Solve 10/y(y^2-1+1/y(1-y)=-1/1+y

$$\frac { 10 } { y ( y ^ { 2 } - 1 } + \frac { 1 } { y ( 1 - y ) } = - \frac { 1 } { 1 + y }$$

Solve for y

$$y=-2\sqrt{2}i+1\approx 1-2.828427125i$$

$$y=1+2\sqrt{2}i\approx 1+2.828427125i$$

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(y^{2}-1\right),y\left(1-y\right),1+y.

$$10-\left(1+y\right)=-y\left(y-1\right)$$

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

$$10-1-y=-y\left(y-1\right)$$

Subtract 1 from 10 to get 9.

$$9-y=-y\left(y-1\right)$$

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

$$9-y=-\left(y^{2}-y\right)$$

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

$$9-y=-y^{2}+y$$

Add y^{2} to both sides.

$$9-y+y^{2}=y$$

Subtract y from both sides.

$$9-y+y^{2}-y=0$$

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

$$9-2y+y^{2}=0$$

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.

$$y^{2}-2y+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times 9.

$$y=\frac{-\left(-2\right)±\sqrt{4-36}}{2}$$

Add 4 to -36.

$$y=\frac{-\left(-2\right)±\sqrt{-32}}{2}$$

Take the square root of -32.

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

The opposite of -2 is 2.

$$y=\frac{2±4\sqrt{2}i}{2}$$

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

$$y=\frac{2+4\sqrt{2}i}{2}$$

Divide 2+4i\sqrt{2} by 2.

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

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

$$y=\frac{-4\sqrt{2}i+2}{2}$$

Divide 2-4i\sqrt{2} by 2.

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

The equation is now solved.

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

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(y^{2}-1\right),y\left(1-y\right),1+y.

$$10-\left(1+y\right)=-y\left(y-1\right)$$

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

$$10-1-y=-y\left(y-1\right)$$

Subtract 1 from 10 to get 9.

$$9-y=-y\left(y-1\right)$$

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

$$9-y=-\left(y^{2}-y\right)$$

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

$$9-y=-y^{2}+y$$

Add y^{2} to both sides.

$$9-y+y^{2}=y$$

Subtract y from both sides.

$$9-y+y^{2}-y=0$$

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

$$9-2y+y^{2}=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-2y+y^{2}=-9$$

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.

$$y^{2}-2y=-9$$

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.

$$y^{2}-2y+1=-9+1$$

Add -9 to 1.

$$y^{2}-2y+1=-8$$

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}.

$$\left(y-1\right)^{2}=-8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-1\right)^{2}}=\sqrt{-8}$$

Simplify.

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

Add 1 to both sides of the equation.

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