Solve 1/12+1/y=3/y+6

$$\frac { 1 } { 12 } + \frac { 1 } { y } = \frac { 3 } { y + 6 }$$

Solve for y

$$y=6$$

$$y=12$$

Steps Using Factoring

Variable y cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by 12y\left(y+6\right), the least common multiple of 12,y,y+6.

$$12y\left(y+6\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y by y+6.

$$\left(12y^{2}+72y\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y^{2}+72y by \frac{1}{12}.

$$y^{2}+6y+12y+72=12y\times 3$$

Combine 6y and 12y to get 18y.

$$y^{2}+18y+72=12y\times 3$$

Multiply 12 and 3 to get 36.

$$y^{2}+18y+72=36y$$

Subtract 36y from both sides.

$$y^{2}+18y+72-36y=0$$

Combine 18y and -36y to get -18y.

$$y^{2}-18y+72=0$$

To solve the equation, factor y^{2}-18y+72 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.

$$a+b=-18 ab=72$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 72.

$$-1,-72 -2,-36 -3,-24 -4,-18 -6,-12 -8,-9$$

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -18.

$$a=-12 b=-6$$

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

$$\left(y-12\right)\left(y-6\right)$$

To find equation solutions, solve y-12=0 and y-6=0.

$$y=12 y=6$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by 12y\left(y+6\right), the least common multiple of 12,y,y+6.

$$12y\left(y+6\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y by y+6.

$$\left(12y^{2}+72y\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y^{2}+72y by \frac{1}{12}.

$$y^{2}+6y+12y+72=12y\times 3$$

Combine 6y and 12y to get 18y.

$$y^{2}+18y+72=12y\times 3$$

Multiply 12 and 3 to get 36.

$$y^{2}+18y+72=36y$$

Subtract 36y from both sides.

$$y^{2}+18y+72-36y=0$$

Combine 18y and -36y to get -18y.

$$y^{2}-18y+72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+72. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 72=72$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 72.

$$-1,-72 -2,-36 -3,-24 -4,-18 -6,-12 -8,-9$$

Calculate the sum for each pair.

$$-1-72=-73 -2-36=-38 -3-24=-27 -4-18=-22 -6-12=-18 -8-9=-17$$

The solution is the pair that gives sum -18.

$$a=-12 b=-6$$

Rewrite y^{2}-18y+72 as \left(y^{2}-12y\right)+\left(-6y+72\right).

$$\left(y^{2}-12y\right)+\left(-6y+72\right)$$

Factor out y in the first and -6 in the second group.

$$y\left(y-12\right)-6\left(y-12\right)$$

Factor out common term y-12 by using distributive property.

$$\left(y-12\right)\left(y-6\right)$$

To find equation solutions, solve y-12=0 and y-6=0.

$$y=12 y=6$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by 12y\left(y+6\right), the least common multiple of 12,y,y+6.

$$12y\left(y+6\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y by y+6.

$$\left(12y^{2}+72y\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y^{2}+72y by \frac{1}{12}.

$$y^{2}+6y+12y+72=12y\times 3$$

Combine 6y and 12y to get 18y.

$$y^{2}+18y+72=12y\times 3$$

Multiply 12 and 3 to get 36.

$$y^{2}+18y+72=36y$$

Subtract 36y from both sides.

$$y^{2}+18y+72-36y=0$$

Combine 18y and -36y to get -18y.

$$y^{2}-18y+72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -18 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 72}}{2}$$

Square -18.

$$y=\frac{-\left(-18\right)±\sqrt{324-4\times 72}}{2}$$

Multiply -4 times 72.

$$y=\frac{-\left(-18\right)±\sqrt{324-288}}{2}$$

Add 324 to -288.

$$y=\frac{-\left(-18\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$y=\frac{-\left(-18\right)±6}{2}$$

The opposite of -18 is 18.

$$y=\frac{18±6}{2}$$

Now solve the equation y=\frac{18±6}{2} when ± is plus. Add 18 to 6.

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

Divide 24 by 2.

$$y=12$$

Now solve the equation y=\frac{18±6}{2} when ± is minus. Subtract 6 from 18.

$$y=\frac{12}{2}$$

Divide 12 by 2.

$$y=6$$

The equation is now solved.

$$y=12 y=6$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by 12y\left(y+6\right), the least common multiple of 12,y,y+6.

$$12y\left(y+6\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y by y+6.

$$\left(12y^{2}+72y\right)\times \frac{1}{12}+12y+72=12y\times 3$$

Use the distributive property to multiply 12y^{2}+72y by \frac{1}{12}.

$$y^{2}+6y+12y+72=12y\times 3$$

Combine 6y and 12y to get 18y.

$$y^{2}+18y+72=12y\times 3$$

Multiply 12 and 3 to get 36.

$$y^{2}+18y+72=36y$$

Subtract 36y from both sides.

$$y^{2}+18y+72-36y=0$$

Combine 18y and -36y to get -18y.

$$y^{2}-18y+72=0$$

Subtract 72 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-18y=-72$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-18y+\left(-9\right)^{2}=-72+\left(-9\right)^{2}$$

Square -9.

$$y^{2}-18y+81=-72+81$$

Add -72 to 81.

$$y^{2}-18y+81=9$$

Factor y^{2}-18y+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-9\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-9\right)^{2}}=\sqrt{9}$$

Simplify.

$$y-9=3 y-9=-3$$

Add 9 to both sides of the equation.

$$y=12 y=6$$