Solve 1/y(1-y)=-1/1+y

$$\frac { 1 } { y ( 1 - y ) } = - \frac { 1 } { 1 + y }$$

Solve for y

$$y=\sqrt{2}+1\approx 2.414213562$$

$$y=1-\sqrt{2}\approx -0.414213562$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right)\left(y+1\right), the least common multiple of y\left(1-y\right),1+y.

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

To find the opposite of 1+y, find the opposite of each term.

$$-1-y=-y\left(y-1\right)$$

Use the distributive property to multiply y by y-1.

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

To find the opposite of y^{2}-y, find the opposite of each term.

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

Add y^{2} to both sides.

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

Subtract y from both sides.

$$-1-y+y^{2}-y=0$$

Combine -y and -y to get -2y.

$$-1-2y+y^{2}=0$$

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.

$$y^{2}-2y-1=0$$

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

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)}}{2}$$

Square -2.

$$y=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)}}{2}$$

Multiply -4 times -1.

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

Add 4 to 4.

$$y=\frac{-\left(-2\right)±\sqrt{8}}{2}$$

Take the square root of 8.

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

The opposite of -2 is 2.

$$y=\frac{2±2\sqrt{2}}{2}$$

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

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

Divide 2+2\sqrt{2} by 2.

$$y=\sqrt{2}+1$$

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

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

Divide 2-2\sqrt{2} by 2.

$$y=1-\sqrt{2}$$

The equation is now solved.

$$y=\sqrt{2}+1 y=1-\sqrt{2}$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right)\left(y+1\right), the least common multiple of y\left(1-y\right),1+y.

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

To find the opposite of 1+y, find the opposite of each term.

$$-1-y=-y\left(y-1\right)$$

Use the distributive property to multiply y by y-1.

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

To find the opposite of y^{2}-y, find the opposite of each term.

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

Add y^{2} to both sides.

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

Subtract y from both sides.

$$-1-y+y^{2}-y=0$$

Combine -y and -y to get -2y.

$$-1-2y+y^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

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

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.

$$y^{2}-2y=1$$

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.

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

Add 1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-1=\sqrt{2} y-1=-\sqrt{2}$$

Add 1 to both sides of the equation.

$$y=\sqrt{2}+1 y=1-\sqrt{2}$$