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

2 Answers
May 3, 2018

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.

May 3, 2018

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