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

2 Answers
Jun 4, 2017

#r=1/2#

Explanation:

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

To check your answer, plug in #1/2# to each equation and solve:
#(1/2+3)/(1/2-4)=(1/2+5)/(1/2+4)#
From there, just simplify: #(7/2)/(-7/2)=(-9/2)/(9/2)#
#-1=-1#

Jun 4, 2017

#r= 1/2#

Explanation:

First cross multiply to give you a linear equation.

#(r+3)(r+4) = (r-4)(r-5)#
#r^2 + 7r + 12 = r^2 -9r + 20#

Then make the equation equal zero.

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

The #r^2# cancel out leaving this equation:

#16r - 8 = 0#
#16r = 8#

#r= 8/16 = 1/2#