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

Jun 4, 2017

$r = \frac{1}{2}$

Explanation:

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

To check your answer, plug in $\frac{1}{2}$ to each equation and solve:
$\frac{\frac{1}{2} + 3}{\frac{1}{2} - 4} = \frac{\frac{1}{2} + 5}{\frac{1}{2} + 4}$
From there, just simplify: $\frac{\frac{7}{2}}{- \frac{7}{2}} = \frac{- \frac{9}{2}}{\frac{9}{2}}$
$- 1 = - 1$

Jun 4, 2017

$r = \frac{1}{2}$

Explanation:

First cross multiply to give you a linear equation.

$\left(r + 3\right) \left(r + 4\right) = \left(r - 4\right) \left(r - 5\right)$
${r}^{2} + 7 r + 12 = {r}^{2} - 9 r + 20$

Then make the equation equal zero.

$\left({r}^{2} + 7 r + 12\right) - \left({r}^{2} - 9 r + 20\right) = 0$

The ${r}^{2}$ cancel out leaving this equation:

$16 r - 8 = 0$
$16 r = 8$

$r = \frac{8}{16} = \frac{1}{2}$