Example

Try It: Solving ∣3x−5∣−1=6|3x - 5| - 1 = 6 and ∣4x−3∣−5=2|4x - 3| - 5 = 2

To actively practice solving absolute value equations requiring prior isolation, solve the given equations ∣3x−5∣−1=6|3x - 5| - 1 = 6 and ∣4x−3∣−5=2|4x - 3| - 5 = 2. For the first equation, isolate the absolute value by adding 1 to both sides, which results in ∣3x−5∣=7|3x - 5| = 7. Generating the equivalent equations yields 3x−5=−73x - 5 = -7 or 3x−5=73x - 5 = 7. Adding 5 and dividing by 3 for each case provides the solutions x=−23x = -\frac{2}{3} or x=4x = 4. For the second equation, adding 5 to both sides isolates the absolute value, producing ∣4x−3∣=7|4x - 3| = 7. The equivalent equations are 4x−3=−74x - 3 = -7 or 4x−3=74x - 3 = 7. Adding 3 and dividing by 4 evaluates to the final solutions x=−1x = -1 or x=52x = \frac{5}{2}.

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Updated 2026-06-29

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Ch.2 Solving Linear Equations - Intermediate Algebra @ OpenStax

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