Solve y=-1/(9-4y)-1+2

$$y = - \frac { 1 } { ( 9 - 4 y ) - 1 } + 2$$

Solve for y

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$y = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Subtract 1 from 9 to get 8.

$$y=-\frac{1}{8-4y}+2$$

Factor 8-4y.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

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

Since -\frac{1}{4\left(-y+2\right)} and \frac{2\times 4\left(-y+2\right)}{4\left(-y+2\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -1+2\times 4\left(-y+2\right).

$$y=\frac{-1-8y+16}{4\left(-y+2\right)}$$

Combine like terms in -1-8y+16.

$$y=\frac{15-8y}{4\left(-y+2\right)}$$

Use the distributive property to multiply 4 by -y+2.

$$y=\frac{15-8y}{-4y+8}$$

Subtract \frac{15-8y}{-4y+8} from both sides.

$$y-\frac{15-8y}{-4y+8}=0$$

Factor -4y+8.

$$y-\frac{15-8y}{4\left(-y+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

$$\frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)}-\frac{15-8y}{4\left(-y+2\right)}=0$$

Since \frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)} and \frac{15-8y}{4\left(-y+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\times 4\left(-y+2\right)-\left(15-8y\right)}{4\left(-y+2\right)}=0$$

Do the multiplications in y\times 4\left(-y+2\right)-\left(15-8y\right).

$$\frac{-4y^{2}+8y-15+8y}{4\left(-y+2\right)}=0$$

Combine like terms in -4y^{2}+8y-15+8y.

$$\frac{-4y^{2}+16y-15}{4\left(-y+2\right)}=0$$

Variable y cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(-y+2\right).

$$-4y^{2}+16y-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4y^{2}+ay+by-15. To find a and b, set up a system to be solved.

$$a+b=16 ab=-4\left(-15\right)=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 16.

$$a=10 b=6$$

Rewrite -4y^{2}+16y-15 as \left(-4y^{2}+10y\right)+\left(6y-15\right).

$$\left(-4y^{2}+10y\right)+\left(6y-15\right)$$

Factor out -2y in the first and 3 in the second group.

$$-2y\left(2y-5\right)+3\left(2y-5\right)$$

Factor out common term 2y-5 by using distributive property.

$$\left(2y-5\right)\left(-2y+3\right)$$

To find equation solutions, solve 2y-5=0 and -2y+3=0.

$$y=\frac{5}{2} y=\frac{3}{2}$$

Steps Using the Quadratic Formula

Subtract 1 from 9 to get 8.

$$y=-\frac{1}{8-4y}+2$$

Factor 8-4y.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

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

Since -\frac{1}{4\left(-y+2\right)} and \frac{2\times 4\left(-y+2\right)}{4\left(-y+2\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -1+2\times 4\left(-y+2\right).

$$y=\frac{-1-8y+16}{4\left(-y+2\right)}$$

Combine like terms in -1-8y+16.

$$y=\frac{15-8y}{4\left(-y+2\right)}$$

Use the distributive property to multiply 4 by -y+2.

$$y=\frac{15-8y}{-4y+8}$$

Subtract \frac{15-8y}{-4y+8} from both sides.

$$y-\frac{15-8y}{-4y+8}=0$$

Factor -4y+8.

$$y-\frac{15-8y}{4\left(-y+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

$$\frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)}-\frac{15-8y}{4\left(-y+2\right)}=0$$

Since \frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)} and \frac{15-8y}{4\left(-y+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\times 4\left(-y+2\right)-\left(15-8y\right)}{4\left(-y+2\right)}=0$$

Do the multiplications in y\times 4\left(-y+2\right)-\left(15-8y\right).

$$\frac{-4y^{2}+8y-15+8y}{4\left(-y+2\right)}=0$$

Combine like terms in -4y^{2}+8y-15+8y.

$$\frac{-4y^{2}+16y-15}{4\left(-y+2\right)}=0$$

Variable y cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(-y+2\right).

$$-4y^{2}+16y-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\left(-15\right)}}{2\left(-4\right)}$$

Square 16.

$$y=\frac{-16±\sqrt{256-4\left(-4\right)\left(-15\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$y=\frac{-16±\sqrt{256+16\left(-15\right)}}{2\left(-4\right)}$$

Multiply 16 times -15.

$$y=\frac{-16±\sqrt{256-240}}{2\left(-4\right)}$$

Add 256 to -240.

$$y=\frac{-16±\sqrt{16}}{2\left(-4\right)}$$

Take the square root of 16.

$$y=\frac{-16±4}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{-16±4}{-8}$$

Now solve the equation y=\frac{-16±4}{-8} when ± is plus. Add -16 to 4.

$$y=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$y=\frac{3}{2}$$

Now solve the equation y=\frac{-16±4}{-8} when ± is minus. Subtract 4 from -16.

$$y=-\frac{20}{-8}$$

Reduce the fraction \frac{-20}{-8} to lowest terms by extracting and canceling out 4.

$$y=\frac{5}{2}$$

The equation is now solved.

$$y=\frac{3}{2} y=\frac{5}{2}$$

Steps for Completing the Square

Subtract 1 from 9 to get 8.

$$y=-\frac{1}{8-4y}+2$$

Factor 8-4y.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

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

Since -\frac{1}{4\left(-y+2\right)} and \frac{2\times 4\left(-y+2\right)}{4\left(-y+2\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in -1+2\times 4\left(-y+2\right).

$$y=\frac{-1-8y+16}{4\left(-y+2\right)}$$

Combine like terms in -1-8y+16.

$$y=\frac{15-8y}{4\left(-y+2\right)}$$

Use the distributive property to multiply 4 by -y+2.

$$y=\frac{15-8y}{-4y+8}$$

Subtract \frac{15-8y}{-4y+8} from both sides.

$$y-\frac{15-8y}{-4y+8}=0$$

Factor -4y+8.

$$y-\frac{15-8y}{4\left(-y+2\right)}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y times \frac{4\left(-y+2\right)}{4\left(-y+2\right)}.

$$\frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)}-\frac{15-8y}{4\left(-y+2\right)}=0$$

Since \frac{y\times 4\left(-y+2\right)}{4\left(-y+2\right)} and \frac{15-8y}{4\left(-y+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{y\times 4\left(-y+2\right)-\left(15-8y\right)}{4\left(-y+2\right)}=0$$

Do the multiplications in y\times 4\left(-y+2\right)-\left(15-8y\right).

$$\frac{-4y^{2}+8y-15+8y}{4\left(-y+2\right)}=0$$

Combine like terms in -4y^{2}+8y-15+8y.

$$\frac{-4y^{2}+16y-15}{4\left(-y+2\right)}=0$$

Variable y cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by 4\left(-y+2\right).

$$-4y^{2}+16y-15=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$-4y^{2}+16y=15$$

Divide both sides by -4.

$$\frac{-4y^{2}+16y}{-4}=\frac{15}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y^{2}+\frac{16}{-4}y=\frac{15}{-4}$$

Divide 16 by -4.

$$y^{2}-4y=\frac{15}{-4}$$

Divide 15 by -4.

$$y^{2}-4y=-\frac{15}{4}$$

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

$$y^{2}-4y+\left(-2\right)^{2}=-\frac{15}{4}+\left(-2\right)^{2}$$

Square -2.

$$y^{2}-4y+4=-\frac{15}{4}+4$$

Add -\frac{15}{4} to 4.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

$$y=\frac{5}{2} y=\frac{3}{2}$$