Solve 1/y-3+1/9=6/y^2-9

$$\frac { 1 } { y - 3 } + \frac { 1 } { 9 } = \frac { 6 } { y ^ { 2 } - 9 }$$

Solve for y

$$y=-12$$

Steps Using Factoring

Variable y cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 9\left(y-3\right)\left(y+3\right), the least common multiple of y-3,9,y^{2}-9.

$$9y+27+9\left(y-3\right)\left(y+3\right)\times \frac{1}{9}=9\times 6$$

Multiply 9 and \frac{1}{9} to get 1.

$$9y+27+\left(y-3\right)\left(y+3\right)=9\times 6$$

Consider \left(y-3\right)\left(y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9y+27+y^{2}-9=9\times 6$$

Subtract 9 from 27 to get 18.

$$9y+18+y^{2}=9\times 6$$

Multiply 9 and 6 to get 54.

$$9y+18+y^{2}=54$$

Subtract 54 from both sides.

$$9y+18+y^{2}-54=0$$

Subtract 54 from 18 to get -36.

$$9y-36+y^{2}=0$$

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

$$y^{2}+9y-36=0$$

To solve the equation, factor y^{2}+9y-36 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.

$$a+b=9 ab=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

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

$$\left(y-3\right)\left(y+12\right)$$

To find equation solutions, solve y-3=0 and y+12=0.

$$y=3 y=-12$$

Variable y cannot be equal to 3.

$$y=-12$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 9\left(y-3\right)\left(y+3\right), the least common multiple of y-3,9,y^{2}-9.

$$9y+27+9\left(y-3\right)\left(y+3\right)\times \frac{1}{9}=9\times 6$$

Multiply 9 and \frac{1}{9} to get 1.

$$9y+27+\left(y-3\right)\left(y+3\right)=9\times 6$$

Consider \left(y-3\right)\left(y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9y+27+y^{2}-9=9\times 6$$

Subtract 9 from 27 to get 18.

$$9y+18+y^{2}=9\times 6$$

Multiply 9 and 6 to get 54.

$$9y+18+y^{2}=54$$

Subtract 54 from both sides.

$$9y+18+y^{2}-54=0$$

Subtract 54 from 18 to get -36.

$$9y-36+y^{2}=0$$

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

$$y^{2}+9y-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-36. To find a and b, set up a system to be solved.

$$a+b=9 ab=1\left(-36\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 9.

$$a=-3 b=12$$

Rewrite y^{2}+9y-36 as \left(y^{2}-3y\right)+\left(12y-36\right).

$$\left(y^{2}-3y\right)+\left(12y-36\right)$$

Factor out y in the first and 12 in the second group.

$$y\left(y-3\right)+12\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(y+12\right)$$

To find equation solutions, solve y-3=0 and y+12=0.

$$y=3 y=-12$$

Variable y cannot be equal to 3.

$$y=-12$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 9\left(y-3\right)\left(y+3\right), the least common multiple of y-3,9,y^{2}-9.

$$9y+27+9\left(y-3\right)\left(y+3\right)\times \frac{1}{9}=9\times 6$$

Multiply 9 and \frac{1}{9} to get 1.

$$9y+27+\left(y-3\right)\left(y+3\right)=9\times 6$$

Consider \left(y-3\right)\left(y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9y+27+y^{2}-9=9\times 6$$

Subtract 9 from 27 to get 18.

$$9y+18+y^{2}=9\times 6$$

Multiply 9 and 6 to get 54.

$$9y+18+y^{2}=54$$

Subtract 54 from both sides.

$$9y+18+y^{2}-54=0$$

Subtract 54 from 18 to get -36.

$$9y-36+y^{2}=0$$

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.

$$y^{2}+9y-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-9±\sqrt{9^{2}-4\left(-36\right)}}{2}$$

Square 9.

$$y=\frac{-9±\sqrt{81-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$y=\frac{-9±\sqrt{81+144}}{2}$$

Add 81 to 144.

$$y=\frac{-9±\sqrt{225}}{2}$$

Take the square root of 225.

$$y=\frac{-9±15}{2}$$

Now solve the equation y=\frac{-9±15}{2} when ± is plus. Add -9 to 15.

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

Divide 6 by 2.

$$y=3$$

Now solve the equation y=\frac{-9±15}{2} when ± is minus. Subtract 15 from -9.

$$y=-\frac{24}{2}$$

Divide -24 by 2.

$$y=-12$$

The equation is now solved.

$$y=3 y=-12$$

Variable y cannot be equal to 3.

$$y=-12$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 9\left(y-3\right)\left(y+3\right), the least common multiple of y-3,9,y^{2}-9.

$$9y+27+9\left(y-3\right)\left(y+3\right)\times \frac{1}{9}=9\times 6$$

Multiply 9 and \frac{1}{9} to get 1.

$$9y+27+\left(y-3\right)\left(y+3\right)=9\times 6$$

Consider \left(y-3\right)\left(y+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9y+27+y^{2}-9=9\times 6$$

Subtract 9 from 27 to get 18.

$$9y+18+y^{2}=9\times 6$$

Multiply 9 and 6 to get 54.

$$9y+18+y^{2}=54$$

Subtract 18 from both sides.

$$9y+y^{2}=54-18$$

Subtract 18 from 54 to get 36.

$$9y+y^{2}=36$$

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.

$$y^{2}+9y=36$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+9y+\left(\frac{9}{2}\right)^{2}=36+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+9y+\frac{81}{4}=36+\frac{81}{4}$$

Add 36 to \frac{81}{4}.

$$y^{2}+9y+\frac{81}{4}=\frac{225}{4}$$

Factor y^{2}+9y+\frac{81}{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}.

$$\left(y+\frac{9}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{9}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$y+\frac{9}{2}=\frac{15}{2} y+\frac{9}{2}=-\frac{15}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$y=3 y=-12$$

Variable y cannot be equal to 3.

$$y=-12$$