## Algebra 1

$\dfrac{1}{a+b}$
(i) $x^m \cdot x^n = x^{m+n}$ (ii) $x^{-m} = \dfrac{1}{x^m}, x \ne 0$ Use rule (i) above to obtain: $=(a+b)^{2+(-3)} \\=(a+b)^{-1}$ Use rule (ii) above to obtain: $=\dfrac{1}{a+b}$