Solve for $r$: $$\frac{r+3}{r-2} = \frac{r-1}{r+1}.$$Express your answer as a fraction.
Ok, let's start by multiplying both sides of the equation by $r-2$. That gives us $$r+3 = \frac{(r-1)(r-2)}{r+1}.$$ Right. Now, let's multiply both sides of the equation by $r+1$. Now we have $$(r+3)(r+1) = (r-1)(r-2).$$ Expanding both sides gives us $$r^2 + 4r + 3 = r^2 - 3r + 2.$$ So, if we subtract $r^2$ from both sides, we get $$4r+3=-3r+2.$$ If we add $3r$ to both sides, we get $$7r+3=2.$$ And if we subtract two from both sides, we get $$7r= -1.$$ So $r$ is equal to $$-\frac{1}{7}.$$ Right. Let's check our answer. Our equation is $$\frac{r+3}{r-2} = \frac{r-1}{r+1}.$$ So if we plug in $r = -\frac{1}{7}$, we get $$\frac{-\frac{1}{7}+3}{-\frac{1}{7}-2} = \frac{-\frac{1}{7}-1}{-\frac{1}{7}+1}.$$ Right. And the right side is $\frac{-\frac{8}{7}}{\frac{6}{7}}$ So the left side is $-\frac{4}{3}$ and the right side is $-\frac{4}{3}$. And since both sides are equal to $-\frac{4}{3}$, we can verify that $r = -\frac{1}{7}$ is the correct answer.



