Question

matk311_mathematics iii semester a_2026 - 080220207 - integ math 3a
polynomials, rational, and radical relationships: part three
what is the solution to the equation \\(\frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{y^{2}-16}\\)?
no solution
\\(y = 4\\)
all real numbers
\\(y = 0\\)
\\(y=-4\\) and \\(y = 4\\)

Explanation:

Step1: Factor the denominator

The denominator \( y^2 - 16 \) can be factored as \( (y - 4)(y + 4) \) using the difference of squares formula \( a^2 - b^2=(a - b)(a + b) \). So the equation is \( \frac{y}{y - 4}-\frac{4}{y + 4}=\frac{32}{(y - 4)(y + 4)} \).

Step2: Find the common denominator

The common denominator for the left - hand side is \( (y - 4)(y + 4) \). Multiply the first fraction \( \frac{y}{y - 4} \) by \( \frac{y + 4}{y + 4} \) and the second fraction \( \frac{4}{y + 4} \) by \( \frac{y - 4}{y - 4} \).
We get \( \frac{y(y + 4)}{(y - 4)(y + 4)}-\frac{4(y - 4)}{(y - 4)(y + 4)}=\frac{32}{(y - 4)(y + 4)} \).

Step3: Simplify the left - hand side

Expand the numerators: \( \frac{y^{2}+4y-4y + 16}{(y - 4)(y + 4)}=\frac{32}{(y - 4)(y + 4)} \).
Simplify the numerator of the left - hand side: \( \frac{y^{2}+16}{(y - 4)(y + 4)}=\frac{32}{(y - 4)(y + 4)} \).
Since the denominators are the same (and \( y
eq4 \) and \( y
eq - 4 \) to avoid division by zero), we can set the numerators equal: \( y^{2}+16 = 32 \).

Step4: Solve the resulting equation

Subtract 32 from both sides: \( y^{2}+16-32=0 \), which simplifies to \( y^{2}-16 = 0 \).
Factor the left - hand side: \( (y - 4)(y + 4)=0 \). The solutions to this equation are \( y = 4 \) and \( y=-4 \). But we must exclude these values because they make the original denominators zero (they are extraneous solutions). So there is no solution to the original rational equation.

Answer:

no solution