review or score improvement is claimed. Sources checked November 10, 2026. This guide is not affiliated with or endorsed by ACT.