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

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

Solve for y

$$y=-3$$

$$y=1$$

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^{2}, the least common multiple of y,y^{2}.

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

Subtract 3 from both sides.

$$y\times 2+y^{2}-3=0$$

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

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

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

$$a+b=2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(y-1\right)\left(y+3\right)$$

To find equation solutions, solve y-1=0 and y+3=0.

$$y=1 y=-3$$

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^{2}, the least common multiple of y,y^{2}.

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

Subtract 3 from both sides.

$$y\times 2+y^{2}-3=0$$

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

Factor out y in the first and 3 in the second group.

$$y\left(y-1\right)+3\left(y-1\right)$$

Factor out common term y-1 by using distributive property.

$$\left(y-1\right)\left(y+3\right)$$

To find equation solutions, solve y-1=0 and y+3=0.

$$y=1 y=-3$$

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^{2}, the least common multiple of y,y^{2}.

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

Subtract 3 from both sides.

$$y\times 2+y^{2}-3=0$$

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.

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

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

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

Square 2.

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

Multiply -4 times -3.

$$y=\frac{-2±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$y=\frac{-2±\sqrt{16}}{2}$$

Take the square root of 16.

$$y=\frac{-2±4}{2}$$

Now solve the equation y=\frac{-2±4}{2} when ± is plus. Add -2 to 4.

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

Divide 2 by 2.

$$y=1$$

Now solve the equation y=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

The equation is now solved.

$$y=1 y=-3$$

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^{2}, the least common multiple of y,y^{2}.

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

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.

$$y^{2}+2y=3$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+2y+1^{2}=3+1^{2}$$

Square 1.

$$y^{2}+2y+1=3+1$$

Add 3 to 1.

$$y^{2}+2y+1=4$$

Factor y^{2}+2y+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$y+1=2 y+1=-2$$

Subtract 1 from both sides of the equation.

$$y=1 y=-3$$