Solve y-4/3y=-1/3

$$y - \frac { 4 } { 3 y } = - \frac { 1 } { 3 }$$

Solve for y

$$y = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$y=1$$

Steps Using Factoring By Grouping

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3y, the least common multiple of 3y,3.

$$3yy-4=-y$$

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

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

Add y to both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3y^{2}+ay+by-4. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-4\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 3y^{2}+y-4 as \left(3y^{2}-3y\right)+\left(4y-4\right).

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

Factor out 3y in the first and 4 in the second group.

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

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

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

To find equation solutions, solve y-1=0 and 3y+4=0.

$$y=1 y=-\frac{4}{3}$$

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 3y, the least common multiple of 3y,3.

$$3yy-4=-y$$

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

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

Add y to both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 3\left(-4\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-1±\sqrt{1-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$y=\frac{-1±\sqrt{1+48}}{2\times 3}$$

Add 1 to 48.

$$y=\frac{-1±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$y=\frac{-1±7}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-1±7}{6}$$

Now solve the equation y=\frac{-1±7}{6} when ± is plus. Add -1 to 7.

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

Divide 6 by 6.

$$y=1$$

Now solve the equation y=\frac{-1±7}{6} when ± is minus. Subtract 7 from -1.

$$y=-\frac{8}{6}$$

Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$y=1 y=-\frac{4}{3}$$

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 3y, the least common multiple of 3y,3.

$$3yy-4=-y$$

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

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

Add y to both sides.

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

Add 4 to both sides. Anything plus zero gives itself.

$$3y^{2}+y=4$$

Divide both sides by 3.

$$\frac{3y^{2}+y}{3}=\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\frac{1}{3}y=\frac{4}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{1}{3}y+\left(\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(\frac{1}{6}\right)^{2}$$

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{1}{3}y+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{1}{3}y+\frac{1}{36}=\frac{49}{36}$$

Factor y^{2}+\frac{1}{3}y+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$y+\frac{1}{6}=\frac{7}{6} y+\frac{1}{6}=-\frac{7}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$y=1 y=-\frac{4}{3}$$