Question

How do you solve `1/8x+3/2=3/4x-1`?

Answer 1

See a solution process below:

Explanation:

First, multiply both sides of the equation by `color(red)(8)` to eliminate the fractions. `color(red)(8)` is the Lowest Common Denominator (LCD) of the three fractions:

`color(red)(8)(1/8x + 3/2) = color(red)(8)(3/4x - 1)`

`(color(red)(8) xx 1/8x) + (color(red)(8) xx 3/2) = (color(red)(8) xx 3/4x) - (color(red)(8) xx 1)`

`(cancel(color(red)(8)) xx 1/color(red)(cancel(color(black)(8)))x) + (cancel(color(red)(8))4 xx 3/color(red)(cancel(color(black)(2)))) = (cancel(color(red)(8))2 xx 3/color(red)(cancel(color(black)(4)))x) - 8`

`1x + 12 = 6x - 8`

Next, subtract `color(red)(1x)` and add `color(red)(8)` from each side of the equation to isolate the `x` term while keeping the equation balanced:

`-color(red)(1x) + 1x + 12 + color(red)(8) = -color(red)(1x) + 6x - 8 + color(red)(8)`

`0 + 20 = (-color(red)(1) + 6)x - 0`

`20 = 5x`

Now, divide each side of the equation by `color(red)(5)` to solve for `x` while keeping the equation balanced:

`20/color(red)(5) = (5x)/color(red)(5)`

`4 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))`

`4 = x`

`x = 4`

Answer 2

`color(orange)(x!=4` is a extraneous solution

Explanation:

`1/8x+3/2=3/4x-1`

multiply L.H.S AND R.H.S by `8`

`:.x+12=6x-8`

`:.x-6x=-8-12`

`:.-5x=-20`

multiply L.H.S AND R.H.S by`-1`

`:.5x=20`

`:.color(orange)(x=4`

substitute `color(orange)(x=4`

`:.1/8(color(orange)4)+3/2=3/4(color(orange)4)-1`

`:.4/8+3/2=12/4-1`

multiply L.H.S AND R.H.S by`8`

`:.32/8+24/2=24/4-8`

`:.4+12=6-8`

`:.color(orange)(16!=-2`

`:.color(orange)(x=4` is a extraneous solution