How do you solve r+3r4=r5r+4?

2 Answers
Jun 4, 2017

r=12

Explanation:

Eliminate the fractions by cross multiplying: r+3r4=r5r+4 becomes (r+3)(r+4)=(r5)(r4)
Distribute each binomial to get two quadratic equations: r2+7r+12=r29r+20
Subtract r2 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=816
Simplify: r=12

To check your answer, plug in 12 to each equation and solve:
12+3124=12+512+4
From there, just simplify: 7272=9292
1=1

Jun 4, 2017

r=12

Explanation:

First cross multiply to give you a linear equation.

(r+3)(r+4)=(r4)(r5)
r2+7r+12=r29r+20

Then make the equation equal zero.

(r2+7r+12)(r29r+20)=0

The r2 cancel out leaving this equation:

16r8=0
16r=8

r=816=12