Question

what is the solution to the equation \\(\frac{y}{y - 4} - \frac{4}{y + 4} = \frac{32}{y^2 - 16}\\)?
\\(y = 0\\)
\\(y = -4\\) and \\(y = 4\\)
all real numbers
no solution

Explanation:

Step1: Factor the denominator

Notice that \(y^2 - 16=(y - 4)(y + 4)\) by the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\). So the equation \(\frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{y^2 - 16}\) can be rewritten as \(\frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{(y - 4)(y + 4)}\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((y - 4)(y + 4)\). Multiply each term in the equation by \((y - 4)(y + 4)\) to eliminate the denominators:

$$
LATEXBLOCK0
$$

Step3: Solve the resulting equation

Subtract 32 from both sides: \(y^2+16 - 32=0\), which simplifies to \(y^2-16 = 0\). Factoring again (difference of squares), we get \((y - 4)(y + 4)=0\). So \(y = 4\) or \(y=-4\).

Step4: Check for extraneous solutions

We need to check if these solutions make the original denominators zero. If \(y = 4\), then \(y - 4=0\) and \(y^2 - 16 = 0\). If \(y=-4\), then \(y + 4=0\) and \(y^2 - 16 = 0\). Since division by zero is undefined, \(y = 4\) and \(y=-4\) are extraneous solutions. So there is no solution that satisfies the original equation.

Answer:

no solution