Question

How do you solve `-\frac { 8} { v - 4} = - \frac { 4} { 5v - 20} + 1`?

Answer 1

See a solution process below: #v = -16/5

Explanation:

First, we need to multiply the fraction on the left side of the equation by the appropriate form of `1` to put both fractions over a common denominator:

`5/5 xx -8/(v - 4) = -4/(5v - 20) + 1`

`(5 xx -8)/(5(v - 4)) = -4/(5v - 20) + 1`

`-40/(5v - 20) = -4/(5v - 20) + 1`

We can next add `color(red)(4/(5v - 20)` to be able to add the fractions while keeping the equation balanced:

`color(red)(4)/(5v - 20) - 40/(5v - 20) = color(red)(4)/(5v - 20) - 4/(5v - 20) + 1`

`(color(red)(4) - 40)/(5v - 20) = 0 + 1`

`-36/(5v - 20) = 1`

Then, we can multiply each side of the equation by `color(red)(5v - 20)` to eliminate the fraction while keeping the equation balanced:

`(color(red)(5v - 20)) xx -36/(5v - 20) = 1 xx (color(red)(5v - 20))`

`cancel((color(red)(5v - 20))) xx -36/color(red)(cancel(color(black)((5v - 20)))) = 5v - 20`

`-36 = 5v - 20`

Next, add `color(red)(20)` to each side of the equation to isolate the `v` term while keeping the equation balanced:

`-36 + color(red)(20) = 5v - 20 + color(red)(20)`

`-16 = 5v - 0`

`-16 = 5v`

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

`-16/color(red)(5) = (5v)/color(red)(5)`

`-16/5 = (color(red)(cancel(color(black)(5)))v)/cancel(color(red)(5))`

`-16/5 = v`

`v = -16/5`