Solve 3y+3/y=20

$$3 y + \frac { 3 } { y } = 20$$

Solve for y

$$y = \frac{\sqrt{91} + 10}{3} \approx 6.513130671$$

$$y=\frac{10-\sqrt{91}}{3}\approx 0.153535995$$

Steps Using the Quadratic Formula

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$3yy+3=20y$$

Multiply y and y to get y^{2}.

$$3y^{2}+3=20y$$

Subtract 20y from both sides.

$$3y^{2}+3-20y=0$$

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.

$$3y^{2}-20y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -20 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\times 3}}{2\times 3}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-20\right)±\sqrt{400-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$y=\frac{-\left(-20\right)±\sqrt{400-36}}{2\times 3}$$

Add 400 to -36.

$$y=\frac{-\left(-20\right)±\sqrt{364}}{2\times 3}$$

Take the square root of 364.

$$y=\frac{-\left(-20\right)±2\sqrt{91}}{2\times 3}$$

The opposite of -20 is 20.

$$y=\frac{20±2\sqrt{91}}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{20±2\sqrt{91}}{6}$$

Now solve the equation y=\frac{20±2\sqrt{91}}{6} when ± is plus. Add 20 to 2\sqrt{91}.

$$y=\frac{2\sqrt{91}+20}{6}$$

Divide 20+2\sqrt{91} by 6.

$$y=\frac{\sqrt{91}+10}{3}$$

Now solve the equation y=\frac{20±2\sqrt{91}}{6} when ± is minus. Subtract 2\sqrt{91} from 20.

$$y=\frac{20-2\sqrt{91}}{6}$$

Divide 20-2\sqrt{91} by 6.

$$y=\frac{10-\sqrt{91}}{3}$$

The equation is now solved.

$$y=\frac{\sqrt{91}+10}{3} y=\frac{10-\sqrt{91}}{3}$$

Steps for Completing the Square

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by y.

$$3yy+3=20y$$

Multiply y and y to get y^{2}.

$$3y^{2}+3=20y$$

Subtract 20y from both sides.

$$3y^{2}+3-20y=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$3y^{2}-20y=-3$$

Divide both sides by 3.

$$\frac{3y^{2}-20y}{3}=-\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}-\frac{20}{3}y=-\frac{3}{3}$$

Divide -3 by 3.

$$y^{2}-\frac{20}{3}y=-1$$

Divide -\frac{20}{3}, the coefficient of the x term, by 2 to get -\frac{10}{3}. Then add the square of -\frac{10}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{20}{3}y+\left(-\frac{10}{3}\right)^{2}=-1+\left(-\frac{10}{3}\right)^{2}$$

Square -\frac{10}{3} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{20}{3}y+\frac{100}{9}=-1+\frac{100}{9}$$

Add -1 to \frac{100}{9}.

$$y^{2}-\frac{20}{3}y+\frac{100}{9}=\frac{91}{9}$$

Factor y^{2}-\frac{20}{3}y+\frac{100}{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}.

$$\left(y-\frac{10}{3}\right)^{2}=\frac{91}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{10}{3}\right)^{2}}=\sqrt{\frac{91}{9}}$$

Simplify.

$$y-\frac{10}{3}=\frac{\sqrt{91}}{3} y-\frac{10}{3}=-\frac{\sqrt{91}}{3}$$

Add \frac{10}{3} to both sides of the equation.

$$y=\frac{\sqrt{91}+10}{3} y=\frac{10-\sqrt{91}}{3}$$