Solve -4/3x=x+7/3

$$- \frac { 4 } { 3 x } = x + \frac { 7 } { 3 }$$

Solve for x

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

$$x=-1$$

Steps Using the Quadratic Formula

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

$$-4=3xx+3x\times \frac{7}{3}$$

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

$$-4=3x^{2}+3x\times \frac{7}{3}$$

Multiply 3 and \frac{7}{3} to get 7.

$$-4=3x^{2}+7x$$

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

$$3x^{2}+7x=-4$$

Add 4 to both sides.

$$3x^{2}+7x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\times 4}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-7±\sqrt{49-48}}{2\times 3}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±1}{6}$$

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

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

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

$$-4=3xx+3x\times \frac{7}{3}$$

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

$$-4=3x^{2}+3x\times \frac{7}{3}$$

Multiply 3 and \frac{7}{3} to get 7.

$$-4=3x^{2}+7x$$

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

$$3x^{2}+7x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=-\frac{4}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{4}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-\frac{4}{3}+\frac{49}{36}$$

Add -\frac{4}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

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

Subtract \frac{7}{6} from both sides of the equation.

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