Solve (8/x+5/9)-(8/x-5/9)

$$( \frac { 8 } { x } + \frac { 5 } { 9 } ) - ( \frac { 8 } { x } - \frac { 5 } { 9 } )$$

Evaluate

$$\frac{10}{9}\approx 1.111111111$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 9 is 9x. Multiply \frac{8}{x} times \frac{9}{9}. Multiply \frac{5}{9} times \frac{x}{x}.

$$\frac{8}{x}+\frac{5}{9}-\left(\frac{8\times 9}{9x}-\frac{5x}{9x}\right)$$

Since \frac{8\times 9}{9x} and \frac{5x}{9x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8}{x}+\frac{5}{9}-\frac{8\times 9-5x}{9x}$$

Do the multiplications in 8\times 9-5x.

$$\frac{8}{x}+\frac{5}{9}-\frac{72-5x}{9x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x and 9 is 9x. Multiply \frac{8}{x} times \frac{9}{9}. Multiply \frac{5}{9} times \frac{x}{x}.

$$\frac{8\times 9}{9x}+\frac{5x}{9x}-\frac{72-5x}{9x}$$

Since \frac{8\times 9}{9x} and \frac{5x}{9x} have the same denominator, add them by adding their numerators.

$$\frac{8\times 9+5x}{9x}-\frac{72-5x}{9x}$$

Do the multiplications in 8\times 9+5x.

$$\frac{72+5x}{9x}-\frac{72-5x}{9x}$$

Since \frac{72+5x}{9x} and \frac{72-5x}{9x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{72+5x-\left(72-5x\right)}{9x}$$

Do the multiplications in 72+5x-\left(72-5x\right).

$$\frac{72+5x-72+5x}{9x}$$

Combine like terms in 72+5x-72+5x.

$$\frac{10x}{9x}$$

Cancel out x in both numerator and denominator.

$$\frac{10}{9}$$

Factor

$$\frac{2 \cdot 5}{3 ^ {2}} = 1\frac{1}{9} = 1.1111111111111112$$