Solve 1-3/x/1+frac{3{x}}=2/3

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

Solve for x

$$x=15$$

Steps Using Factoring

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, subtract them by subtracting their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, add them by adding their numerators.

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

Variable x cannot be equal to 0 since division by zero is not defined. Divide \frac{x-3}{x} by \frac{x+3}{x} by multiplying \frac{x-3}{x} by the reciprocal of \frac{x+3}{x}.

$$\frac{\left(x-3\right)x}{x\left(x+3\right)}=\frac{2}{3}$$

Use the distributive property to multiply x-3 by x.

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

Use the distributive property to multiply x by x+3.

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

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+3\right), the least common multiple of x^{2}+3x,3.

$$3\left(x^{2}-3x\right)=2x\left(x+3\right)$$

Use the distributive property to multiply 3 by x^{2}-3x.

$$3x^{2}-9x=2x\left(x+3\right)$$

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

$$3x^{2}-9x=2x^{2}+6x$$

Subtract 2x^{2} from both sides.

$$3x^{2}-9x-2x^{2}=6x$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-9x=6x$$

Subtract 6x from both sides.

$$x^{2}-9x-6x=0$$

Combine -9x and -6x to get -15x.

$$x^{2}-15x=0$$

Factor out x.

$$x\left(x-15\right)=0$$

To find equation solutions, solve x=0 and x-15=0.

$$x=0 x=15$$

Variable x cannot be equal to 0.

$$x=15$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, subtract them by subtracting their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, add them by adding their numerators.

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

Variable x cannot be equal to 0 since division by zero is not defined. Divide \frac{x-3}{x} by \frac{x+3}{x} by multiplying \frac{x-3}{x} by the reciprocal of \frac{x+3}{x}.

$$\frac{\left(x-3\right)x}{x\left(x+3\right)}=\frac{2}{3}$$

Use the distributive property to multiply x-3 by x.

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

Use the distributive property to multiply x by x+3.

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

Subtract \frac{2}{3} from both sides.

$$\frac{x^{2}-3x}{x^{2}+3x}-\frac{2}{3}=0$$

Factor x^{2}+3x.

$$\frac{x^{2}-3x}{x\left(x+3\right)}-\frac{2}{3}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x\left(x+3\right) and 3 is 3x\left(x+3\right). Multiply \frac{x^{2}-3x}{x\left(x+3\right)} times \frac{3}{3}. Multiply \frac{2}{3} times \frac{x\left(x+3\right)}{x\left(x+3\right)}.

$$\frac{3\left(x^{2}-3x\right)}{3x\left(x+3\right)}-\frac{2x\left(x+3\right)}{3x\left(x+3\right)}=0$$

Since \frac{3\left(x^{2}-3x\right)}{3x\left(x+3\right)} and \frac{2x\left(x+3\right)}{3x\left(x+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(x^{2}-3x\right)-2x\left(x+3\right)}{3x\left(x+3\right)}=0$$

Do the multiplications in 3\left(x^{2}-3x\right)-2x\left(x+3\right).

$$\frac{3x^{2}-9x-2x^{2}-6x}{3x\left(x+3\right)}=0$$

Combine like terms in 3x^{2}-9x-2x^{2}-6x.

$$\frac{x^{2}-15x}{3x\left(x+3\right)}=0$$

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+3\right).

$$x^{2}-15x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}}}{2}$$

Take the square root of \left(-15\right)^{2}.

$$x=\frac{-\left(-15\right)±15}{2}$$

The opposite of -15 is 15.

$$x=\frac{15±15}{2}$$

Now solve the equation x=\frac{15±15}{2} when ± is plus. Add 15 to 15.

$$x=\frac{30}{2}$$

Divide 30 by 2.

$$x=15$$

Now solve the equation x=\frac{15±15}{2} when ± is minus. Subtract 15 from 15.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=15 x=0$$

Variable x cannot be equal to 0.

$$x=15$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, subtract them by subtracting their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{3}{x} have the same denominator, add them by adding their numerators.

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

Variable x cannot be equal to 0 since division by zero is not defined. Divide \frac{x-3}{x} by \frac{x+3}{x} by multiplying \frac{x-3}{x} by the reciprocal of \frac{x+3}{x}.

$$\frac{\left(x-3\right)x}{x\left(x+3\right)}=\frac{2}{3}$$

Use the distributive property to multiply x-3 by x.

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

Use the distributive property to multiply x by x+3.

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

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+3\right), the least common multiple of x^{2}+3x,3.

$$3\left(x^{2}-3x\right)=2x\left(x+3\right)$$

Use the distributive property to multiply 3 by x^{2}-3x.

$$3x^{2}-9x=2x\left(x+3\right)$$

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

$$3x^{2}-9x=2x^{2}+6x$$

Subtract 2x^{2} from both sides.

$$3x^{2}-9x-2x^{2}=6x$$

Combine 3x^{2} and -2x^{2} to get x^{2}.

$$x^{2}-9x=6x$$

Subtract 6x from both sides.

$$x^{2}-9x-6x=0$$

Combine -9x and -6x to get -15x.

$$x^{2}-15x=0$$

Divide -15, the coefficient of the x term, by 2 to get -\frac{15}{2}. Then add the square of -\frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-15x+\left(-\frac{15}{2}\right)^{2}=\left(-\frac{15}{2}\right)^{2}$$

Square -\frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-15x+\frac{225}{4}=\frac{225}{4}$$

Factor x^{2}-15x+\frac{225}{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}.

$$\left(x-\frac{15}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{15}{2}=\frac{15}{2} x-\frac{15}{2}=-\frac{15}{2}$$

Add \frac{15}{2} to both sides of the equation.

$$x=15 x=0$$

Variable x cannot be equal to 0.

$$x=15$$