Question

$\frac{8}{6 + x} - \frac{8}{x - 7}$
$\frac{9 {a }^ {2} + {y}^{2} }{3a - y} + \frac{6ay}{y - 3a}$
​

Answer 1

$\dfrac{8}{6+x} -\dfrac{8}{x-7} =\dfrac{8(x-7)}{(6+x)(x-7)} -\dfrac{8(6+x)}{(x-7)(6+x)} =$

$=\dfrac{8(x-7)-8(6+x)}{(6+x)(x-7)}=\dfrac{8x-56-48-8x}{6x-42+x^2-7x}=\dfrac{-104}{x^2-x-42}=-\dfrac{104}{x^2-x-42}$

$\dfrac{9a^2+y^2}{3a-y} +\dfrac{6ay}{y-3a} =\dfrac{9a^2+y^2}{3a-y} -\dfrac{6ay}{3a-y} =$

$=\dfrac{9a^2+y^2-6ay}{3a-y}= \dfrac{(3a-y)^2}{3a-y} =3a-y$