what is the solution to the equation $\frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{y^{2}-16}$?\n$y=-4$ and $y =…

Explanation:

Step1: Factor the denominator

Note that $y^{2}-16=(y - 4)(y + 4)$. The given equation is $\frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{y^{2}-16}$.

Step2: Find the common - denominator

The common denominator of the left - hand side is $(y - 4)(y + 4)$. Rewrite the left - hand side: $\frac{y(y + 4)-4(y - 4)}{(y - 4)(y + 4)}=\frac{y^{2}+4y-4y + 16}{(y - 4)(y + 4)}=\frac{y^{2}+16}{(y - 4)(y + 4)}$.

Step3: Set up the new equation

The equation becomes $\frac{y^{2}+16}{(y - 4)(y + 4)}=\frac{32}{(y - 4)(y + 4)}$. Since the denominators are non - zero (we will check for extraneous solutions later), we can set the numerators equal: $y^{2}+16 = 32$.

Step4: Solve for y

Subtract 16 from both sides: $y^{2}=32 - 16=16$. Then $y=\pm4$.

Step5: Check for extraneous solutions

If $y = 4$, the original denominators $y - 4=0$ and $y^{2}-16 = 0$. If $y=-4$, the original denominators $y + 4 = 0$ and $y^{2}-16 = 0$. So, $y = 4$ and $y=-4$ are extraneous solutions.

Answer:

no solution