View Full Version : Absolute-value equations: |2x - 1| = 7, etc

:shock: I'm having some trouble with solving absolute-value equations and finding the solution sets. Here are three with which I'm needing some help:

:arrow: Find the solution sets of the following:

1) |2x - 1| = 7

2) |3 + 4x = 1

3) 4x - |xt5| = 10

stapel

05-25-2006, 06:24 PM

Since the absolute value of an expression returns the positive (or at least non-negative) value, you have to account for the possible cancellation of a "minus" sign when you take off the absolute-value bars. For instance:

. . . . .|5x + 7| = 2

. . . . .5x + 7 = ±2

. . . . .5x + 7 = -2 or 5x + 7 = 2

. . . . .5x = -9 or 5x = -5

. . . . .x = -9/5 or x = -1

. . . . .Checking:

. . . . .x = -9/5: |5(-9/5) + 7| = |-9 + 7| = |-2| = 2

. . . . .x = -1: |5(-1) + 7] = |-5 + 7| = |2| = 2

Follow this process with your examples. If you get stuck, please reply showing (as I did above) all of your steps. In particular, please include a corrected version of the second exercise (where is the other absolute-value bar?) and an explanation of the third exercise (what does "xt5" mean?).

Thank you.

Eliz.

:arrow: xl3-2xl=11

x=7

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:arrow: l3x-1l=5

x=2

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:arrow: 3lx-2l=15

x=7

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:idea: I HAVE NO IDEA ON HOW 2 PUT DA PLUS N MINUS ON THE EQUATION!!!

Denis

05-26-2006, 09:30 PM

BB, you obviously need your teacher to help you out...

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