Solve by Factoring -12/y=y-7

$\frac{- 12}{y} = y - 7$

Step 1

Move all the expressions to the left side of the equation.

Subtract $y$ from both sides of the equation.

$\frac{- 12}{y} - y = - 7$

Add $7$ to both sides of the equation.

$\frac{- 12}{y} - y + 7 = 0$

Step 2

Move the negative in front of the fraction.

$- \frac{12}{y} - y + 7 = 0$

Step 3

Reorder terms.

$- y - \frac{12}{y} + 7 = 0$

Step 4

To write $- y$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$- y \cdot \frac{y}{y} - \frac{12}{y} + 7 = 0$

Step 5

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

$\frac{- y \cdot y}{y} - \frac{12}{y} + 7 = 0$

Step 6

Combine the numerators over the common denominator.

$\frac{- y \cdot y - 12}{y} + 7 = 0$

Step 7

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

Move $y$ .

$\frac{- (y \cdot y) - 12}{y} + 7 = 0$

Multiply $y$ by $y$ .

$\frac{- y^{2} - 12}{y} + 7 = 0$

Step 8

To write $7$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{- y^{2} - 12}{y} + \frac{7 y}{y} = 0$

Step 9

Combine the numerators over the common denominator.

$\frac{- y^{2} - 12 + 7 y}{y} = 0$

Step 10

Factor by grouping.

Reorder terms.

$\frac{- y^{2} + 7 y - 12}{y} = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 12 = 12$ and whose sum is $b = 7$ .

Factor $7$ out of $7 y$ .

$\frac{- y^{2} + 7 (y) - 12}{y} = 0$

Rewrite $7$ as $3$ plus $4$

$\frac{- y^{2} + (3 + 4) y - 12}{y} = 0$

Apply the distributive property.

$\frac{- y^{2} + 3 y + 4 y - 12}{y} = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- y^{2} + 3 y) + 4 y - 12}{y} = 0$

Factor out the greatest common factor (GCF) from each group.

$\frac{y (- y + 3) - 4 (- y + 3)}{y} = 0$

Factor the polynomial by factoring out the greatest common factor, $- y + 3$ .

$\frac{(- y + 3) (y - 4)}{y} = 0$

Step 11

Set the numerator equal to zero.

$(- y + 3) (y - 4) = 0$

Step 12

Solve the equation for $y$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$- y + 3 = 0$

$y - 4 = 0$

Set $- y + 3$ equal to $0$ and solve for $y$ .

Set $- y + 3$ equal to $0$ .

$- y + 3 = 0$

Solve $- y + 3 = 0$ for $y$ .

Subtract $3$ from both sides of the equation.

$- y = - 3$

Divide each term in $- y = - 3$ by $- 1$ and simplify.

Divide each term in $- y = - 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{- 3}{- 1}$

Divide $y$ by $1$ .

$y = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$y = 3$

Set $y - 4$ equal to $0$ and solve for $y$ .

Set $y - 4$ equal to $0$ .

$y - 4 = 0$

Add $4$ to both sides of the equation.

$y = 4$

The final solution is all the values that make $(- y + 3) (y - 4) = 0$ true.

$y = 3, 4$

Solve by Factoring -12/y=y-7

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