Solve 1/40*5+(1/40+1/x)*(40-14-5)=1

$$\frac { 1 } { 40 } \cdot 5 + ( \frac { 1 } { 40 } + \frac { 1 } { x } ) \cdot ( 40 - 14 - 5 ) = 1$$

Solve for x

$$x=60$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 40x, the least common multiple of 40,x.

$$\frac{1}{40}\times 5\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{40} and 5 to get \frac{5}{40}.

$$\frac{5}{40}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Reduce the fraction \frac{5}{40} to lowest terms by extracting and canceling out 5.

$$\frac{1}{8}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{8} and 40 to get \frac{40}{8}.

$$\frac{40}{8}x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Divide 40 by 8 to get 5.

$$5x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 40 and x is 40x. Multiply \frac{1}{40} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{40}{40}.

$$5x+\left(\frac{x}{40x}+\frac{40}{40x}\right)\left(40-14-5\right)\times 40x=40x$$

Since \frac{x}{40x} and \frac{40}{40x} have the same denominator, add them by adding their numerators.

$$5x+\frac{x+40}{40x}\left(40-14-5\right)\times 40x=40x$$

Subtract 14 from 40 to get 26.

$$5x+\frac{x+40}{40x}\left(26-5\right)\times 40x=40x$$

Subtract 5 from 26 to get 21.

$$5x+\frac{x+40}{40x}\times 21\times 40x=40x$$

Multiply 21 and 40 to get 840.

$$5x+\frac{x+40}{40x}\times 840x=40x$$

Express \frac{x+40}{40x}\times 840 as a single fraction.

$$5x+\frac{\left(x+40\right)\times 840}{40x}x=40x$$

Cancel out 40 in both numerator and denominator.

$$5x+\frac{21\left(x+40\right)}{x}x=40x$$

Express \frac{21\left(x+40\right)}{x}x as a single fraction.

$$5x+\frac{21\left(x+40\right)x}{x}=40x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5x times \frac{x}{x}.

$$\frac{5xx}{x}+\frac{21\left(x+40\right)x}{x}=40x$$

Since \frac{5xx}{x} and \frac{21\left(x+40\right)x}{x} have the same denominator, add them by adding their numerators.

$$\frac{5xx+21\left(x+40\right)x}{x}=40x$$

Do the multiplications in 5xx+21\left(x+40\right)x.

$$\frac{5x^{2}+21x^{2}+840x}{x}=40x$$

Combine like terms in 5x^{2}+21x^{2}+840x.

$$\frac{26x^{2}+840x}{x}=40x$$

Subtract 40x from both sides.

$$\frac{26x^{2}+840x}{x}-40x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -40x times \frac{x}{x}.

$$\frac{26x^{2}+840x}{x}+\frac{-40xx}{x}=0$$

Since \frac{26x^{2}+840x}{x} and \frac{-40xx}{x} have the same denominator, add them by adding their numerators.

$$\frac{26x^{2}+840x-40xx}{x}=0$$

Do the multiplications in 26x^{2}+840x-40xx.

$$\frac{26x^{2}+840x-40x^{2}}{x}=0$$

Combine like terms in 26x^{2}+840x-40x^{2}.

$$\frac{-14x^{2}+840x}{x}=0$$

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

$$-14x^{2}+840x=0$$

Factor out x.

$$x\left(-14x+840\right)=0$$

To find equation solutions, solve x=0 and -14x+840=0.

$$x=0 x=60$$

Variable x cannot be equal to 0.

$$x=60$$

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 40x, the least common multiple of 40,x.

$$\frac{1}{40}\times 5\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{40} and 5 to get \frac{5}{40}.

$$\frac{5}{40}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Reduce the fraction \frac{5}{40} to lowest terms by extracting and canceling out 5.

$$\frac{1}{8}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{8} and 40 to get \frac{40}{8}.

$$\frac{40}{8}x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Divide 40 by 8 to get 5.

$$5x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 40 and x is 40x. Multiply \frac{1}{40} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{40}{40}.

$$5x+\left(\frac{x}{40x}+\frac{40}{40x}\right)\left(40-14-5\right)\times 40x=40x$$

Since \frac{x}{40x} and \frac{40}{40x} have the same denominator, add them by adding their numerators.

$$5x+\frac{x+40}{40x}\left(40-14-5\right)\times 40x=40x$$

Subtract 14 from 40 to get 26.

$$5x+\frac{x+40}{40x}\left(26-5\right)\times 40x=40x$$

Subtract 5 from 26 to get 21.

$$5x+\frac{x+40}{40x}\times 21\times 40x=40x$$

Multiply 21 and 40 to get 840.

$$5x+\frac{x+40}{40x}\times 840x=40x$$

Express \frac{x+40}{40x}\times 840 as a single fraction.

$$5x+\frac{\left(x+40\right)\times 840}{40x}x=40x$$

Cancel out 40 in both numerator and denominator.

$$5x+\frac{21\left(x+40\right)}{x}x=40x$$

Express \frac{21\left(x+40\right)}{x}x as a single fraction.

$$5x+\frac{21\left(x+40\right)x}{x}=40x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5x times \frac{x}{x}.

$$\frac{5xx}{x}+\frac{21\left(x+40\right)x}{x}=40x$$

Since \frac{5xx}{x} and \frac{21\left(x+40\right)x}{x} have the same denominator, add them by adding their numerators.

$$\frac{5xx+21\left(x+40\right)x}{x}=40x$$

Do the multiplications in 5xx+21\left(x+40\right)x.

$$\frac{5x^{2}+21x^{2}+840x}{x}=40x$$

Combine like terms in 5x^{2}+21x^{2}+840x.

$$\frac{26x^{2}+840x}{x}=40x$$

Subtract 40x from both sides.

$$\frac{26x^{2}+840x}{x}-40x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -40x times \frac{x}{x}.

$$\frac{26x^{2}+840x}{x}+\frac{-40xx}{x}=0$$

Since \frac{26x^{2}+840x}{x} and \frac{-40xx}{x} have the same denominator, add them by adding their numerators.

$$\frac{26x^{2}+840x-40xx}{x}=0$$

Do the multiplications in 26x^{2}+840x-40xx.

$$\frac{26x^{2}+840x-40x^{2}}{x}=0$$

Combine like terms in 26x^{2}+840x-40x^{2}.

$$\frac{-14x^{2}+840x}{x}=0$$

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

$$-14x^{2}+840x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -14 for a, 840 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-840±\sqrt{840^{2}}}{2\left(-14\right)}$$

Take the square root of 840^{2}.

$$x=\frac{-840±840}{2\left(-14\right)}$$

Multiply 2 times -14.

$$x=\frac{-840±840}{-28}$$

Now solve the equation x=\frac{-840±840}{-28} when ± is plus. Add -840 to 840.

$$x=\frac{0}{-28}$$

Divide 0 by -28.

$$x=0$$

Now solve the equation x=\frac{-840±840}{-28} when ± is minus. Subtract 840 from -840.

$$x=-\frac{1680}{-28}$$

Divide -1680 by -28.

$$x=60$$

The equation is now solved.

$$x=0 x=60$$

Variable x cannot be equal to 0.

$$x=60$$

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 40x, the least common multiple of 40,x.

$$\frac{1}{40}\times 5\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{40} and 5 to get \frac{5}{40}.

$$\frac{5}{40}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Reduce the fraction \frac{5}{40} to lowest terms by extracting and canceling out 5.

$$\frac{1}{8}\times 40x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Multiply \frac{1}{8} and 40 to get \frac{40}{8}.

$$\frac{40}{8}x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

Divide 40 by 8 to get 5.

$$5x+\left(\frac{1}{40}+\frac{1}{x}\right)\left(40-14-5\right)\times 40x=40x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 40 and x is 40x. Multiply \frac{1}{40} times \frac{x}{x}. Multiply \frac{1}{x} times \frac{40}{40}.

$$5x+\left(\frac{x}{40x}+\frac{40}{40x}\right)\left(40-14-5\right)\times 40x=40x$$

Since \frac{x}{40x} and \frac{40}{40x} have the same denominator, add them by adding their numerators.

$$5x+\frac{x+40}{40x}\left(40-14-5\right)\times 40x=40x$$

Subtract 14 from 40 to get 26.

$$5x+\frac{x+40}{40x}\left(26-5\right)\times 40x=40x$$

Subtract 5 from 26 to get 21.

$$5x+\frac{x+40}{40x}\times 21\times 40x=40x$$

Multiply 21 and 40 to get 840.

$$5x+\frac{x+40}{40x}\times 840x=40x$$

Express \frac{x+40}{40x}\times 840 as a single fraction.

$$5x+\frac{\left(x+40\right)\times 840}{40x}x=40x$$

Cancel out 40 in both numerator and denominator.

$$5x+\frac{21\left(x+40\right)}{x}x=40x$$

Express \frac{21\left(x+40\right)}{x}x as a single fraction.

$$5x+\frac{21\left(x+40\right)x}{x}=40x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5x times \frac{x}{x}.

$$\frac{5xx}{x}+\frac{21\left(x+40\right)x}{x}=40x$$

Since \frac{5xx}{x} and \frac{21\left(x+40\right)x}{x} have the same denominator, add them by adding their numerators.

$$\frac{5xx+21\left(x+40\right)x}{x}=40x$$

Do the multiplications in 5xx+21\left(x+40\right)x.

$$\frac{5x^{2}+21x^{2}+840x}{x}=40x$$

Combine like terms in 5x^{2}+21x^{2}+840x.

$$\frac{26x^{2}+840x}{x}=40x$$

Subtract 40x from both sides.

$$\frac{26x^{2}+840x}{x}-40x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -40x times \frac{x}{x}.

$$\frac{26x^{2}+840x}{x}+\frac{-40xx}{x}=0$$

Since \frac{26x^{2}+840x}{x} and \frac{-40xx}{x} have the same denominator, add them by adding their numerators.

$$\frac{26x^{2}+840x-40xx}{x}=0$$

Do the multiplications in 26x^{2}+840x-40xx.

$$\frac{26x^{2}+840x-40x^{2}}{x}=0$$

Combine like terms in 26x^{2}+840x-40x^{2}.

$$\frac{-14x^{2}+840x}{x}=0$$

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

$$-14x^{2}+840x=0$$

Divide both sides by -14.

$$\frac{-14x^{2}+840x}{-14}=\frac{0}{-14}$$

Dividing by -14 undoes the multiplication by -14.

$$x^{2}+\frac{840}{-14}x=\frac{0}{-14}$$

Divide 840 by -14.

$$x^{2}-60x=\frac{0}{-14}$$

Divide 0 by -14.

$$x^{2}-60x=0$$

Divide -60, the coefficient of the x term, by 2 to get -30. Then add the square of -30 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-60x+\left(-30\right)^{2}=\left(-30\right)^{2}$$

Square -30.

$$x^{2}-60x+900=900$$

Factor x^{2}-60x+900. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-30\right)^{2}=900$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-30\right)^{2}}=\sqrt{900}$$

Simplify.

$$x-30=30 x-30=-30$$

Add 30 to both sides of the equation.

$$x=60 x=0$$

Variable x cannot be equal to 0.

$$x=60$$