Solve 3x/x-3-4/x=12/x^2-3x

$$\frac { 3 x } { x - 3 } - \frac { 4 } { x } = \frac { 12 } { x ^ { 2 } - 3 x }$$

Solve for x

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

Steps Using Factoring

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

$$x\times 3x-\left(x-3\right)\times 4=12$$

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

$$x^{2}\times 3-\left(x-3\right)\times 4=12$$

Use the distributive property to multiply x-3 by 4.

$$x^{2}\times 3-\left(4x-12\right)=12$$

To find the opposite of 4x-12, find the opposite of each term.

$$x^{2}\times 3-4x+12=12$$

Subtract 12 from both sides.

$$x^{2}\times 3-4x+12-12=0$$

Subtract 12 from 12 to get 0.

$$x^{2}\times 3-4x=0$$

Factor out x.

$$x\left(3x-4\right)=0$$

To find equation solutions, solve x=0 and 3x-4=0.

$$x=0 x=\frac{4}{3}$$

Variable x cannot be equal to 0.

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

Steps Using the Quadratic Formula

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

$$x\times 3x-\left(x-3\right)\times 4=12$$

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

$$x^{2}\times 3-\left(x-3\right)\times 4=12$$

Use the distributive property to multiply x-3 by 4.

$$x^{2}\times 3-\left(4x-12\right)=12$$

To find the opposite of 4x-12, find the opposite of each term.

$$x^{2}\times 3-4x+12=12$$

Subtract 12 from both sides.

$$x^{2}\times 3-4x+12-12=0$$

Subtract 12 from 12 to get 0.

$$x^{2}\times 3-4x=0$$

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.

$$3x^{2}-4x=0$$

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

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

Take the square root of \left(-4\right)^{2}.

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

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±4}{6}$$

Now solve the equation x=\frac{4±4}{6} when ± is plus. Add 4 to 4.

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

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

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

Now solve the equation x=\frac{4±4}{6} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{4}{3} x=0$$

Variable x cannot be equal to 0.

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

Steps for Completing the Square

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

$$x\times 3x-\left(x-3\right)\times 4=12$$

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

$$x^{2}\times 3-\left(x-3\right)\times 4=12$$

Use the distributive property to multiply x-3 by 4.

$$x^{2}\times 3-\left(4x-12\right)=12$$

To find the opposite of 4x-12, find the opposite of each term.

$$x^{2}\times 3-4x+12=12$$

Subtract 12 from both sides.

$$x^{2}\times 3-4x=12-12$$

Subtract 12 from 12 to get 0.

$$x^{2}\times 3-4x=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}-4x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-4x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{4}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{4}{3}x=0$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{4}{3} x=0$$

Variable x cannot be equal to 0.

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