Simplify a rational identity

Question

Simplify: $$\frac{\dfrac{a}{b}-\dfrac{b}{a}}{1+\dfrac{b}{a}}$$

I have a feeling the solution has to do with factoring, but I'm really not sure, and would appreciate any help.

Answer 1

Multiply numerator and denominator by $ab$ to get $\frac {a^{2}-b^{2}} {b(a+b)}$. Then use the fact that $a^{2}-b^{2} =(a-b)(a+b)$. Cancel $a+b$ to get $\frac {a-b} b$.

Answer 2

Multiply with $\frac{ab}{ab}$ as first step: $$\frac{\frac ab-\frac ba}{1+\frac ba}=\frac{a^2-b^2}{ab+b^2}=\frac{(a+b)(a-b)}{(a+b)b}=\frac{a-b}b=\frac ab-1 $$ (provided $a+b\ne0$, but in that case the original fraction would not be defined)

Answer 3

Well, in view of addition the rule is $$\frac{a}{b}+\frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}$$ i.e., the fractions need to be extended first to obtain the same denominator and then the addition can be performed (same for subtraction).

In view of multiplication the rule is $$\frac{a}{b}\cdot\frac{c}{d} = \frac{ac}{bd}$$ i.e., numerator and denominator are multiplied separately.

In view of division the rule is $$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b}:\frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}$$ i.e., in the fraction (denominator) $c/d$ the numerator and denominator are flipped $d/c$ and then multiplied with $a/b$.

You can easily use these rules to simplify your fraction: $$\frac{\frac{a}{b}-\frac{b}{a}}{1+\frac{b}{a}} = \frac{\frac{a^2}{ab}-\frac{b^2}{ab}}{\frac{a}{a}+\frac{b}{a}} = \frac{\frac{a^2-b^2}{ab}}{\frac{a+b}{a}} = \frac{a^2-b^2}{ab}\cdot \frac{a}{a+b} = \frac{(a^2-b^2)a}{ab(a+b)} = \frac{a^2-b^2}{b(a+b)} =\frac{(a-b)(a+b)}{b(a+b)} =\frac{a-b}{b}.$$