$3x\:+2y\:-3z\:=\:0$
$\left(x^2-4y^4\right)^2$
$\frac{2x}{x-3}+\frac{x}{x+4}=\frac{3}{x+4}$
$\left(-7\right).\left(-2\right).2\left(-3\right).\left(5\right).2$
$\frac{tan^2x+1}{tanx+1}=sec^2x-tanx$
$\:16\:a^2\:+\:88\:ab\:+\:121\:b^2\:$
$tan^2x+tan\left(x\right)-2=0$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!