Solve x-4/4+x/3x=6

$$\frac{ x-4 }{ 4 } + \frac{ x }{ 3x } = 6$$

Solve for x

$$x = \frac{80}{3} = 26\frac{2}{3} \approx 26.666666667$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12x, the least common multiple of 4,3x.

$$3x\left(x-4\right)+4x=72x$$

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

$$3x^{2}-12x+4x=72x$$

Combine -12x and 4x to get -8x.

$$3x^{2}-8x=72x$$

Subtract 72x from both sides.

$$3x^{2}-8x-72x=0$$

Combine -8x and -72x to get -80x.

$$3x^{2}-80x=0$$

Factor out x.

$$x\left(3x-80\right)=0$$

To find equation solutions, solve x=0 and 3x-80=0.

$$x=0 x=\frac{80}{3}$$

Variable x cannot be equal to 0.

$$x=\frac{80}{3}$$

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 12x, the least common multiple of 4,3x.

$$3x\left(x-4\right)+4x=72x$$

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

$$3x^{2}-12x+4x=72x$$

Combine -12x and 4x to get -8x.

$$3x^{2}-8x=72x$$

Subtract 72x from both sides.

$$3x^{2}-8x-72x=0$$

Combine -8x and -72x to get -80x.

$$3x^{2}-80x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -80 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}}}{2\times 3}$$

Take the square root of \left(-80\right)^{2}.

$$x=\frac{-\left(-80\right)±80}{2\times 3}$$

The opposite of -80 is 80.

$$x=\frac{80±80}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{80±80}{6}$$

Now solve the equation x=\frac{80±80}{6} when ± is plus. Add 80 to 80.

$$x=\frac{160}{6}$$

Reduce the fraction \frac{160}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{80}{3}$$

Now solve the equation x=\frac{80±80}{6} when ± is minus. Subtract 80 from 80.

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

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{80}{3} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{80}{3}$$

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 12x, the least common multiple of 4,3x.

$$3x\left(x-4\right)+4x=72x$$

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

$$3x^{2}-12x+4x=72x$$

Combine -12x and 4x to get -8x.

$$3x^{2}-8x=72x$$

Subtract 72x from both sides.

$$3x^{2}-8x-72x=0$$

Combine -8x and -72x to get -80x.

$$3x^{2}-80x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-80x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{80}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{80}{3}x=0$$

Divide -\frac{80}{3}, the coefficient of the x term, by 2 to get -\frac{40}{3}. Then add the square of -\frac{40}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{80}{3}x+\left(-\frac{40}{3}\right)^{2}=\left(-\frac{40}{3}\right)^{2}$$

Square -\frac{40}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{80}{3}x+\frac{1600}{9}=\frac{1600}{9}$$

Factor x^{2}-\frac{80}{3}x+\frac{1600}{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}.

$$\left(x-\frac{40}{3}\right)^{2}=\frac{1600}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{40}{3}\right)^{2}}=\sqrt{\frac{1600}{9}}$$

Simplify.

$$x-\frac{40}{3}=\frac{40}{3} x-\frac{40}{3}=-\frac{40}{3}$$

Add \frac{40}{3} to both sides of the equation.

$$x=\frac{80}{3} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{80}{3}$$