ACT Math Absolute Value: 25 Practice Problems with Step-by-Step Explanations

Evaluate each absolute value separately: |8−5| = |3| = 3, and |4−8| = |−4| = 4.

Subtract: 3 − 4 = −1.

Simplify inside the bars first: 5(−3)+4 = −15+4 = −11.

Take the absolute value: |−11| = 11.

Substitute x=−7: |3(−7)−4|+5 = |−21−4|+5 = |−25|+5.

= 25+5 = 30.

Split into two cases: x+4=8, which gives x=4; and x+4=−8, which gives x=−12.

Both solutions are valid: x=4 and x=−12.

The equation splits into two cases: 9−x=6 (which gives the known solution x=3), and 9−x=−6.

Solve the second case: 9−x=−6, so −x=−15, giving x=15.

Substitute x=4 into each term: |3(4)−6| = |12−6| = |6| = 6, and |−4(4)+8| = |−16+8| = |−8| = 8.

Add: 6+8 = 14.

Simplify inside the first bars: (−2+4)²−2(5) = (2)²−10 = 4−10 = −6, so |−6| = 6.

Simplify inside the second bars: −10(5−2) = −10(3) = −30, so |−30| = 30.

Subtract: 6−30 = −24.

Split into two cases. From 36−3x=18: −3x=−18, so x=6. From 36−3x=−18: −3x=−54, so x=18.

Sum: 6+18 = 24.

Isolate the absolute value: |4a+4|=16. Split into two cases: 4a+4=16 gives a=3, and 4a+4=−16 gives a=−5.

These two solutions are x and y: |xy| = |3×(−5)| = |−15| = 15.

Absolute value is always non-negative, so |a| = −a can only hold when −a is itself non-negative — in other words, when a is non-positive.

So a≤0.

Since c < b < 0, both b and c are negative, and c is more negative than b, so |c| > |b| — statement III is always true.

For statements I and II, a's sign isn't pinned down (a > b just means a is greater than a negative number, so a could still be negative). Testing a=−1, b=−3, c=−5: |a|=1 and |b|=3, so |a|>|b| is false, ruling out I; and |a|=1, |c|=5, so |a|>|c| is also false, ruling out II. Only III always holds.

Solve for x: |4x−3|=13, giving x=4 or x=−2.5. Solve for y: |5y+3|=17, giving y=2.8 or y=−4.

Test all four combinations of x and y to find the smallest product: 4(2.8)=11.2, 4(−4)=−16, (−2.5)(2.8)=−7, (−2.5)(−4)=10. The smallest value is −16.

The integers from −p to p include 0 (which contributes nothing) and each of 1 through p appears twice — once as itself and once as its negative, both with the same absolute value. So the sum is 2(1+2+…+p) = p(p+1).

Set p(p+1)=30. Testing p=5: 5×6=30. ✓

A quantity and its negative always have the same absolute value: |A| = |−A| for any expression A.

Here, −(x−y−z) = −x+y+z, so |x−y−z| = |−x+y+z|.

Test a=1, b=2, c=3 (all positive, satisfying a<b<c and c>0): |a|=1, |b|=2, so |a|>|b| is false (I fails), and since a/c and b/c share the same relationship as a and b, III fails too.

Test a=−5, b=−1, c=1: |c|=1, |a|=5, so |c|>|a| is false (II fails). Since each statement fails in at least one valid case, none of them is always true.

Since x>0, |x|=x. Since y<0, |y|=−y.

Substitute: |x|−|y| = x−(−y) = x+y.

Isolate the absolute value: (5/3)|−3x+1| < −3, so |−3x+1| < −9/5.

An absolute value can never be less than a negative number, since absolute values are always non-negative. So there is no solution.

First isolate the absolute value: |2x+5|=3. This splits into two equivalent cases: 2x+5=3, and the expression's negative equal to 3, written as −(2x+5)=3 (which is algebraically the same as 2x+5=−3).

Only option C correctly represents both cases.

Since b<a, a−b is positive. And |b−a| = |−(a−b)| = |a−b|, which equals a−b exactly, since a−b is already positive.

So |b−a| = a−b.

An absolute value equals the original (unmodified) expression only when that expression is already non-negative: |A| = A exactly when A≥0.

Here, that means 3x−2≥0, so 3x≥2, giving x≥2/3.

Since concrete numbers make abstract sign-based comparisons much easier, test a=1, b=−10 (satisfying a>0, b<0, |a|<|b|): option A gives |11/−10|=1.1; option B gives |11/1|=11; option C gives |−9/−11|≈0.82; option D gives |−9/−10|=0.9.

Option B is clearly the largest. Testing a second set of values (a=2, b=−5) confirms the same pattern holds: option B remains the largest.

Distance from a point p is written as |x−p|. "8 units from −2" means |x−(−2)| = 8, which simplifies to |x+2| = 8.

Test a=−2, b=0.5 (satisfying |a|>1, 0<b<1): option A gives |0.5−(−2)|=2.5; option B gives |−1.5/0.5|=3; option C gives |−2.5/0.5|=5; option D gives |0.5|=0.5.

Option C is clearly the largest here. Testing a second set (a=−3, b=0.2) confirms option C remains largest (16 vs. 3.2, 14, and 0.2).

Check each sign region separately. For x≥0: the equation becomes x−1=x+1, which simplifies to −1=1, never true — no solutions here.

For −1≤x<0: the equation becomes −x−1=x+1, giving x=−1, which is valid at that boundary. For x<−1: the equation becomes −x−1=−x−1, which is always true, so every x<−1 works. Combining both valid regions gives x≤−1.

Equation I is linear in x, so it has exactly one solution — solving it gives a positive value (approximately x≈16.1). Equation II is identical except it uses |x| in place of x.

Since |x| = 16.1 has two solutions, x=16.1 and x=−16.1, equation II has 2 solutions. One of those two solutions (the positive one) exactly matches the single solution to equation I. So one of the 2 solutions to II is also a solution to I.