How do you solve $12/(v^2-16) - 24/(v-4) = 3$ ?

Question

1630 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-19T02:25:06

You have to put on an equal denominator.

First, factor the denominator to find the LCD (Least Common Denominator).

$12/((v -4)(v + 4)) - 24/(v - 4) = 3$

The LCD is $(v - 4)(v + 4)$

$12/((v - 4)(v + 4)) - (24(v + 4))/((v - 4)(v + 4)) = (3(v - 4)(v + 4))/((v - 4)(v + 4))$

We can now eliminate the denominator since all the fractions are now equivalent.

$12 - 24v + 96 = 3(v^2 - 16)$

$12 - 24v + 96 = 3v^2 - 48$

$0 = 3v^2 + 24v - 156$

$0 = 3(v^2 + 8v - 52)$

Solving by completing the square since factoring isn't possible:

$0 + 52= 1(v^2 + 8v + 16 - 16)$

$52 = (v + 4)^2$

$+-sqrt(52) = v + 4$

$+-sqrt(52) - 4 = v$

$+-2sqrt(13) - 4 = v$

Hopefully this helps!

How do you solve 12/(v^2-16) - 24/(v-4) = 312/(v^2-16) - 24/(v-4) = 3?