Solve 64/x^2-9/3+frac{8{x}}

$$\frac{ \frac{ 64 }{ { x }^{ 2 } } -9 }{ 3+ \frac{ 8 }{ x } }$$

Evaluate

$$-3+\frac{8}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{x^{2}}{x^{2}}.

$$\frac{\frac{64}{x^{2}}-\frac{9x^{2}}{x^{2}}}{3+\frac{8}{x}}$$

Since \frac{64}{x^{2}} and \frac{9x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{3+\frac{8}{x}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{\frac{3x}{x}+\frac{8}{x}}$$

Since \frac{3x}{x} and \frac{8}{x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{\frac{3x+8}{x}}$$

Divide \frac{64-9x^{2}}{x^{2}} by \frac{3x+8}{x} by multiplying \frac{64-9x^{2}}{x^{2}} by the reciprocal of \frac{3x+8}{x}.

$$\frac{\left(64-9x^{2}\right)x}{x^{2}\left(3x+8\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{-9x^{2}+64}{x\left(3x+8\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(-3x-8\right)\left(3x-8\right)}{x\left(3x+8\right)}$$

Extract the negative sign in -8-3x.

$$\frac{-\left(3x-8\right)\left(3x+8\right)}{x\left(3x+8\right)}$$

Cancel out 3x+8 in both numerator and denominator.

$$\frac{-\left(3x-8\right)}{x}$$

Expand the expression.

$$\frac{-3x+8}{x}$$

Expand

$$-3+\frac{8}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 9 times \frac{x^{2}}{x^{2}}.

$$\frac{\frac{64}{x^{2}}-\frac{9x^{2}}{x^{2}}}{3+\frac{8}{x}}$$

Since \frac{64}{x^{2}} and \frac{9x^{2}}{x^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{3+\frac{8}{x}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{\frac{3x}{x}+\frac{8}{x}}$$

Since \frac{3x}{x} and \frac{8}{x} have the same denominator, add them by adding their numerators.

$$\frac{\frac{64-9x^{2}}{x^{2}}}{\frac{3x+8}{x}}$$

Divide \frac{64-9x^{2}}{x^{2}} by \frac{3x+8}{x} by multiplying \frac{64-9x^{2}}{x^{2}} by the reciprocal of \frac{3x+8}{x}.

$$\frac{\left(64-9x^{2}\right)x}{x^{2}\left(3x+8\right)}$$

Cancel out x in both numerator and denominator.

$$\frac{-9x^{2}+64}{x\left(3x+8\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(-3x-8\right)\left(3x-8\right)}{x\left(3x+8\right)}$$

Extract the negative sign in -8-3x.

$$\frac{-\left(3x-8\right)\left(3x+8\right)}{x\left(3x+8\right)}$$

Cancel out 3x+8 in both numerator and denominator.

$$\frac{-\left(3x-8\right)}{x}$$

Expand the expression.

$$\frac{-3x+8}{x}$$