Example

Try It: Solving 3x51=6|3x - 5| - 1 = 6 and 4x35=2|4x - 3| - 5 = 2

To actively practice solving absolute value equations requiring prior isolation, solve the given equations 3x51=6|3x - 5| - 1 = 6 and 4x35=2|4x - 3| - 5 = 2. For the first equation, isolate the absolute value by adding 11 to both sides, which results in 3x5=7|3x - 5| = 7. Generating the equivalent equations yields 3x5=73x - 5 = -7 or 3x5=73x - 5 = 7. Adding 55 and dividing by 33 for each case provides the solutions x=23x = -\frac{2}{3} or x=4x = 4. For the second equation, adding 55 to both sides isolates the absolute value, producing 4x3=7|4x - 3| = 7. The equivalent equations are 4x3=74x - 3 = -7 or 4x3=74x - 3 = 7. Adding 33 and dividing by 44 evaluates to the final solutions x=1x = -1 or x=52x = \frac{5}{2}.

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Updated 2026-05-03

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Ch.2 Solving Linear Equations - Intermediate Algebra @ OpenStax

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