Solve 1/8-y-1/y=2/15

$$\frac { 1 } { 8 - y } - \frac { 1 } { y } = \frac { 2 } { 15 }$$

Solve for y

$$y=-12$$

$$y=5$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 15y\left(y-8\right), the least common multiple of 8-y,y,15.

$$-15y-\left(15y-120\right)=2y\left(y-8\right)$$

To find the opposite of 15y-120, find the opposite of each term.

$$-15y-15y+120=2y\left(y-8\right)$$

Combine -15y and -15y to get -30y.

$$-30y+120=2y\left(y-8\right)$$

Use the distributive property to multiply 2y by y-8.

$$-30y+120=2y^{2}-16y$$

Subtract 2y^{2} from both sides.

$$-30y+120-2y^{2}=-16y$$

Add 16y to both sides.

$$-30y+120-2y^{2}+16y=0$$

Combine -30y and 16y to get -14y.

$$-14y+120-2y^{2}=0$$

Divide both sides by 2.

$$-7y+60-y^{2}=0$$

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

$$-y^{2}-7y+60=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by+60. To find a and b, set up a system to be solved.

$$a+b=-7 ab=-60=-60$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -7.

$$a=5 b=-12$$

Rewrite -y^{2}-7y+60 as \left(-y^{2}+5y\right)+\left(-12y+60\right).

$$\left(-y^{2}+5y\right)+\left(-12y+60\right)$$

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

$$y\left(-y+5\right)+12\left(-y+5\right)$$

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

$$\left(-y+5\right)\left(y+12\right)$$

To find equation solutions, solve -y+5=0 and y+12=0.

$$y=5 y=-12$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 15y\left(y-8\right), the least common multiple of 8-y,y,15.

$$-15y-\left(15y-120\right)=2y\left(y-8\right)$$

To find the opposite of 15y-120, find the opposite of each term.

$$-15y-15y+120=2y\left(y-8\right)$$

Combine -15y and -15y to get -30y.

$$-30y+120=2y\left(y-8\right)$$

Use the distributive property to multiply 2y by y-8.

$$-30y+120=2y^{2}-16y$$

Subtract 2y^{2} from both sides.

$$-30y+120-2y^{2}=-16y$$

Add 16y to both sides.

$$-30y+120-2y^{2}+16y=0$$

Combine -30y and 16y to get -14y.

$$-14y+120-2y^{2}=0$$

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.

$$-2y^{2}-14y+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -14 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-2\right)\times 120}}{2\left(-2\right)}$$

Square -14.

$$y=\frac{-\left(-14\right)±\sqrt{196-4\left(-2\right)\times 120}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-\left(-14\right)±\sqrt{196+8\times 120}}{2\left(-2\right)}$$

Multiply 8 times 120.

$$y=\frac{-\left(-14\right)±\sqrt{196+960}}{2\left(-2\right)}$$

Add 196 to 960.

$$y=\frac{-\left(-14\right)±\sqrt{1156}}{2\left(-2\right)}$$

Take the square root of 1156.

$$y=\frac{-\left(-14\right)±34}{2\left(-2\right)}$$

The opposite of -14 is 14.

$$y=\frac{14±34}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{14±34}{-4}$$

Now solve the equation y=\frac{14±34}{-4} when ± is plus. Add 14 to 34.

$$y=\frac{48}{-4}$$

Divide 48 by -4.

$$y=-12$$

Now solve the equation y=\frac{14±34}{-4} when ± is minus. Subtract 34 from 14.

$$y=-\frac{20}{-4}$$

Divide -20 by -4.

$$y=5$$

The equation is now solved.

$$y=-12 y=5$$

Steps for Completing the Square

Variable y cannot be equal to any of the values 0,8 since division by zero is not defined. Multiply both sides of the equation by 15y\left(y-8\right), the least common multiple of 8-y,y,15.

$$-15y-\left(15y-120\right)=2y\left(y-8\right)$$

To find the opposite of 15y-120, find the opposite of each term.

$$-15y-15y+120=2y\left(y-8\right)$$

Combine -15y and -15y to get -30y.

$$-30y+120=2y\left(y-8\right)$$

Use the distributive property to multiply 2y by y-8.

$$-30y+120=2y^{2}-16y$$

Subtract 2y^{2} from both sides.

$$-30y+120-2y^{2}=-16y$$

Add 16y to both sides.

$$-30y+120-2y^{2}+16y=0$$

Combine -30y and 16y to get -14y.

$$-14y+120-2y^{2}=0$$

Subtract 120 from both sides. Anything subtracted from zero gives its negation.

$$-14y-2y^{2}=-120$$

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.

$$-2y^{2}-14y=-120$$

Divide both sides by -2.

$$\frac{-2y^{2}-14y}{-2}=-\frac{120}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$y^{2}+\left(-\frac{14}{-2}\right)y=-\frac{120}{-2}$$

Divide -14 by -2.

$$y^{2}+7y=-\frac{120}{-2}$$

Divide -120 by -2.

$$y^{2}+7y=60$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+7y+\left(\frac{7}{2}\right)^{2}=60+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+7y+\frac{49}{4}=60+\frac{49}{4}$$

Add 60 to \frac{49}{4}.

$$y^{2}+7y+\frac{49}{4}=\frac{289}{4}$$

Factor y^{2}+7y+\frac{49}{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}.

$$\left(y+\frac{7}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{7}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$y+\frac{7}{2}=\frac{17}{2} y+\frac{7}{2}=-\frac{17}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$y=5 y=-12$$