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

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

Solve for y

$$y=-\frac{1}{3}\approx -0.333333333$$

Steps Using the Quadratic Formula

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

$$-4=\left(y+1\right)\left(y+3\right)\left(-3\right)+\left(y+1\right)\times 2$$

Use the distributive property to multiply y+1 by y+3 and combine like terms.

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

Use the distributive property to multiply y^{2}+4y+3 by -3.

$$-4=-3y^{2}-12y-9+\left(y+1\right)\times 2$$

Use the distributive property to multiply y+1 by 2.

$$-4=-3y^{2}-12y-9+2y+2$$

Combine -12y and 2y to get -10y.

$$-4=-3y^{2}-10y-9+2$$

Add -9 and 2 to get -7.

$$-4=-3y^{2}-10y-7$$

Swap sides so that all variable terms are on the left hand side.

$$-3y^{2}-10y-7=-4$$

Add 4 to both sides.

$$-3y^{2}-10y-7+4=0$$

Add -7 and 4 to get -3.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)\left(-3\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-\left(-10\right)±\sqrt{100+12\left(-3\right)}}{2\left(-3\right)}$$

Multiply 12 times -3.

$$y=\frac{-\left(-10\right)±\sqrt{100-36}}{2\left(-3\right)}$$

Add 100 to -36.

$$y=\frac{-\left(-10\right)±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$y=\frac{-\left(-10\right)±8}{2\left(-3\right)}$$

The opposite of -10 is 10.

$$y=\frac{10±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{10±8}{-6}$$

Now solve the equation y=\frac{10±8}{-6} when ± is plus. Add 10 to 8.

$$y=\frac{18}{-6}$$

Divide 18 by -6.

$$y=-3$$

Now solve the equation y=\frac{10±8}{-6} when ± is minus. Subtract 8 from 10.

$$y=\frac{2}{-6}$$

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

$$y=-\frac{1}{3}$$

The equation is now solved.

$$y=-3 y=-\frac{1}{3}$$

Variable y cannot be equal to -3.

$$y=-\frac{1}{3}$$

Steps for Completing the Square

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

$$-4=\left(y+1\right)\left(y+3\right)\left(-3\right)+\left(y+1\right)\times 2$$

Use the distributive property to multiply y+1 by y+3 and combine like terms.

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

Use the distributive property to multiply y^{2}+4y+3 by -3.

$$-4=-3y^{2}-12y-9+\left(y+1\right)\times 2$$

Use the distributive property to multiply y+1 by 2.

$$-4=-3y^{2}-12y-9+2y+2$$

Combine -12y and 2y to get -10y.

$$-4=-3y^{2}-10y-9+2$$

Add -9 and 2 to get -7.

$$-4=-3y^{2}-10y-7$$

Swap sides so that all variable terms are on the left hand side.

$$-3y^{2}-10y-7=-4$$

Add 7 to both sides.

$$-3y^{2}-10y=-4+7$$

Add -4 and 7 to get 3.

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

Divide both sides by -3.

$$\frac{-3y^{2}-10y}{-3}=\frac{3}{-3}$$

Dividing by -3 undoes the multiplication by -3.

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

Divide -10 by -3.

$$y^{2}+\frac{10}{3}y=\frac{3}{-3}$$

Divide 3 by -3.

$$y^{2}+\frac{10}{3}y=-1$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{10}{3}y+\left(\frac{5}{3}\right)^{2}=-1+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{10}{3}y+\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$y^{2}+\frac{10}{3}y+\frac{25}{9}=\frac{16}{9}$$

Factor y^{2}+\frac{10}{3}y+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$y+\frac{5}{3}=\frac{4}{3} y+\frac{5}{3}=-\frac{4}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$y=-\frac{1}{3} y=-3$$

Variable y cannot be equal to -3.

$$y=-\frac{1}{3}$$