## An algebra exercise

As I worked on a problem involving the exponential distribution, I had to carry out an algebra derivation, which is a nice stand alone algebra problem. Here it is:

$\displaystyle \frac{\beta}{\beta - \alpha} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma}=\frac{\alpha}{\alpha-\gamma} \ \frac{\beta}{\beta-\gamma}$

The exercise is to show that the left hand side equals to the right hand side. For anyone taking algebra or preparing for SAT or ACT, this would be an excellent exercise. You may want to try it on your own first. You can find the derivation beow.

\displaystyle \begin{aligned}\frac{\beta}{\beta - \alpha} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma}&=\frac{-\beta}{\alpha-\beta} \ \frac{\alpha}{\alpha-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma} \\& \text{ } \\&=\frac{-\beta}{\alpha-\beta} \ \frac{\alpha}{\alpha-\gamma} \ \frac{\beta-\gamma}{\beta-\gamma}+\frac{\alpha}{\alpha-\beta} \ \frac{\beta}{\beta-\gamma} \ \frac{\alpha-\gamma}{\alpha-\gamma} \\& \text{ } \\&=\frac{-\alpha \ \beta \ (\beta-\gamma)+\alpha \ \beta \ (\alpha-\gamma)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta \ (\gamma-\beta+\alpha - \gamma)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta \ (\alpha-\beta)}{(\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha \ \beta}{(\alpha-\gamma)(\beta-\gamma)} \\& \text{ } \\&=\frac{\alpha}{\alpha-\gamma} \ \frac{\beta}{\beta-\gamma} \end{aligned}