To solve the given equation, we need to simplify and factor each term. First, factor the numerators and denominators where possible. Then, find a common denominator to combine the fractions on the left side of the equation. Finally, simplify the equation and solve for \( y \).
We start with the left-hand side of the equation:
\[
\frac{y^{2}-9}{y}+\frac{y+3}{5y-15}
\]
Factoring \(y^2 - 9\) gives us \((y - 3)(y + 3)\), and \(5y - 15\) can be factored as \(5(y - 3)\). Thus, we rewrite the left-hand side as:
\[
\frac{(y - 3)(y + 3)}{y} + \frac{y + 3}{5(y - 3)}
\]
The common denominator for the left-hand side is \(5y(y - 3)\). Rewriting each term with this common denominator, we have:
\[
\frac{5(y - 3)(y + 3)}{5y(y - 3)} + \frac{(y + 3)y}{5y(y - 3)}
\]
Combining these fractions results in:
\[
\frac{(y + 3)(5y^2 - 29y + 45)}{5y(y - 3)}
\]
The right-hand side of the original equation is:
\[
\frac{3(2y - 5)(y + 3)}{5y(y - 3)}
\]
Setting the left-hand side equal to the right-hand side gives us:
\[
\frac{(y + 3)(5y^2 - 29y + 45)}{5y(y - 3)} = \frac{3(y + 3)(2y - 5)}{5y(y - 3)}
\]
Since \(y + 3\) is common in both sides, we can cancel it (assuming \(y \neq -3\)). This leads us to solve:
\[
5y^2 - 29y + 45 = 3(2y - 5)
\]
Expanding and rearranging gives us:
\[
5y^2 - 29y + 45 - 6y + 15 = 0 \implies 5y^2 - 35y + 60 = 0
\]
Dividing through by 5 simplifies to:
\[
y^2 - 7y + 12 = 0
\]
Factoring this quadratic yields:
\[
(y - 3)(y - 4) = 0
\]
Thus, the solutions are:
\[
y = 3 \quad \text{and} \quad y = 4
\]
The solutions to the equation are:
\[
\boxed{y = -3} \quad \text{and} \quad \boxed{y = 4}
\]
Answered
