Solve (10/x+1)=3(2/x+2)

$$( \frac { 10 } { x } + 1 ) = 3 ( \frac { 2 } { x } + 2 )$$

Solve for x

$$x=\frac{4}{5}=0.8$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$10+x=3\left(\frac{2}{x}+2\right)x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$10+x=3\left(\frac{2}{x}+\frac{2x}{x}\right)x$$

Since \frac{2}{x} and \frac{2x}{x} have the same denominator, add them by adding their numerators.

$$10+x=3\times \frac{2+2x}{x}x$$

Express 3\times \frac{2+2x}{x} as a single fraction.

$$10+x=\frac{3\left(2+2x\right)}{x}x$$

Express \frac{3\left(2+2x\right)}{x}x as a single fraction.

$$10+x=\frac{3\left(2+2x\right)x}{x}$$

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

$$10+x=\frac{\left(6+6x\right)x}{x}$$

Use the distributive property to multiply 6+6x by x.

$$10+x=\frac{6x+6x^{2}}{x}$$

Subtract \frac{6x+6x^{2}}{x} from both sides.

$$10+x-\frac{6x+6x^{2}}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 10+x times \frac{x}{x}.

$$\frac{\left(10+x\right)x}{x}-\frac{6x+6x^{2}}{x}=0$$

Since \frac{\left(10+x\right)x}{x} and \frac{6x+6x^{2}}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(10+x\right)x-\left(6x+6x^{2}\right)}{x}=0$$

Do the multiplications in \left(10+x\right)x-\left(6x+6x^{2}\right).

$$\frac{10x+x^{2}-6x-6x^{2}}{x}=0$$

Combine like terms in 10x+x^{2}-6x-6x^{2}.

$$\frac{4x-5x^{2}}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$4x-5x^{2}=0$$

Factor out x.

$$x\left(4-5x\right)=0$$

To find equation solutions, solve x=0 and 4-5x=0.

$$x=0 x=\frac{4}{5}$$

Variable x cannot be equal to 0.

$$x=\frac{4}{5}$$

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 x.

$$10+x=3\left(\frac{2}{x}+2\right)x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$10+x=3\left(\frac{2}{x}+\frac{2x}{x}\right)x$$

Since \frac{2}{x} and \frac{2x}{x} have the same denominator, add them by adding their numerators.

$$10+x=3\times \frac{2+2x}{x}x$$

Express 3\times \frac{2+2x}{x} as a single fraction.

$$10+x=\frac{3\left(2+2x\right)}{x}x$$

Express \frac{3\left(2+2x\right)}{x}x as a single fraction.

$$10+x=\frac{3\left(2+2x\right)x}{x}$$

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

$$10+x=\frac{\left(6+6x\right)x}{x}$$

Use the distributive property to multiply 6+6x by x.

$$10+x=\frac{6x+6x^{2}}{x}$$

Subtract \frac{6x+6x^{2}}{x} from both sides.

$$10+x-\frac{6x+6x^{2}}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 10+x times \frac{x}{x}.

$$\frac{\left(10+x\right)x}{x}-\frac{6x+6x^{2}}{x}=0$$

Since \frac{\left(10+x\right)x}{x} and \frac{6x+6x^{2}}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(10+x\right)x-\left(6x+6x^{2}\right)}{x}=0$$

Do the multiplications in \left(10+x\right)x-\left(6x+6x^{2}\right).

$$\frac{10x+x^{2}-6x-6x^{2}}{x}=0$$

Combine like terms in 10x+x^{2}-6x-6x^{2}.

$$\frac{4x-5x^{2}}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$4x-5x^{2}=0$$

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.

$$-5x^{2}+4x=0$$

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

$$x=\frac{-4±\sqrt{4^{2}}}{2\left(-5\right)}$$

Take the square root of 4^{2}.

$$x=\frac{-4±4}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-4±4}{-10}$$

Now solve the equation x=\frac{-4±4}{-10} when ± is plus. Add -4 to 4.

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

Divide 0 by -10.

$$x=0$$

Now solve the equation x=\frac{-4±4}{-10} when ± is minus. Subtract 4 from -4.

$$x=-\frac{8}{-10}$$

Reduce the fraction \frac{-8}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{5}$$

The equation is now solved.

$$x=0 x=\frac{4}{5}$$

Variable x cannot be equal to 0.

$$x=\frac{4}{5}$$

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 x.

$$10+x=3\left(\frac{2}{x}+2\right)x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{x}{x}.

$$10+x=3\left(\frac{2}{x}+\frac{2x}{x}\right)x$$

Since \frac{2}{x} and \frac{2x}{x} have the same denominator, add them by adding their numerators.

$$10+x=3\times \frac{2+2x}{x}x$$

Express 3\times \frac{2+2x}{x} as a single fraction.

$$10+x=\frac{3\left(2+2x\right)}{x}x$$

Express \frac{3\left(2+2x\right)}{x}x as a single fraction.

$$10+x=\frac{3\left(2+2x\right)x}{x}$$

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

$$10+x=\frac{\left(6+6x\right)x}{x}$$

Use the distributive property to multiply 6+6x by x.

$$10+x=\frac{6x+6x^{2}}{x}$$

Subtract \frac{6x+6x^{2}}{x} from both sides.

$$10+x-\frac{6x+6x^{2}}{x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 10+x times \frac{x}{x}.

$$\frac{\left(10+x\right)x}{x}-\frac{6x+6x^{2}}{x}=0$$

Since \frac{\left(10+x\right)x}{x} and \frac{6x+6x^{2}}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(10+x\right)x-\left(6x+6x^{2}\right)}{x}=0$$

Do the multiplications in \left(10+x\right)x-\left(6x+6x^{2}\right).

$$\frac{10x+x^{2}-6x-6x^{2}}{x}=0$$

Combine like terms in 10x+x^{2}-6x-6x^{2}.

$$\frac{4x-5x^{2}}{x}=0$$

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$4x-5x^{2}=0$$

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.

$$-5x^{2}+4x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}+4x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{4}{-5}x=\frac{0}{-5}$$

Divide 4 by -5.

$$x^{2}-\frac{4}{5}x=\frac{0}{-5}$$

Divide 0 by -5.

$$x^{2}-\frac{4}{5}x=0$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{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}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{2}{5} x-\frac{2}{5}=-\frac{2}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{4}{5} x=0$$

Variable x cannot be equal to 0.

$$x=\frac{4}{5}$$