Simplify a sum of fractions

Question

I am stuck trying to get from:

$$\frac{pZ(a)}{pZ(a) - (1-p)Z(b)} - \frac{p(pZ(a) - (1-p)Z(b))}{pZ(a) - (1-p)Z(b)} $$

to

$$\frac{p(1-p)(Z(a) - Z(b))}{pZ(a) - (1-p)Z(b)} $$

Obviously my problem is the numerator. Could some suggest the route I should take to get from first expression to the last?

Answer 1

First, you can factor out a $p$ in the numerator to give

$$p(Z(a)-pZ(a)-(1-p)Z(b))$$

Next, factor out a $Z(a)$ from the first two terms in the parentheses

$$p((1-p)Z(a)-(1-p)Z(b))$$

Factor out a $(1-p)$ and you're there!

Answer 2

Putting the numerators over a common denominator and expanding we have $pZ(a) - p^{2}Z(a) - p(1-p)Z(b)$. But $pZ(a) - p^{2}Z(a)$ simplifies to $p(1-p)Z(a)$, so the total simplifies to $p(1-p)(Z(a) - Z(b))$.

Answer 3

Sometimes rearranging solves the problem without making a common denominator. This is particularly in the cases where the denominator is common as in our example problem

$$\frac{pZ(a)}{pZ(a) - (1-p)Z(b)} - \frac{p(pZ(a) - (1-p)Z(b))}{pZ(a) - (1-p)Z(b)} $$ $$=\frac{pZ(a)}{pZ(a) - (1-p)Z(b)} - \frac{p\cdot pZ(a) }{pZ(a) - (1-p)Z(b)} + \frac{p(1-p)Z(b))}{pZ(a) - (1-p)Z(b)}$$ $$=\frac{pZ(a)(1-p)}{pZ(a) - (1-p)Z(b)} + \frac{p(1-p)Z(b))}{pZ(a) - (1-p)Z(b)}$$ $$=\frac{p(1-p)(Z(a)-Z(b))}{pZ(a) - (1-p)Z(b)}$$ $$=R.H.S$$