Chalk−1

Math · College algebra · Worked example

Solve an absolute value equation

Solve |2x − 3| + 4 = 11 and check both solutions.

∣2⁢x−3⁢∣+4=11

Isolate the absolute value

Subtract 4 from both sides. The right side, 7, is positive, so there will be two solutions.

∣2⁢x−3⁢∣=7

Split into two equations

2x − 3 is 7 units from zero, so it equals 7 or −7.

2⁢x−3=7 or 2⁢x−3=−7

Solve each equation

Add 3 and divide by 2 in each: 2x = 10 gives x = 5, and 2x = −4 gives x = −2.

x=5 or ⁢x=−2

Check both in the original equation

For x = 5, |10 − 3| + 4 = 7 + 4 = 11. For x = −2, |−4 − 3| + 4 = 7 + 4 = 11. Both work.

∣2⁢(5)−3⁢∣+4=11∣2⁢(−2)−3⁢∣+4=11

See the two solutions on a graph

The V-shaped graph of y = |2x − 3| + 4 meets the line y = 11 at x = −2 and x = 5, once on each arm.

Result

x = 5 or x = −2. Both satisfy the original equation.

x=5 or ⁢x=−2

Your turn

Solve 3|x + 1| − 2 = 10.

Show the answer and explanation

x = 3 or x = −5.

Add 2 and divide by 3: |x + 1| = 4. So x + 1 = 4 or x + 1 = −4, which gives x = 3 or x = −5. Check: 3|4| − 2 = 10 and 3|−4| − 2 = 10.

3⁢∣⁢x+1⁢∣−2=10∣⁢x+1⁢∣=4x+1=4 or ⁢x+1=−4x=3 or ⁢x=−5

Keep exploring

Open the steps in Math and change 11 to 3: isolating gives |2x − 3| = −1, and no number makes a distance negative.

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