# 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 \((15, -12)\).