Solve 35/y+y=12

$$\frac{ 35 }{ y } +y=12$$

Solve for y

$$y=5$$

$$y=7$$

Steps Using Factoring

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$35+yy=12y$$

Multiply y and y to get y^{2}.

$$35+y^{2}=12y$$

Subtract 12y from both sides.

$$35+y^{2}-12y=0$$

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

$$y^{2}-12y+35=0$$

To solve the equation, factor y^{2}-12y+35 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.

$$a+b=-12 ab=35$$

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 35.

$$-1,-35 -5,-7$$

Calculate the sum for each pair.

$$-1-35=-36 -5-7=-12$$

The solution is the pair that gives sum -12.

$$a=-7 b=-5$$

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

$$\left(y-7\right)\left(y-5\right)$$

To find equation solutions, solve y-7=0 and y-5=0.

$$y=7 y=5$$

Steps Using Factoring By Grouping

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$35+yy=12y$$

Multiply y and y to get y^{2}.

$$35+y^{2}=12y$$

Subtract 12y from both sides.

$$35+y^{2}-12y=0$$

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

$$y^{2}-12y+35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+35. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 35=35$$

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 35.

$$-1,-35 -5,-7$$

Calculate the sum for each pair.

$$-1-35=-36 -5-7=-12$$

The solution is the pair that gives sum -12.

$$a=-7 b=-5$$

Rewrite y^{2}-12y+35 as \left(y^{2}-7y\right)+\left(-5y+35\right).

$$\left(y^{2}-7y\right)+\left(-5y+35\right)$$

Factor out y in the first and -5 in the second group.

$$y\left(y-7\right)-5\left(y-7\right)$$

Factor out common term y-7 by using distributive property.

$$\left(y-7\right)\left(y-5\right)$$

To find equation solutions, solve y-7=0 and y-5=0.

$$y=7 y=5$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$35+yy=12y$$

Multiply y and y to get y^{2}.

$$35+y^{2}=12y$$

Subtract 12y from both sides.

$$35+y^{2}-12y=0$$

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.

$$y^{2}-12y+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 35}}{2}$$

Square -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 35}}{2}$$

Multiply -4 times 35.

$$y=\frac{-\left(-12\right)±\sqrt{144-140}}{2}$$

Add 144 to -140.

$$y=\frac{-\left(-12\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$y=\frac{-\left(-12\right)±2}{2}$$

The opposite of -12 is 12.

$$y=\frac{12±2}{2}$$

Now solve the equation y=\frac{12±2}{2} when ± is plus. Add 12 to 2.

$$y=\frac{14}{2}$$

Divide 14 by 2.

$$y=7$$

Now solve the equation y=\frac{12±2}{2} when ± is minus. Subtract 2 from 12.

$$y=\frac{10}{2}$$

Divide 10 by 2.

$$y=5$$

The equation is now solved.

$$y=7 y=5$$

Steps for Completing the Square

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$35+yy=12y$$

Multiply y and y to get y^{2}.

$$35+y^{2}=12y$$

Subtract 12y from both sides.

$$35+y^{2}-12y=0$$

Subtract 35 from both sides. Anything subtracted from zero gives its negation.

$$y^{2}-12y=-35$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-12y+\left(-6\right)^{2}=-35+\left(-6\right)^{2}$$

Square -6.

$$y^{2}-12y+36=-35+36$$

Add -35 to 36.

$$y^{2}-12y+36=1$$

Factor y^{2}-12y+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-6=1 y-6=-1$$

Add 6 to both sides of the equation.

$$y=7 y=5$$