Solve for y y/(5y-10)-5/(3y+6)=(2y^2-19y+54)/(15y^2-60)

$\frac{y}{5 y - 10} - \frac{5}{3 y + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor each term.

Factor $5$ out of $5 y - 10$ .

Factor $5$ out of $5 y$ .

$\frac{y}{5 (y) - 10} - \frac{5}{3 y + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $5$ out of $- 10$ .

$\frac{y}{5 y + 5 \cdot - 2} - \frac{5}{3 y + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $5$ out of $5 y + 5 \cdot - 2$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 y + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $3$ out of $3 y + 6$ .

Factor $3$ out of $3 y$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y) + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $3$ out of $6$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 y + 3 \cdot 2} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $3$ out of $3 y + 3 \cdot 2$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$

Factor $15$ out of $15 y^{2} - 60$ .

Factor $15$ out of $15 y^{2}$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y^{2}) - 60}$

Factor $15$ out of $- 60$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} + 15 \cdot - 4}$

Factor $15$ out of $15 y^{2} + 15 \cdot - 4$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y^{2} - 4)}$

Rewrite $4$ as $2^{2}$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y^{2} - 2^{2})}$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 2$ .

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 ((y + 2) (y - 2))}$

Remove unnecessary parentheses.

$\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)}$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$5 (y - 2), 3 (y + 2), 15 (y + 2) (y - 2)$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Since $5$ has no factors besides $1$ and $5$ .

$5$ is a prime number

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

$15$ has factors of $3$ and $5$ .

$3 \cdot 5$

Multiply $3$ by $5$ .

$15$

The factor for $y - 2$ is $y - 2$ itself.

$(y - 2) = y - 2$

$(y - 2)$ occurs $1$ time.

The factor for $y + 2$ is $y + 2$ itself.

$(y + 2) = y + 2$

$(y + 2)$ occurs $1$ time.

The factor for $y - 2$ is $y - 2$ itself.

$(y - 2) = y - 2$

$(y - 2)$ occurs $1$ time.

The LCM of $y - 2, y + 2, y + 2, y - 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(y - 2) (y + 2)$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$15 (y - 2) (y + 2)$

Multiply each term in $\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)}$ by $15 (y - 2) (y + 2)$ to eliminate the fractions.

Multiply each term in $\frac{y}{5 (y - 2)} - \frac{5}{3 (y + 2)} = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)}$ by $15 (y - 2) (y + 2)$ .

$\frac{y}{5 (y - 2)} (15 (y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$15 \frac{y}{5 (y - 2)} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$5 (3) \frac{y}{5 (y - 2)} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Cancel the common factor.

$5 \cdot 3 \frac{y}{5 (y - 2)} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Rewrite the expression.

$3 \frac{y}{y - 2} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Combine $3$ and $\frac{y}{y - 2}$ .

$\frac{3 y}{y - 2} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Cancel the common factor of $y - 2$ .

Cancel the common factor.

$\frac{3 y}{y - 2} ((y - 2) (y + 2)) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Rewrite the expression.

$3 y (y + 2) - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Apply the distributive property.

$3 y \cdot y + 3 y \cdot 2 - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$3 (y \cdot y) + 3 y \cdot 2 - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Multiply $y$ by $y$ .

$3 y^{2} + 3 y \cdot 2 - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Multiply $2$ by $3$ .

$3 y^{2} + 6 y - \frac{5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Cancel the common factor of $3 (y + 2)$ .

Move the leading negative in $- \frac{5}{3 (y + 2)}$ into the numerator.

$3 y^{2} + 6 y + \frac{- 5}{3 (y + 2)} (15 (y - 2) (y + 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Factor $3 (y + 2)$ out of $15 (y - 2) (y + 2)$ .

$3 y^{2} + 6 y + \frac{- 5}{3 (y + 2)} (3 (y + 2) (5 (y - 2))) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Cancel the common factor.

$3 y^{2} + 6 y + \frac{- 5}{3 (y + 2)} (3 (y + 2) (5 (y - 2))) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Rewrite the expression.

$3 y^{2} + 6 y - 5 (5 (y - 2)) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Multiply $5$ by $- 5$ .

$3 y^{2} + 6 y - 25 (y - 2) = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Apply the distributive property.

$3 y^{2} + 6 y - 25 y - 25 \cdot - 2 = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Multiply $- 25$ by $- 2$ .

$3 y^{2} + 6 y - 25 y + 50 = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Subtract $25 y$ from $6 y$ .

$3 y^{2} - 19 y + 50 = \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} (15 (y - 2) (y + 2))$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$3 y^{2} - 19 y + 50 = 15 \frac{2 y^{2} - 19 y + 54}{15 (y + 2) (y - 2)} ((y - 2) (y + 2))$

Cancel the common factor of $15$ .

Factor $15$ out of $15 (y + 2) (y - 2)$ .

$3 y^{2} - 19 y + 50 = 15 \frac{2 y^{2} - 19 y + 54}{15 ((y + 2) (y - 2))} ((y - 2) (y + 2))$

Cancel the common factor.

$3 y^{2} - 19 y + 50 = 15 \frac{2 y^{2} - 19 y + 54}{15 ((y + 2) (y - 2))} ((y - 2) (y + 2))$

Rewrite the expression.

$3 y^{2} - 19 y + 50 = \frac{2 y^{2} - 19 y + 54}{(y + 2) (y - 2)} ((y - 2) (y + 2))$

Cancel the common factor of $(y - 2) (y + 2)$ .

Factor $(y - 2) (y + 2)$ out of $(y + 2) (y - 2)$ .

$3 y^{2} - 19 y + 50 = \frac{2 y^{2} - 19 y + 54}{(y - 2) (y + 2) (1)} ((y - 2) (y + 2))$

Cancel the common factor.

$3 y^{2} - 19 y + 50 = \frac{2 y^{2} - 19 y + 54}{(y - 2) (y + 2) \cdot 1} ((y - 2) (y + 2))$

Rewrite the expression.

$3 y^{2} - 19 y + 50 = 2 y^{2} - 19 y + 54$

Solve the equation.

Move all terms containing $y$ to the left side of the equation.

Subtract $2 y^{2}$ from both sides of the equation.

$3 y^{2} - 19 y + 50 - 2 y^{2} = - 19 y + 54$

Add $19 y$ to both sides of the equation.

$3 y^{2} - 19 y + 50 - 2 y^{2} + 19 y = 54$

Combine the opposite terms in $3 y^{2} - 19 y + 50 - 2 y^{2} + 19 y$ .

Add $- 19 y$ and $19 y$ .

$3 y^{2} + 50 - 2 y^{2} + 0 = 54$

Add $3 y^{2} + 50 - 2 y^{2}$ and $0$ .

$3 y^{2} + 50 - 2 y^{2} = 54$

Subtract $2 y^{2}$ from $3 y^{2}$ .

$y^{2} + 50 = 54$

Move all terms not containing $y$ to the right side of the equation.

Subtract $50$ from both sides of the equation.

$y^{2} = 54 - 50$

Subtract $50$ from $54$ .

$y^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$y = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 2, - 2$

Exclude the solutions that do not make $\frac{y}{5 y - 10} - \frac{5}{3 y + 6} = \frac{2 y^{2} - 19 y + 54}{15 y^{2} - 60}$ true.

$No solution$

Solve for y y/(5y-10)-5/(3y+6)=(2y^2-19y+54)/(15y^2-60)

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