Question

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

Answer 1

`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`

Answer 2

`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`