# Problem of the Week Problem D and Solution Fraction Distraction

## Problem

Find all ordered pairs, $$(a,b)$$, that satisfy $$\dfrac{a - b}{a + b} = 9$$ and $$\dfrac{ab}{a+b} = -60$$.

## Solution

Multiplying both sides of the first equation, $$\dfrac{a - b}{a + b} = 9$$, by $$a+b$$ gives $$a - b = 9a + 9b$$ and so $$-8a = 10b$$ or $$-4a = 5b$$. Thus, $$a = -\dfrac{5}{4} b$$.

Multiplying both sides of the second equation, $$\dfrac{ab}{a+b} = -60$$, by $$a+b$$ gives $$ab = -60a -60b$$. Substituting $$a = -\dfrac{5}{4} b$$ into $$ab = -60a -60b$$, we get \begin{aligned} ab &= -60a -60b\\ \left(-\frac{5}{4}b\right)(b) & = -60\left(-\frac{5}{4}b\right) - 60b\\ -\frac{5}{4}b^2 &= 75b - 60b\\ -\frac{5}{4}b^2 &= 15b\\ b^2 &= - 12b\\ b^2 + 12b&= 0\end{aligned}

Notice that $$b=0$$ satisfies this equation. Thus $$b=0$$ is one possibility. When $$b\neq 0$$, we can divide both sides of the equation by $$b$$ to get $$b+12 = 0$$, or $$b=-12$$. Thus, $$b=0$$ or $$b=-12$$.

If $$b = 0$$, then $$a = -\dfrac{5}{4}(0) = 0$$. But this gives us a denominator of $$0$$ in each of the original equations. Therefore, $$b \neq 0$$.

If $$b =-12$$, then $$a= -\dfrac{5}{4} (-12) = 15$$.

Therefore, the only ordered pair that satisfies both equations is $$(15, -12)$$.