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

Question

1112 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T21:49:10

x=7, -4

$(7x+7-6x+18)/((x-3)(x+1))$ (Don't forget to distribute the negative)

$(x+25)/((x-3)(x+1))$

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

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

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

ANSWER: x=7, -4

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