Solve 6/2+y-5/y=-1

$$\frac{ 6 }{ 2+y } - \frac{ 5 }{ y } =-1$$

Solve for y

$$y=-5$$

$$y=2$$

Steps Using Factoring

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

$$y\times 6-\left(y+2\right)\times 5=-y\left(y+2\right)$$

Use the distributive property to multiply y+2 by 5.

$$y\times 6-\left(5y+10\right)=-y\left(y+2\right)$$

To find the opposite of 5y+10, find the opposite of each term.

$$y\times 6-5y-10=-y\left(y+2\right)$$

Combine y\times 6 and -5y to get y.

$$y-10=-y\left(y+2\right)$$

Use the distributive property to multiply -y by y+2.

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

Add y^{2} to both sides.

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

Add 2y to both sides.

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

Combine y and 2y to get 3y.

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-2 b=5$$

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

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

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

$$y=2 y=-5$$

Steps Using Factoring By Grouping

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

$$y\times 6-\left(y+2\right)\times 5=-y\left(y+2\right)$$

Use the distributive property to multiply y+2 by 5.

$$y\times 6-\left(5y+10\right)=-y\left(y+2\right)$$

To find the opposite of 5y+10, find the opposite of each term.

$$y\times 6-5y-10=-y\left(y+2\right)$$

Combine y\times 6 and -5y to get y.

$$y-10=-y\left(y+2\right)$$

Use the distributive property to multiply -y by y+2.

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

Add y^{2} to both sides.

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

Add 2y to both sides.

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

Combine y and 2y to get 3y.

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-2 b=5$$

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

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

Factor out y in the first and 5 in the second group.

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

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

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

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

$$y=2 y=-5$$

Steps Using the Quadratic Formula

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

$$y\times 6-\left(y+2\right)\times 5=-y\left(y+2\right)$$

Use the distributive property to multiply y+2 by 5.

$$y\times 6-\left(5y+10\right)=-y\left(y+2\right)$$

To find the opposite of 5y+10, find the opposite of each term.

$$y\times 6-5y-10=-y\left(y+2\right)$$

Combine y\times 6 and -5y to get y.

$$y-10=-y\left(y+2\right)$$

Use the distributive property to multiply -y by y+2.

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

Add y^{2} to both sides.

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

Add 2y to both sides.

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

Combine y and 2y to get 3y.

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

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.

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

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}.

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

Square 3.

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

Multiply -4 times -10.

$$y=\frac{-3±\sqrt{9+40}}{2}$$

Add 9 to 40.

$$y=\frac{-3±\sqrt{49}}{2}$$

Take the square root of 49.

$$y=\frac{-3±7}{2}$$

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

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

Divide 4 by 2.

$$y=2$$

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

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

Divide -10 by 2.

$$y=-5$$

The equation is now solved.

$$y=2 y=-5$$

Steps for Completing the Square

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

$$y\times 6-\left(y+2\right)\times 5=-y\left(y+2\right)$$

Use the distributive property to multiply y+2 by 5.

$$y\times 6-\left(5y+10\right)=-y\left(y+2\right)$$

To find the opposite of 5y+10, find the opposite of each term.

$$y\times 6-5y-10=-y\left(y+2\right)$$

Combine y\times 6 and -5y to get y.

$$y-10=-y\left(y+2\right)$$

Use the distributive property to multiply -y by y+2.

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

Add y^{2} to both sides.

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

Add 2y to both sides.

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

Combine y and 2y to get 3y.

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

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

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

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.

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

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.

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

Subtract \frac{3}{2} from both sides of the equation.

$$y=2 y=-5$$