Solve 1/18+1/y=2/y+3

$$\frac { 1 } { 18 } + \frac { 1 } { y } = \frac { 2 } { y + 3 }$$

Solve for y

$$y=6$$

$$y=9$$

Steps Using Factoring

Variable y cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 18y\left(y+3\right), the least common multiple of 18,y,y+3.

$$18y\left(y+3\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y by y+3.

$$\left(18y^{2}+54y\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y^{2}+54y by \frac{1}{18}.

$$y^{2}+3y+18y+54=18y\times 2$$

Combine 3y and 18y to get 21y.

$$y^{2}+21y+54=18y\times 2$$

Multiply 18 and 2 to get 36.

$$y^{2}+21y+54=36y$$

Subtract 36y from both sides.

$$y^{2}+21y+54-36y=0$$

Combine 21y and -36y to get -15y.

$$y^{2}-15y+54=0$$

To solve the equation, factor y^{2}-15y+54 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.

$$a+b=-15 ab=54$$

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 54.

$$-1,-54 -2,-27 -3,-18 -6,-9$$

Calculate the sum for each pair.

$$-1-54=-55 -2-27=-29 -3-18=-21 -6-9=-15$$

The solution is the pair that gives sum -15.

$$a=-9 b=-6$$

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

$$\left(y-9\right)\left(y-6\right)$$

To find equation solutions, solve y-9=0 and y-6=0.

$$y=9 y=6$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 18y\left(y+3\right), the least common multiple of 18,y,y+3.

$$18y\left(y+3\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y by y+3.

$$\left(18y^{2}+54y\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y^{2}+54y by \frac{1}{18}.

$$y^{2}+3y+18y+54=18y\times 2$$

Combine 3y and 18y to get 21y.

$$y^{2}+21y+54=18y\times 2$$

Multiply 18 and 2 to get 36.

$$y^{2}+21y+54=36y$$

Subtract 36y from both sides.

$$y^{2}+21y+54-36y=0$$

Combine 21y and -36y to get -15y.

$$y^{2}-15y+54=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+54. To find a and b, set up a system to be solved.

$$a+b=-15 ab=1\times 54=54$$

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 54.

$$-1,-54 -2,-27 -3,-18 -6,-9$$

Calculate the sum for each pair.

$$-1-54=-55 -2-27=-29 -3-18=-21 -6-9=-15$$

The solution is the pair that gives sum -15.

$$a=-9 b=-6$$

Rewrite y^{2}-15y+54 as \left(y^{2}-9y\right)+\left(-6y+54\right).

$$\left(y^{2}-9y\right)+\left(-6y+54\right)$$

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

$$y\left(y-9\right)-6\left(y-9\right)$$

Factor out common term y-9 by using distributive property.

$$\left(y-9\right)\left(y-6\right)$$

To find equation solutions, solve y-9=0 and y-6=0.

$$y=9 y=6$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 18y\left(y+3\right), the least common multiple of 18,y,y+3.

$$18y\left(y+3\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y by y+3.

$$\left(18y^{2}+54y\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y^{2}+54y by \frac{1}{18}.

$$y^{2}+3y+18y+54=18y\times 2$$

Combine 3y and 18y to get 21y.

$$y^{2}+21y+54=18y\times 2$$

Multiply 18 and 2 to get 36.

$$y^{2}+21y+54=36y$$

Subtract 36y from both sides.

$$y^{2}+21y+54-36y=0$$

Combine 21y and -36y to get -15y.

$$y^{2}-15y+54=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and 54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 54}}{2}$$

Square -15.

$$y=\frac{-\left(-15\right)±\sqrt{225-4\times 54}}{2}$$

Multiply -4 times 54.

$$y=\frac{-\left(-15\right)±\sqrt{225-216}}{2}$$

Add 225 to -216.

$$y=\frac{-\left(-15\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$y=\frac{-\left(-15\right)±3}{2}$$

The opposite of -15 is 15.

$$y=\frac{15±3}{2}$$

Now solve the equation y=\frac{15±3}{2} when ± is plus. Add 15 to 3.

$$y=\frac{18}{2}$$

Divide 18 by 2.

$$y=9$$

Now solve the equation y=\frac{15±3}{2} when ± is minus. Subtract 3 from 15.

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

Divide 12 by 2.

$$y=6$$

The equation is now solved.

$$y=9 y=6$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 18y\left(y+3\right), the least common multiple of 18,y,y+3.

$$18y\left(y+3\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y by y+3.

$$\left(18y^{2}+54y\right)\times \frac{1}{18}+18y+54=18y\times 2$$

Use the distributive property to multiply 18y^{2}+54y by \frac{1}{18}.

$$y^{2}+3y+18y+54=18y\times 2$$

Combine 3y and 18y to get 21y.

$$y^{2}+21y+54=18y\times 2$$

Multiply 18 and 2 to get 36.

$$y^{2}+21y+54=36y$$

Subtract 36y from both sides.

$$y^{2}+21y+54-36y=0$$

Combine 21y and -36y to get -15y.

$$y^{2}-15y+54=0$$

Subtract 54 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-15y=-54$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-15y+\left(-\frac{15}{2}\right)^{2}=-54+\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-15y+\frac{225}{4}=-54+\frac{225}{4}$$

Add -54 to \frac{225}{4}.

$$y^{2}-15y+\frac{225}{4}=\frac{9}{4}$$

Factor y^{2}-15y+\frac{225}{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}.

$$\left(y-\frac{15}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{15}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y-\frac{15}{2}=\frac{3}{2} y-\frac{15}{2}=-\frac{3}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$y=9 y=6$$