Solve 9/y-8+y/y-2=-12/y^2-10y+16

$$\frac { 9 } { y - 8 } + \frac { y } { y - 2 } = \frac { - 12 } { y ^ { 2 } - 10 y + 16 }$$

Solve for y

$$y=-3$$

Steps Using Factoring

Variable y cannot be equal to any of the values 2,8 since division by zero is not defined. Multiply both sides of the equation by \left(y-8\right)\left(y-2\right), the least common multiple of y-8,y-2,y^{2}-10y+16.

$$\left(y-2\right)\times 9+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-2 by 9.

$$9y-18+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-8 by y.

$$9y-18+y^{2}-8y=-12$$

Combine 9y and -8y to get y.

$$y-18+y^{2}=-12$$

Add 12 to both sides.

$$y-18+y^{2}+12=0$$

Add -18 and 12 to get -6.

$$y-6+y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}+y-6=0$$

To solve the equation, factor y^{2}+y-6 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.

$$a+b=1 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

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

$$\left(y-2\right)\left(y+3\right)$$

To find equation solutions, solve y-2=0 and y+3=0.

$$y=2 y=-3$$

Variable y cannot be equal to 2.

$$y=-3$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values 2,8 since division by zero is not defined. Multiply both sides of the equation by \left(y-8\right)\left(y-2\right), the least common multiple of y-8,y-2,y^{2}-10y+16.

$$\left(y-2\right)\times 9+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-2 by 9.

$$9y-18+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-8 by y.

$$9y-18+y^{2}-8y=-12$$

Combine 9y and -8y to get y.

$$y-18+y^{2}=-12$$

Add 12 to both sides.

$$y-18+y^{2}+12=0$$

Add -18 and 12 to get -6.

$$y-6+y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}+y-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 1.

$$a=-2 b=3$$

Rewrite y^{2}+y-6 as \left(y^{2}-2y\right)+\left(3y-6\right).

$$\left(y^{2}-2y\right)+\left(3y-6\right)$$

Factor out y in the first and 3 in the second group.

$$y\left(y-2\right)+3\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(y+3\right)$$

To find equation solutions, solve y-2=0 and y+3=0.

$$y=2 y=-3$$

Variable y cannot be equal to 2.

$$y=-3$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values 2,8 since division by zero is not defined. Multiply both sides of the equation by \left(y-8\right)\left(y-2\right), the least common multiple of y-8,y-2,y^{2}-10y+16.

$$\left(y-2\right)\times 9+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-2 by 9.

$$9y-18+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-8 by y.

$$9y-18+y^{2}-8y=-12$$

Combine 9y and -8y to get y.

$$y-18+y^{2}=-12$$

Add 12 to both sides.

$$y-18+y^{2}+12=0$$

Add -18 and 12 to get -6.

$$y-6+y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y^{2}+y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$y=\frac{-1±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$y=\frac{-1±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-1±5}{2}$$

Now solve the equation y=\frac{-1±5}{2} when ± is plus. Add -1 to 5.

$$y=\frac{4}{2}$$

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-1±5}{2} when ± is minus. Subtract 5 from -1.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

The equation is now solved.

$$y=2 y=-3$$

Variable y cannot be equal to 2.

$$y=-3$$

Steps for Completing the Square

Variable y cannot be equal to any of the values 2,8 since division by zero is not defined. Multiply both sides of the equation by \left(y-8\right)\left(y-2\right), the least common multiple of y-8,y-2,y^{2}-10y+16.

$$\left(y-2\right)\times 9+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-2 by 9.

$$9y-18+\left(y-8\right)y=-12$$

Use the distributive property to multiply y-8 by y.

$$9y-18+y^{2}-8y=-12$$

Combine 9y and -8y to get y.

$$y-18+y^{2}=-12$$

Add 18 to both sides.

$$y+y^{2}=-12+18$$

Add -12 and 18 to get 6.

$$y+y^{2}=6$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$y^{2}+y=6$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+y+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$y^{2}+y+\frac{1}{4}=\frac{25}{4}$$

Factor y^{2}+y+\frac{1}{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}.

$$\left(y+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$y+\frac{1}{2}=\frac{5}{2} y+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$y=2 y=-3$$

Variable y cannot be equal to 2.

$$y=-3$$