Solve y-10/y=3

$$y - \frac { 10 } { y } = 3$$

Solve for y

$$y=-2$$

$$y=5$$

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.

$$yy-10=3y$$

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

$$y^{2}-10=3y$$

Subtract 3y from both sides.

$$y^{2}-10-3y=0$$

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

$$y^{2}-3y-10=0$$

To solve the equation, factor y^{2}-3y-10 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.

$$a+b=-3 ab=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

$$\left(y-5\right)\left(y+2\right)$$

To find equation solutions, solve y-5=0 and y+2=0.

$$y=5 y=-2$$

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.

$$yy-10=3y$$

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

$$y^{2}-10=3y$$

Subtract 3y from both sides.

$$y^{2}-10-3y=0$$

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

$$y^{2}-3y-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite y^{2}-3y-10 as \left(y^{2}-5y\right)+\left(2y-10\right).

$$\left(y^{2}-5y\right)+\left(2y-10\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(y-5\right)+2\left(y-5\right)$$

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

$$\left(y-5\right)\left(y+2\right)$$

To find equation solutions, solve y-5=0 and y+2=0.

$$y=5 y=-2$$

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.

$$yy-10=3y$$

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

$$y^{2}-10=3y$$

Subtract 3y from both sides.

$$y^{2}-10-3y=0$$

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.

$$y^{2}-3y-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-10\right)}}{2}$$

Square -3.

$$y=\frac{-\left(-3\right)±\sqrt{9-4\left(-10\right)}}{2}$$

Multiply -4 times -10.

$$y=\frac{-\left(-3\right)±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$y=\frac{-\left(-3\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$y=\frac{-\left(-3\right)±7}{2}$$

The opposite of -3 is 3.

$$y=\frac{3±7}{2}$$

Now solve the equation y=\frac{3±7}{2} when ± is plus. Add 3 to 7.

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

Divide 10 by 2.

$$y=5$$

Now solve the equation y=\frac{3±7}{2} when ± is minus. Subtract 7 from 3.

$$y=-\frac{4}{2}$$

Divide -4 by 2.

$$y=-2$$

The equation is now solved.

$$y=5 y=-2$$

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.

$$yy-10=3y$$

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

$$y^{2}-10=3y$$

Subtract 3y from both sides.

$$y^{2}-10-3y=0$$

Add 10 to both sides. Anything plus zero gives itself.

$$y^{2}-3y=10$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$y^{2}-3y+\frac{9}{4}=\frac{49}{4}$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$y-\frac{3}{2}=\frac{7}{2} y-\frac{3}{2}=-\frac{7}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$y=5 y=-2$$