Solve (x-4)=(4-x)/3x

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

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

$$x=4$$

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.

$$3xx+3x\left(-4\right)=4-x$$

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

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

Multiply 3 and -4 to get -12.

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

Subtract 4 from both sides.

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

Add x to both sides.

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

Combine -12x and x to get -11x.

$$3x^{2}-11x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\left(-4\right)}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\left(-4\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 3}$$

Add 121 to 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-\left(-11\right)±13}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±13}{6}$$

Now solve the equation x=\frac{11±13}{6} when ± is plus. Add 11 to 13.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{11±13}{6} when ± is minus. Subtract 13 from 11.

$$x=-\frac{2}{6}$$

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

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

The equation is now solved.

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

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.

$$3xx+3x\left(-4\right)=4-x$$

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

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

Multiply 3 and -4 to get -12.

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

Add x to both sides.

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

Combine -12x and x to get -11x.

$$3x^{2}-11x=4$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=\frac{4}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Add \frac{4}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{169}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{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}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{13}{6} x-\frac{11}{6}=-\frac{13}{6}$$

Add \frac{11}{6} to both sides of the equation.

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