Solve 1/2(1-y)+5/4y^2+4y+4=3y/y^3-1

$$\frac { 1 } { 2 ( 1 - y ) } + \frac { 5 } { 4 y ^ { 2 } + 4 y + 4 } = \frac { 3 y } { y ^ { 3 } - 1 }$$

Solve for y

$$y = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$y=-1$$

Steps Using Factoring By Grouping

Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 4\left(y-1\right)\left(y^{2}+y+1\right), the least common multiple of 2\left(1-y\right),4y^{2}+4y+4,y^{3}-1.

$$-2-2y-2y^{2}+\left(y-1\right)\times 5=4\times 3y$$

Use the distributive property to multiply y-1 by 5.

$$-2-2y-2y^{2}+5y-5=4\times 3y$$

Combine -2y and 5y to get 3y.

$$-2+3y-2y^{2}-5=4\times 3y$$

Subtract 5 from -2 to get -7.

$$-7+3y-2y^{2}=4\times 3y$$

Multiply 4 and 3 to get 12.

$$-7+3y-2y^{2}=12y$$

Subtract 12y from both sides.

$$-7+3y-2y^{2}-12y=0$$

Combine 3y and -12y to get -9y.

$$-7-9y-2y^{2}=0$$

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

$$-2y^{2}-9y-7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2y^{2}+ay+by-7. To find a and b, set up a system to be solved.

$$a+b=-9 ab=-2\left(-7\right)=14$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-2 b=-7$$

Rewrite -2y^{2}-9y-7 as \left(-2y^{2}-2y\right)+\left(-7y-7\right).

$$\left(-2y^{2}-2y\right)+\left(-7y-7\right)$$

Factor out 2y in the first and 7 in the second group.

$$2y\left(-y-1\right)+7\left(-y-1\right)$$

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

$$\left(-y-1\right)\left(2y+7\right)$$

To find equation solutions, solve -y-1=0 and 2y+7=0.

$$y=-1 y=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 4\left(y-1\right)\left(y^{2}+y+1\right), the least common multiple of 2\left(1-y\right),4y^{2}+4y+4,y^{3}-1.

$$-2-2y-2y^{2}+\left(y-1\right)\times 5=4\times 3y$$

Use the distributive property to multiply y-1 by 5.

$$-2-2y-2y^{2}+5y-5=4\times 3y$$

Combine -2y and 5y to get 3y.

$$-2+3y-2y^{2}-5=4\times 3y$$

Subtract 5 from -2 to get -7.

$$-7+3y-2y^{2}=4\times 3y$$

Multiply 4 and 3 to get 12.

$$-7+3y-2y^{2}=12y$$

Subtract 12y from both sides.

$$-7+3y-2y^{2}-12y=0$$

Combine 3y and -12y to get -9y.

$$-7-9y-2y^{2}=0$$

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.

$$-2y^{2}-9y-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -9 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\left(-2\right)\left(-7\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-\left(-9\right)±\sqrt{81+8\left(-7\right)}}{2\left(-2\right)}$$

Multiply 8 times -7.

$$y=\frac{-\left(-9\right)±\sqrt{81-56}}{2\left(-2\right)}$$

Add 81 to -56.

$$y=\frac{-\left(-9\right)±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$y=\frac{-\left(-9\right)±5}{2\left(-2\right)}$$

The opposite of -9 is 9.

$$y=\frac{9±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{9±5}{-4}$$

Now solve the equation y=\frac{9±5}{-4} when ± is plus. Add 9 to 5.

$$y=\frac{14}{-4}$$

Reduce the fraction \frac{14}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{9±5}{-4} when ± is minus. Subtract 5 from 9.

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

Divide 4 by -4.

$$y=-1$$

The equation is now solved.

$$y=-\frac{7}{2} y=-1$$

Steps for Completing the Square

Variable y cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 4\left(y-1\right)\left(y^{2}+y+1\right), the least common multiple of 2\left(1-y\right),4y^{2}+4y+4,y^{3}-1.

$$-2-2y-2y^{2}+\left(y-1\right)\times 5=4\times 3y$$

Use the distributive property to multiply y-1 by 5.

$$-2-2y-2y^{2}+5y-5=4\times 3y$$

Combine -2y and 5y to get 3y.

$$-2+3y-2y^{2}-5=4\times 3y$$

Subtract 5 from -2 to get -7.

$$-7+3y-2y^{2}=4\times 3y$$

Multiply 4 and 3 to get 12.

$$-7+3y-2y^{2}=12y$$

Subtract 12y from both sides.

$$-7+3y-2y^{2}-12y=0$$

Combine 3y and -12y to get -9y.

$$-7-9y-2y^{2}=0$$

Add 7 to both sides. Anything plus zero gives itself.

$$-9y-2y^{2}=7$$

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.

$$-2y^{2}-9y=7$$

Divide both sides by -2.

$$\frac{-2y^{2}-9y}{-2}=\frac{7}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$y^{2}+\left(-\frac{9}{-2}\right)y=\frac{7}{-2}$$

Divide -9 by -2.

$$y^{2}+\frac{9}{2}y=\frac{7}{-2}$$

Divide 7 by -2.

$$y^{2}+\frac{9}{2}y=-\frac{7}{2}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{9}{2}y+\left(\frac{9}{4}\right)^{2}=-\frac{7}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{9}{2}y+\frac{81}{16}=-\frac{7}{2}+\frac{81}{16}$$

Add -\frac{7}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{9}{2}y+\frac{81}{16}=\frac{25}{16}$$

Factor y^{2}+\frac{9}{2}y+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{9}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{9}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$y+\frac{9}{4}=\frac{5}{4} y+\frac{9}{4}=-\frac{5}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$y=-1 y=-\frac{7}{2}$$