Equations & inequalities
Absolute-value equations
ALEKS placement
Isolate the bars, then split into two equations — one positive, one negative.
What this covers
∣u∣=k⇒u=koru=−k(k≥0)- Start small: |x| = 7 has two answers, 7 and −7, because both land on 7 once the bars strip the sign.
- Get the absolute-value expression alone on one side. Undo whatever sits outside the bars first.
- If the bars equal a negative number there is no solution — a distance from zero is never negative.
- Write two equations: inside = k and inside = −k. The bars hide which sign the inside had.
- Solve each, then check both in the original: |8 − 3| is 5 and |−2 − 3| is also 5.
Worked example
Worked example
Solve 2|x - 3| + 1 = 11.
- ∣x−3∣=5Subtract 1, then divide by 2 to isolate the bars.
- x−3=5orx−3=−5The inside can be 5 units away in either direction.
- x=8orx=−2Solve each branch.
Another worked example
Solve |x − 3/2| = 5/2.
- x−23=25orx−23=−25The bars are already alone. x − 3/2 is 5/2 above zero or 5/2 below zero.
- x=4orx=−1Add 3/2 in each branch: 3/2 + 5/2 = 8/2 = 4 and 3/2 − 5/2 = −2/2 = −1.
A common mistake
A mistake Lemma catches
From 2∣x−3∣+1=11
∣x−3∣=10∣x−3∣=5
“Isolate the bars before splitting: subtract 1 from both sides (11 − 1 = 10), then divide both sides by 2: 10 ÷ 2 = 5, so |x − 3| = 5. Stopping at 10 leaves the 2 in front of the bars on the left.”
Try one
Sample problem
Solve for every value of x.
∣x−7∣=6Practice absolute value free
No account, no email. Every line you type is checked by the same computer algebra system as the worked example above.
Where this fits
Before this