Solve x-4/4+x/5x=6

$$\frac { x - 4 } { 4 } + \frac { x } { 5 x } = 6$$

Solve for x

$$x = \frac{136}{5} = 27\frac{1}{5} = 27.2$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 20x, the least common multiple of 4,5x.

$$5x\left(x-4\right)+4x=120x$$

Use the distributive property to multiply 5x by x-4.

$$5x^{2}-20x+4x=120x$$

Combine -20x and 4x to get -16x.

$$5x^{2}-16x=120x$$

Subtract 120x from both sides.

$$5x^{2}-16x-120x=0$$

Combine -16x and -120x to get -136x.

$$5x^{2}-136x=0$$

Factor out x.

$$x\left(5x-136\right)=0$$

To find equation solutions, solve x=0 and 5x-136=0.

$$x=0 x=\frac{136}{5}$$

Variable x cannot be equal to 0.

$$x=\frac{136}{5}$$

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 20x, the least common multiple of 4,5x.

$$5x\left(x-4\right)+4x=120x$$

Use the distributive property to multiply 5x by x-4.

$$5x^{2}-20x+4x=120x$$

Combine -20x and 4x to get -16x.

$$5x^{2}-16x=120x$$

Subtract 120x from both sides.

$$5x^{2}-16x-120x=0$$

Combine -16x and -120x to get -136x.

$$5x^{2}-136x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -136 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-136\right)±\sqrt{\left(-136\right)^{2}}}{2\times 5}$$

Take the square root of \left(-136\right)^{2}.

$$x=\frac{-\left(-136\right)±136}{2\times 5}$$

The opposite of -136 is 136.

$$x=\frac{136±136}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{136±136}{10}$$

Now solve the equation x=\frac{136±136}{10} when ± is plus. Add 136 to 136.

$$x=\frac{272}{10}$$

Reduce the fraction \frac{272}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{136}{5}$$

Now solve the equation x=\frac{136±136}{10} when ± is minus. Subtract 136 from 136.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

The equation is now solved.

$$x=\frac{136}{5} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{136}{5}$$

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 20x, the least common multiple of 4,5x.

$$5x\left(x-4\right)+4x=120x$$

Use the distributive property to multiply 5x by x-4.

$$5x^{2}-20x+4x=120x$$

Combine -20x and 4x to get -16x.

$$5x^{2}-16x=120x$$

Subtract 120x from both sides.

$$5x^{2}-16x-120x=0$$

Combine -16x and -120x to get -136x.

$$5x^{2}-136x=0$$

Divide both sides by 5.

$$\frac{5x^{2}-136x}{5}=\frac{0}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{136}{5}x=\frac{0}{5}$$

Divide 0 by 5.

$$x^{2}-\frac{136}{5}x=0$$

Divide -\frac{136}{5}, the coefficient of the x term, by 2 to get -\frac{68}{5}. Then add the square of -\frac{68}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{136}{5}x+\left(-\frac{68}{5}\right)^{2}=\left(-\frac{68}{5}\right)^{2}$$

Square -\frac{68}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{136}{5}x+\frac{4624}{25}=\frac{4624}{25}$$

Factor x^{2}-\frac{136}{5}x+\frac{4624}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{68}{5}\right)^{2}=\frac{4624}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{68}{5}\right)^{2}}=\sqrt{\frac{4624}{25}}$$

Simplify.

$$x-\frac{68}{5}=\frac{68}{5} x-\frac{68}{5}=-\frac{68}{5}$$

Add \frac{68}{5} to both sides of the equation.

$$x=\frac{136}{5} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{136}{5}$$