How do you solve \frac { 7} { x - 3} - \frac { 6} { x + 1} = 1?

Mar 18, 2017

x=7, -4

Explanation:

1. Find the gcf of the fractions which is (x-3)(x+1)
7(x+1) and 6(x-3)

2. Subtract the two fractions.
$\frac{7 x + 7 - \left(6 x - 18\right)}{\left(x - 3\right) \left(x + 1\right)}$

$\frac{7 x + 7 - 6 x + 18}{\left(x - 3\right) \left(x + 1\right)}$ (Don't forget to distribute the negative)

$\frac{x + 25}{\left(x - 3\right) \left(x + 1\right)}$

3.Multiply (x-3)(x+1) to get rid of the fractions.

x+25= (x-3)(x+1)
x+25= ${x}^{2}$-2x-3

1. Bring all the terms to one side
${x}^{2}$-3x-28=0

5.Solve the equation (I'm assuming you know how to solve quadratic equations)

(x-7)(x+4)=0
x-7=0 x+4=0
x=7 x= -4