# Problem of the Week

## Updated at Jun 12, 2023 2:07 PM

How would you solve the equation $$\frac{5}{3-4{x}^{2}}=-\frac{5}{97}$$?

Below is the solution.

$\frac{5}{3-4{x}^{2}}=-\frac{5}{97}$

 1 Multiply both sides by $$3-4{x}^{2}$$.$5=-\frac{5}{97}(3-4{x}^{2})$2 Simplify  $$\frac{5}{97}(3-4{x}^{2})$$  to  $$\frac{5(3-4{x}^{2})}{97}$$.$5=-\frac{5(3-4{x}^{2})}{97}$3 Multiply both sides by $$97$$.$5\times 97=-5(3-4{x}^{2})$4 Simplify  $$5\times 97$$  to  $$485$$.$485=-5(3-4{x}^{2})$5 Divide both sides by $$-5$$.$-\frac{485}{5}=3-4{x}^{2}$6 Simplify  $$\frac{485}{5}$$  to  $$97$$.$-97=3-4{x}^{2}$7 Subtract $$3$$ from both sides.$-97-3=-4{x}^{2}$8 Simplify  $$-97-3$$  to  $$-100$$.$-100=-4{x}^{2}$9 Divide both sides by $$-4$$.$\frac{-100}{-4}={x}^{2}$10 Two negatives make a positive.$\frac{100}{4}={x}^{2}$11 Simplify  $$\frac{100}{4}$$  to  $$25$$.$25={x}^{2}$12 Take the square root of both sides.$\pm \sqrt{25}=x$13 Since $$5\times 5=25$$, the square root of $$25$$ is $$5$$.$\pm 5=x$14 Switch sides.$x=\pm 5$Donex=5,-5