Question

How do you solve `\frac { 4} { v ^ { 2} - 3v } = \frac { 1} { 5- v }`?

Answer 1

`v = -5, 4`

Explanation:

We start by cross-multiplying to get rid of the fractions.

`4 / (v^2 - 3v) = 1/(5-v) -> 4(5-v) = v^2 - 3v`

This can be rearranged:

`4(5-v) = v^2 - 3v -> 20 - 4v = v^2 - 3v -> v^2 + v - 20 = 0`
`-> (v + 5)(v - 4) = 0 -> v = -5, 4`.

We've found two solutions, `v = -5` and `v = 4`.

Answer 2

`v = 4 or -5`

Explanation:

You'll want to start by cross multiplying.

`4(5-v) = 1(v^2-3v)`
`20 - 4v = v^2 - 3v`

Have like terms on the same side.
`20 = v^2 + v`

Now you can figure out in your head that `v = 4`, but if you use the quadratic formula, you'll get another possible solution.

`v^2 + v - 20 = 0`

`v = (-b +- sqrt(b^2 - 4ac))/(2a)`

Plug in your values (a = 1, b = 1, c = -20), and you get:

`v = (-1 +- sqrt(1^2 - (4 * 1 * -20)))/(2 * 1)`

Now let's simplify.

`v = (-1 +- sqrt(1 + 80))/2`

`v = (-1 +- sqrt81)/2`

`v = (-1 +- 9)/2`

`(-1 + 9)/2 = 8/2 = 4`

`(-1 - 9)/2 = 8/2 = -5`

So `v` can be either `4` or `-5`