what is the solution to the equation y over y minus four minus four over y plus for equals thirty two over y to the power of two minus sixteen?



Answer :

Step-by-step explanation:

To solve the equation

\[

\frac{y}{y - 4} - \frac{4}{y + 4} = \frac{32}{y^2 - 16},

\]

we first recognize that \( y^2 - 16 \) can be factored using the difference of squares:

\[

y^2 - 16 = (y - 4)(y + 4).

\]

Thus, the equation becomes

\[

\frac{y}{y - 4} - \frac{4}{y + 4} = \frac{32}{(y - 4)(y + 4)}.

\]

To combine the fractions on the left-hand side, we need a common denominator. The common denominator is \((y - 4)(y + 4)\). We rewrite each term with this common denominator:

\[

\frac{y(y + 4)}{(y - 4)(y + 4)} - \frac{4(y - 4)}{(y - 4)(y + 4)} = \frac{32}{(y - 4)(y + 4)}.

\]

Simplify the numerators:

\[

\frac{y^2 + 4y}{(y - 4)(y + 4)} - \frac{4y - 16}{(y - 4)(y + 4)} = \frac{32}{(y - 4)(y + 4)}.

\]

Combine the fractions:

\[

\frac{y^2 + 4y - 4y + 16}{(y - 4)(y + 4)} = \frac{32}{(y - 4)(y + 4)}.

\]

This simplifies to:

\[

\frac{y^2 + 16}{(y - 4)(y + 4)} = \frac{32}{(y - 4)(y + 4)}.

\]

Since the denominators are the same, we equate the numerators:

\[

y^2 + 16 = 32.

\]

Solve for \(y\):

\[

y^2 + 16 = 32,

\]

\[

y^2 = 16,

\]

\[

y = \pm 4.

\]

However, we need to check for any restrictions. The original denominators \(y - 4\) and \(y + 4\) cannot be zero:

- \(y - 4 \neq 0 \implies y \neq 4\)

- \(y + 4 \neq 0 \implies y \neq -4\)

Thus, \(y = 4\) and \(y = -4\) are both invalid solutions. Therefore, the equation has no valid solutions.

\[

\boxed{\text{No solution}}

\]