absolute values: | 1/4x - 1/6 | = | 1/3x + 1/8 |

[ccachu]

Hello! I am having some problems with my practice problems concerning absolute values.
I just want to make sure I am on the right track with what I am doing because I honestly don't really know.

| 1/4x - 1/6 | = | 1/3x + 1/8 |

what I did:

1/4x - 1/6 = 1/3 + 1/8

1/12x = 7/24

x = -7/2

and also:

- ( 1/4x - 1/6 ) = 1/3 + 1/8

-1/4x + 1/6 = 1/3 + 1/8

-7/12x = -7/24

x = 1/2

for the second part I am not too sure if I did it correctly.
I was thinking to make both sides negative, but when I tried that I got -7/2 as an answer too.
It is just the problem asks for a smaller value and a higher value.
It also says that there may be no answer or any real number can be an answer.
So that is what confuses me.

Thank you!!

--
i edited a typo in the problem i presented.

 

[skeeter]

Re: absolute value questions!

Hello! I am having some problems with my practice problems concerning absolute values.
I just want to make sure I am on the right track with what I am doing because I honestly don't really know.

| 1/4x - 1/6 | = | 1/3 + 1/8 |

what I did:

1/4x - 1/6 = 1/3 + 1/8

(1/4)x and 1/6 are not like terms ... next step should be (1/4)x = 1/3 + 1/8 + 1/6


and also:

- ( 1/4x - 1/6 ) = 1/3 + 1/8

-1/4x + 1/6 = 1/3 + 1/8

-(1/4)x = 1/3 + 1/8 - 1/6

 

[ccachu]

Re: absolute value questions!

oh no! i made a typo

it's | 1/4x - 1/6 | = | 1/3x + 1/8 |


sorry for the typo!

 

[mmm4444bot]

Re: absolute value question!

Hi CC:

You know that |x + a| means (x + a) or -(x + a).

When we solve an equation with absolute values on both sides, we could consider all four possibilities.

|x + a| = |x + b|

x + a = x + b

x + a= -x - b

-x - a = x + b

-x - a = -x - b

But solving these four equations would be twice the work. (Notice that multiplying both sides by -1 changes one equation into another.)

So, we only need to solve two equations when we take the absolute value signs away. In other words, we only need the plus and minus on the right side.

\(\displaystyle \mbox{Eqn 1:} \; \; \frac{x}{4} - \frac{1}{6} \;=\; \frac{x}{3} + \frac{1}{8}\)

\(\displaystyle \mbox{Eqn 2:} \; \; \frac{x}{4} - \frac{1}{6} \;=\; -\left( \frac{x}{3} + \frac{1}{8} \right)\)

To get rid of the parentheses on the second equation above, distribute the -1 in front.

\(\displaystyle \mbox{Eqn 2:} \; \; \frac{x}{4} - \frac{1}{6} \;=\; -\frac{x}{3} - \frac{1}{8} \right)\)

Solve Eqn 1 and Eqn 2 to get two solutions for x.

Do you know how to add or subtract fractions like x/4+x/3 and -1/8+1/6?

If you need more help, then please post your work and tell us why you think you're stuck.

Cheers,

~ Mark