Question

How do you solve `-27= \frac { 3v - 6} { 8} - \frac { 8v + 9} { 2}`?

Answer 1

See a solution process below:

Explanation:

First, multiply each side of the equation by the appropriate form of `1` to eliminate the fractions while keeping the equation balanced:

`color(red)(16) xx -27 = color(red)(16)((3v - 6)/8 - (8v + 9)/2)`

`-432 = (color(red)(16) xx (3v - 6)/8) - (color(red)(16) xx (8v + 9)/2)`

`-432 = (cancel(color(red)(16))color(red)(2) xx (3v - 6)/color(red)(cancel(color(black)(8)))) - (cancel(color(red)(16))color(red)(8) xx (8v + 9)/color(red)(cancel(color(black)(2))))`

`-432 = color(red)(2)(3v - 6) - color(red)(8)(8v + 9)`

`-432 = (color(red)(2) xx 3v) - (color(red)(2) xx 6) - (color(red)(8) xx 8v) + (color(red)(8) xx 9)`

`-432 = 6v - 12 - 64v - 72`

Next, group and combine like terms on the right side of the equation:

`-432 = 6v - 64v - 12 - 72`

`-432 = (6 - 64)v + (-12 - 72)`

`-432 = -58v + (-84)`

`-432 = -58v - 84`

Then, add `color(red)(84)` to each side of the equation to isolate the `v` term while keeping the equation balanced:

`-432 + color(red)(84) = -58v - 84 + color(red)(84)`

`-348 = -58v - 0`

`-348 = -58v`

Now, divide each side of the equation by `color(red)(-58)` to solve for `v` while keeping the equation balanced:

`(-348)/color(red)(-58) = (-58v)/color(red)(-58)`

`6 = (color(red)(cancel(color(black)(-58)))v)/cancel(color(red)(-58))`

`6 = v`

`v = 6`

Answer 2

`v=6`

Explanation:

`(3v-6)/8-(8v+9)/2=-27`

`(3v-6-32v-36)/8=-27`

`(-29v-42)/8=-27`

`-29v-42=-216`

`-29v=-174`

`v=(-174)/(-29)=6`

Answer 3

Multiply throughout by 8

`-216=(3v-6) - 4(8v+9)`

Remove the brackets

`-216=3v-6-32v-36`

Collect like terms

`-216=-29v-42`

Add `29v and 216` to both sides

`29v=174`

Divide by 29

`v=174/29`

`v=6`