How do you solve (r+3)/(r-4) = (r-5)/(r+4)r+3r4=r5r+4?

2 Answers
Jun 4, 2017

r=1/2r=12

Explanation:

Eliminate the fractions by cross multiplying: (r+3)/(r-4)=(r-5)/(r+4)r+3r4=r5r+4 becomes (r+3)(r+4)=(r-5)(r-4)(r+3)(r+4)=(r5)(r4)
Distribute each binomial to get two quadratic equations: r^2+7r+12=r^2-9r+20r2+7r+12=r29r+20
Subtract r^2r2 from both sides: 7r+12=-9r+207r+12=9r+20
Add 9 to both sides: 16r+12=2016r+12=20
Subtract 12 from both sides: 16r=816r=8
Divide both sides by 16: r=8/16r=816
Simplify: r=1/2r=12

To check your answer, plug in 1/212 to each equation and solve:
(1/2+3)/(1/2-4)=(1/2+5)/(1/2+4)12+3124=12+512+4
From there, just simplify: (7/2)/(-7/2)=(-9/2)/(9/2)7272=9292
-1=-11=1

Jun 4, 2017

r= 1/2r=12

Explanation:

First cross multiply to give you a linear equation.

(r+3)(r+4) = (r-4)(r-5)(r+3)(r+4)=(r4)(r5)
r^2 + 7r + 12 = r^2 -9r + 20r2+7r+12=r29r+20

Then make the equation equal zero.

(r^2 +7r+12)-(r^2 -9r+20) = 0(r2+7r+12)(r29r+20)=0

The r^2r2 cancel out leaving this equation:

16r - 8 = 016r8=0
16r = 816r=8

r= 8/16 = 1/2r=816=12