Solve y+6/y-1-y-4/y^2-y=1/y-1

$$\frac { y + 6 } { y - 1 } - \frac { y - 4 } { y ^ { 2 } - y } = \frac { 1 } { y - 1 }$$

Solve for y

$$y=-2$$

Steps Using Factoring

Variable y cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right), the least common multiple of y-1,y^{2}-y.

$$y\left(y+6\right)-\left(y-4\right)=y$$

Use the distributive property to multiply y by y+6.

$$y^{2}+6y-\left(y-4\right)=y$$

To find the opposite of y-4, find the opposite of each term.

$$y^{2}+6y-y+4=y$$

Combine 6y and -y to get 5y.

$$y^{2}+5y+4=y$$

Subtract y from both sides.

$$y^{2}+5y+4-y=0$$

Combine 5y and -y to get 4y.

$$y^{2}+4y+4=0$$

To solve the equation, factor y^{2}+4y+4 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.

$$a+b=4 ab=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

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

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

Rewrite as a binomial square.

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

To find equation solution, solve y+2=0.

$$y=-2$$

Steps Using Factoring By Grouping

Variable y cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right), the least common multiple of y-1,y^{2}-y.

$$y\left(y+6\right)-\left(y-4\right)=y$$

Use the distributive property to multiply y by y+6.

$$y^{2}+6y-\left(y-4\right)=y$$

To find the opposite of y-4, find the opposite of each term.

$$y^{2}+6y-y+4=y$$

Combine 6y and -y to get 5y.

$$y^{2}+5y+4=y$$

Subtract y from both sides.

$$y^{2}+5y+4-y=0$$

Combine 5y and -y to get 4y.

$$y^{2}+4y+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite y^{2}+4y+4 as \left(y^{2}+2y\right)+\left(2y+4\right).

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

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

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

Factor out common term y+2 by using distributive property.

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

Rewrite as a binomial square.

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

To find equation solution, solve y+2=0.

$$y=-2$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right), the least common multiple of y-1,y^{2}-y.

$$y\left(y+6\right)-\left(y-4\right)=y$$

Use the distributive property to multiply y by y+6.

$$y^{2}+6y-\left(y-4\right)=y$$

To find the opposite of y-4, find the opposite of each term.

$$y^{2}+6y-y+4=y$$

Combine 6y and -y to get 5y.

$$y^{2}+5y+4=y$$

Subtract y from both sides.

$$y^{2}+5y+4-y=0$$

Combine 5y and -y to get 4y.

$$y^{2}+4y+4=0$$

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

$$y=\frac{-4±\sqrt{4^{2}-4\times 4}}{2}$$

Square 4.

$$y=\frac{-4±\sqrt{16-4\times 4}}{2}$$

Multiply -4 times 4.

$$y=\frac{-4±\sqrt{16-16}}{2}$$

Add 16 to -16.

$$y=\frac{-4±\sqrt{0}}{2}$$

Take the square root of 0.

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

Divide -4 by 2.

$$y=-2$$

Steps for Completing the Square

Variable y cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by y\left(y-1\right), the least common multiple of y-1,y^{2}-y.

$$y\left(y+6\right)-\left(y-4\right)=y$$

Use the distributive property to multiply y by y+6.

$$y^{2}+6y-\left(y-4\right)=y$$

To find the opposite of y-4, find the opposite of each term.

$$y^{2}+6y-y+4=y$$

Combine 6y and -y to get 5y.

$$y^{2}+5y+4=y$$

Subtract y from both sides.

$$y^{2}+5y+4-y=0$$

Combine 5y and -y to get 4y.

$$y^{2}+4y+4=0$$

Factor y^{2}+4y+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}.

$$\left(y+2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+2\right)^{2}}=\sqrt{0}$$

Simplify.

$$y+2=0 y+2=0$$

Subtract 2 from both sides of the equation.

$$y=-2 y=-2$$

The equation is now solved. Solutions are the same.

$$y=-2$$