Absolute Value

Absolute Value

The absolute value of a number is its distance from zero, so it is never negative. Absolute value questions on the ACT come in three flavors: equations, inequalities, and equations where a variable also appears outside the bars. Each has a short, reliable procedure, and the common mistakes are all about forgetting a case or forgetting a check.

Absolute value equations

An absolute value equation splits into two separate cases, since the expression inside the bars could be positive or negative before the bars were applied. $|A| = B$, where $B$ is positive, becomes $A = B$ or $A = -B$. Both cases must be solved, and both solutions are typically valid, unless the right side of the original equation contains a variable, in which case each candidate solution must be checked.

For $|x - 5| = 8$: case 1 is $x - 5 = 8$, so $x = 13$; case 2 is $x - 5 = -8$, so $x = -3$. A frequent error is to drop the negative sign in the second case and report $x = 3$. Put the negative on the entire right side: $-8$, not $+8$ with the sign moved somewhere else. For $|4x + 1| = 9$, case 2 is $4x + 1 = -9$, so $x = -2.5$.

Absolute value inequalities

An absolute value inequality splits differently depending on the inequality symbol.

• $|A| < B$ becomes a single compound inequality: $-B < A < B$. Think "less than" means "between."
• $|A| > B$ becomes two separate inequalities joined by "or": $A > B$ or $A < -B$. Think "greater than" means "outside."

Confusing which form goes with which symbol is one of the most common errors on this topic. The same split applies to $\le$ and $\ge$, with the boundaries included.

Worked Example. Solve $|3x - 1| \ge 11$.

A $\ge$ form splits into two separate inequalities joined by "or." Case 1: $3x - 1 \ge 11$, so $x \ge 4$. Case 2: $3x - 1 \le -11$, so $x \le -\frac{10}{3}$. The solution is $x \le -\frac{10}{3}$ or $x \ge 4$.

For $|x + 3| < 6$, write $-6 < x + 3 < 6$ and subtract $3$ from every part: $-9 < x < 3$. Subtract, do not add; the boundaries move opposite to the number inside the bars. For $|x - 4| < 10$ the center is $4$ and the boundaries are $-6$ and $14$.

Variables on both sides

When the right side of an absolute value equation contains a variable, that side must be nonnegative for a solution to be valid, since an absolute value can never equal a negative number. Every candidate solution from both cases must be substituted back into the original equation to confirm it actually works.

Worked Example. Solve $|x + 2| = 3x - 6$.

Case 1: $x + 2 = 3x - 6$ gives $x = 4$. Case 2: $x + 2 = -(3x - 6)$ gives $x = 1$.

Check $x = 4$: $|6| = 6$, and $3(4) - 6 = 6$. Valid. Check $x = 1$: $|3| = 3$, and $3(1) - 6 = -3$. Since $3 \ne -3$, it is rejected. Only $x = 4$ is valid.

Two absolute values

When absolute values appear on both sides, as in $|x - 1| = |2x + 4|$, the two expressions are either equal or opposites: $x - 1 = 2x + 4$ gives $x = -5$, and $x - 1 = -(2x + 4)$ gives $x = -1$. Here no variable stands outside the bars, so both solutions check out automatically (verify: $|-6| = |-6|$ and $|-2| = |2|$).

Where this appears as a model

Absolute value describes a tolerance. A machined part required to be $10$ centimeters long, within $0.05$ centimeters, is described by $|x - 10| \le 0.05$, giving an acceptable range from $9.95$ to $10.05$ centimeters. In general, $|x - c| \le d$ means "$x$ is within $d$ of the center $c$."

Strategy checklist

• Equation: write two cases, with the negative on the whole right side.
• "Less than": one compound inequality. "Greater than": two inequalities joined by "or."
• Variable outside the bars: check every candidate in the original equation.
• Two absolute values: set the insides equal, then set one inside equal to the opposite of the other.