Solve y/2y+3=2/y+5

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

Solve for y

$$y=-3$$

$$y=2$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -5,-\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(y+5\right)\left(2y+3\right), the least common multiple of 2y+3,y+5.

$$\left(y+5\right)y=\left(2y+3\right)\times 2$$

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

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

Use the distributive property to multiply 2y+3 by 2.

$$y^{2}+5y=4y+6$$

Subtract 4y from both sides.

$$y^{2}+5y-4y=6$$

Combine 5y and -4y to get y.

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

Subtract 6 from both sides.

$$y^{2}+y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\left(-6\right)}}{2}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$y=\frac{-1±\sqrt{1+24}}{2}$$

Add 1 to 24.

$$y=\frac{-1±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-1±5}{2}$$

Now solve the equation y=\frac{-1±5}{2} when ± is plus. Add -1 to 5.

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

Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{-1±5}{2} when ± is minus. Subtract 5 from -1.

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

Divide -6 by 2.

$$y=-3$$

The equation is now solved.

$$y=2 y=-3$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -5,-\frac{3}{2} since division by zero is not defined. Multiply both sides of the equation by \left(y+5\right)\left(2y+3\right), the least common multiple of 2y+3,y+5.

$$\left(y+5\right)y=\left(2y+3\right)\times 2$$

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

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

Use the distributive property to multiply 2y+3 by 2.

$$y^{2}+5y=4y+6$$

Subtract 4y from both sides.

$$y^{2}+5y-4y=6$$

Combine 5y and -4y to get y.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+y+\left(\frac{1}{2}\right)^{2}=6+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 6 to \frac{1}{4}.

$$y^{2}+y+\frac{1}{4}=\frac{25}{4}$$

Factor y^{2}+y+\frac{1}{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}.

$$\left(y+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$y=2 y=-3$$