Solve 2/x=-1/3x+7/3

$$\frac { 2 } { x } = - \frac { 1 } { 3 } x + \frac { 7 } { 3 }$$

Solve for x

$$x=1$$

$$x=6$$

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 x,3.

$$3\times 2=-\frac{1}{3}x\times 3x+3x\times \frac{7}{3}$$

Multiply 3 and 2 to get 6.

$$6=-\frac{1}{3}x\times 3x+3x\times \frac{7}{3}$$

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

$$6=-\frac{1}{3}x^{2}\times 3+3x\times \frac{7}{3}$$

Cancel out 3 and 3.

$$6=-x^{2}+3x\times \frac{7}{3}$$

Cancel out 3 and 3.

$$6=-x^{2}+7x$$

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

$$-x^{2}+7x=6$$

Subtract 6 from both sides.

$$-x^{2}+7x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\left(-6\right)}}{2\left(-1\right)}$$

Multiply 4 times -6.

$$x=\frac{-7±\sqrt{49-24}}{2\left(-1\right)}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±5}{-2}$$

Now solve the equation x=\frac{-7±5}{-2} when ± is plus. Add -7 to 5.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-7±5}{-2} when ± is minus. Subtract 5 from -7.

$$x=-\frac{12}{-2}$$

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=1 x=6$$

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 x,3.

$$3\times 2=-\frac{1}{3}x\times 3x+3x\times \frac{7}{3}$$

Multiply 3 and 2 to get 6.

$$6=-\frac{1}{3}x\times 3x+3x\times \frac{7}{3}$$

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

$$6=-\frac{1}{3}x^{2}\times 3+3x\times \frac{7}{3}$$

Cancel out 3 and 3.

$$6=-x^{2}+3x\times \frac{7}{3}$$

Cancel out 3 and 3.

$$6=-x^{2}+7x$$

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

$$-x^{2}+7x=6$$

Divide both sides by -1.

$$\frac{-x^{2}+7x}{-1}=\frac{6}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{7}{-1}x=\frac{6}{-1}$$

Divide 7 by -1.

$$x^{2}-7x=\frac{6}{-1}$$

Divide 6 by -1.

$$x^{2}-7x=-6$$

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

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

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-6+\frac{49}{4}$$

Add -6 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{25}{4}$$

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

$$\left(x-\frac{7}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{5}{2} x-\frac{7}{2}=-\frac{5}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=6 x=1$$