Solve y/y+3+9/y+4=y-8/y^2+7y+12

$$\frac { y } { y + 3 } + \frac { 9 } { y + 4 } = \frac { y - 8 } { y ^ { 2 } + 7 y + 12 }$$

Solve for y

$$y=-7$$

$$y=-5$$

Steps Using Factoring

Variable y cannot be equal to any of the values -4,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+4\right), the least common multiple of y+3,y+4,y^{2}+7y+12.

$$\left(y+4\right)y+\left(y+3\right)\times 9=y-8$$

Use the distributive property to multiply y+4 by y.

$$y^{2}+4y+\left(y+3\right)\times 9=y-8$$

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

$$y^{2}+4y+9y+27=y-8$$

Combine 4y and 9y to get 13y.

$$y^{2}+13y+27=y-8$$

Subtract y from both sides.

$$y^{2}+13y+27-y=-8$$

Combine 13y and -y to get 12y.

$$y^{2}+12y+27=-8$$

Add 8 to both sides.

$$y^{2}+12y+27+8=0$$

Add 27 and 8 to get 35.

$$y^{2}+12y+35=0$$

To solve the equation, factor y^{2}+12y+35 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.

$$a+b=12 ab=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

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

$$\left(y+5\right)\left(y+7\right)$$

To find equation solutions, solve y+5=0 and y+7=0.

$$y=-5 y=-7$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values -4,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+4\right), the least common multiple of y+3,y+4,y^{2}+7y+12.

$$\left(y+4\right)y+\left(y+3\right)\times 9=y-8$$

Use the distributive property to multiply y+4 by y.

$$y^{2}+4y+\left(y+3\right)\times 9=y-8$$

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

$$y^{2}+4y+9y+27=y-8$$

Combine 4y and 9y to get 13y.

$$y^{2}+13y+27=y-8$$

Subtract y from both sides.

$$y^{2}+13y+27-y=-8$$

Combine 13y and -y to get 12y.

$$y^{2}+12y+27=-8$$

Add 8 to both sides.

$$y^{2}+12y+27+8=0$$

Add 27 and 8 to get 35.

$$y^{2}+12y+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+35. To find a and b, set up a system to be solved.

$$a+b=12 ab=1\times 35=35$$

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 35.

$$1,35 5,7$$

Calculate the sum for each pair.

$$1+35=36 5+7=12$$

The solution is the pair that gives sum 12.

$$a=5 b=7$$

Rewrite y^{2}+12y+35 as \left(y^{2}+5y\right)+\left(7y+35\right).

$$\left(y^{2}+5y\right)+\left(7y+35\right)$$

Factor out y in the first and 7 in the second group.

$$y\left(y+5\right)+7\left(y+5\right)$$

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

$$\left(y+5\right)\left(y+7\right)$$

To find equation solutions, solve y+5=0 and y+7=0.

$$y=-5 y=-7$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -4,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+4\right), the least common multiple of y+3,y+4,y^{2}+7y+12.

$$\left(y+4\right)y+\left(y+3\right)\times 9=y-8$$

Use the distributive property to multiply y+4 by y.

$$y^{2}+4y+\left(y+3\right)\times 9=y-8$$

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

$$y^{2}+4y+9y+27=y-8$$

Combine 4y and 9y to get 13y.

$$y^{2}+13y+27=y-8$$

Subtract y from both sides.

$$y^{2}+13y+27-y=-8$$

Combine 13y and -y to get 12y.

$$y^{2}+12y+27=-8$$

Add 8 to both sides.

$$y^{2}+12y+27+8=0$$

Add 27 and 8 to get 35.

$$y^{2}+12y+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\times 35}}{2}$$

Square 12.

$$y=\frac{-12±\sqrt{144-4\times 35}}{2}$$

Multiply -4 times 35.

$$y=\frac{-12±\sqrt{144-140}}{2}$$

Add 144 to -140.

$$y=\frac{-12±\sqrt{4}}{2}$$

Take the square root of 4.

$$y=\frac{-12±2}{2}$$

Now solve the equation y=\frac{-12±2}{2} when ± is plus. Add -12 to 2.

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

Divide -10 by 2.

$$y=-5$$

Now solve the equation y=\frac{-12±2}{2} when ± is minus. Subtract 2 from -12.

$$y=-\frac{14}{2}$$

Divide -14 by 2.

$$y=-7$$

The equation is now solved.

$$y=-5 y=-7$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -4,-3 since division by zero is not defined. Multiply both sides of the equation by \left(y+3\right)\left(y+4\right), the least common multiple of y+3,y+4,y^{2}+7y+12.

$$\left(y+4\right)y+\left(y+3\right)\times 9=y-8$$

Use the distributive property to multiply y+4 by y.

$$y^{2}+4y+\left(y+3\right)\times 9=y-8$$

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

$$y^{2}+4y+9y+27=y-8$$

Combine 4y and 9y to get 13y.

$$y^{2}+13y+27=y-8$$

Subtract y from both sides.

$$y^{2}+13y+27-y=-8$$

Combine 13y and -y to get 12y.

$$y^{2}+12y+27=-8$$

Subtract 27 from both sides.

$$y^{2}+12y=-8-27$$

Subtract 27 from -8 to get -35.

$$y^{2}+12y=-35$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+12y+6^{2}=-35+6^{2}$$

Square 6.

$$y^{2}+12y+36=-35+36$$

Add -35 to 36.

$$y^{2}+12y+36=1$$

Factor y^{2}+12y+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+6\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+6\right)^{2}}=\sqrt{1}$$

Simplify.

$$y+6=1 y+6=-1$$

Subtract 6 from both sides of the equation.

$$y=-5 y=-7$$