Solve y/y+3+9/y+5=y-13/y^2+8y+15

$$\frac { y } { y + 3 } + \frac { 9 } { y + 5 } = \frac { y - 13 } { y ^ { 2 } + 8 y + 15 }$$

Solve for y

$$y=-8$$

Steps Using Factoring

Variable y cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+5\right), the least common multiple of y+3,y+5,y^{2}+8y+15.

$$\left(y+5\right)y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+5 by y.

$$y^{2}+5y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+3 by 9.

$$y^{2}+5y+9y+27=y-13$$

Combine 5y and 9y to get 14y.

$$y^{2}+14y+27=y-13$$

Subtract y from both sides.

$$y^{2}+14y+27-y=-13$$

Combine 14y and -y to get 13y.

$$y^{2}+13y+27=-13$$

Add 13 to both sides.

$$y^{2}+13y+27+13=0$$

Add 27 and 13 to get 40.

$$y^{2}+13y+40=0$$

To solve the equation, factor y^{2}+13y+40 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=13 ab=40$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 13.

$$a=5 b=8$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y+5\right)\left(y+8\right)$$

To find equation solutions, solve y+5=0 and y+8=0.

$$y=-5 y=-8$$

Variable y cannot be equal to -5.

$$y=-8$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+5\right), the least common multiple of y+3,y+5,y^{2}+8y+15.

$$\left(y+5\right)y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+5 by y.

$$y^{2}+5y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+3 by 9.

$$y^{2}+5y+9y+27=y-13$$

Combine 5y and 9y to get 14y.

$$y^{2}+14y+27=y-13$$

Subtract y from both sides.

$$y^{2}+14y+27-y=-13$$

Combine 14y and -y to get 13y.

$$y^{2}+13y+27=-13$$

Add 13 to both sides.

$$y^{2}+13y+27+13=0$$

Add 27 and 13 to get 40.

$$y^{2}+13y+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+40. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\times 40=40$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 13.

$$a=5 b=8$$

Rewrite y^{2}+13y+40 as \left(y^{2}+5y\right)+\left(8y+40\right).

$$\left(y^{2}+5y\right)+\left(8y+40\right)$$

Factor out y in the first and 8 in the second group.

$$y\left(y+5\right)+8\left(y+5\right)$$

Factor out common term y+5 by using distributive property.

$$\left(y+5\right)\left(y+8\right)$$

To find equation solutions, solve y+5=0 and y+8=0.

$$y=-5 y=-8$$

Variable y cannot be equal to -5.

$$y=-8$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+5\right), the least common multiple of y+3,y+5,y^{2}+8y+15.

$$\left(y+5\right)y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+5 by y.

$$y^{2}+5y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+3 by 9.

$$y^{2}+5y+9y+27=y-13$$

Combine 5y and 9y to get 14y.

$$y^{2}+14y+27=y-13$$

Subtract y from both sides.

$$y^{2}+14y+27-y=-13$$

Combine 14y and -y to get 13y.

$$y^{2}+13y+27=-13$$

Add 13 to both sides.

$$y^{2}+13y+27+13=0$$

Add 27 and 13 to get 40.

$$y^{2}+13y+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 13 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-13±\sqrt{13^{2}-4\times 40}}{2}$$

Square 13.

$$y=\frac{-13±\sqrt{169-4\times 40}}{2}$$

Multiply -4 times 40.

$$y=\frac{-13±\sqrt{169-160}}{2}$$

Add 169 to -160.

$$y=\frac{-13±\sqrt{9}}{2}$$

Take the square root of 9.

$$y=\frac{-13±3}{2}$$

Now solve the equation y=\frac{-13±3}{2} when ± is plus. Add -13 to 3.

$$y=-\frac{10}{2}$$

Divide -10 by 2.

$$y=-5$$

Now solve the equation y=\frac{-13±3}{2} when ± is minus. Subtract 3 from -13.

$$y=-\frac{16}{2}$$

Divide -16 by 2.

$$y=-8$$

The equation is now solved.

$$y=-5 y=-8$$

Variable y cannot be equal to -5.

$$y=-8$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -5,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+5\right), the least common multiple of y+3,y+5,y^{2}+8y+15.

$$\left(y+5\right)y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+5 by y.

$$y^{2}+5y+\left(y+3\right)\times 9=y-13$$

Use the distributive property to multiply y+3 by 9.

$$y^{2}+5y+9y+27=y-13$$

Combine 5y and 9y to get 14y.

$$y^{2}+14y+27=y-13$$

Subtract y from both sides.

$$y^{2}+14y+27-y=-13$$

Combine 14y and -y to get 13y.

$$y^{2}+13y+27=-13$$

Subtract 27 from both sides.

$$y^{2}+13y=-13-27$$

Subtract 27 from -13 to get -40.

$$y^{2}+13y=-40$$

Divide 13, the coefficient of the x term, by 2 to get \frac{13}{2}. Then add the square of \frac{13}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+13y+\left(\frac{13}{2}\right)^{2}=-40+\left(\frac{13}{2}\right)^{2}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+13y+\frac{169}{4}=-40+\frac{169}{4}$$

Add -40 to \frac{169}{4}.

$$y^{2}+13y+\frac{169}{4}=\frac{9}{4}$$

Factor y^{2}+13y+\frac{169}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{13}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{13}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y+\frac{13}{2}=\frac{3}{2} y+\frac{13}{2}=-\frac{3}{2}$$

Subtract \frac{13}{2} from both sides of the equation.

$$y=-5 y=-8$$

Variable y cannot be equal to -5.

$$y=-8$$