$10a+10a+6a-4a$

$\frac{1}{2x}+\frac{1}{4}=\frac{1}{10x}+\frac{1}{5}$

$42x^2+30y^2-10=0$

$\left(m-2\left(m+2\right)\right)$

$r^2+12r+36$

$\frac{\left(1-\cos^2\left(x\right)\right)}{\sec^2\left(x\right)-1}=1-\sin^2\left(x\right)$

$\frac{2x-5}{12}=\frac{x}{4}-\frac{5}{3}$