How do I do this problem for my college algebra class?

May 2021 2 52 Report

How do I do this problem for my college algebra class?

First, write the value(s) that make the denominator(s) zero. Then solve the equation.

x - 6 over 3x + 2 = x + 8 over x

OR written like this without brackets...

[x - 6] / 3x + 2 = [x + 8] / x

(x - 6) * x = (x + 8) * (3x + 2)

x^2 - 6x = 3x^2 + 26x + 16

0 = 2x^2 + 32x + 16

0 = x^2 + 16x + 8

0 = (x+4)^2

x = -4

x cannot equal 0 or -2/3

x=0 will make the denominator become 0.

3x*[(x-6)/3x+2]=[(x+8)/x]*3x (multiply both sides by 3x)

(x-6)+2(3x)=3(x+8) (distribute 3x into the equation)

x-6+6x=3x+24

7x-6=3x+24

7x-3x=24+6

4x=30

x=30/4

x=15/2