Solve 5x^-1-2y^-1/25x^-2-4y^-2

$$\frac { 5 x ^ { - 1 } - 2 y ^ { - 1 } } { 25 x ^ { - 2 } - 4 y ^ { - 2 } }$$

Evaluate

$$\frac{xy}{2x+5y}$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{\left(-2\times \frac{1}{y}x+5\right)\times \frac{1}{x}}{\left(-4y^{-2}x^{2}+25\right)x^{-2}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{\left(-2\times \frac{1}{y}x+5\right)x^{1}}{-4y^{-2}x^{2}+25}$$

Expand the expression.

$$\frac{-2\times \frac{1}{y}x^{2}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express -2\times \frac{1}{y} as a single fraction.

$$\frac{\frac{-2}{y}x^{2}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express \frac{-2}{y}x^{2} as a single fraction.

$$\frac{\frac{-2x^{2}}{y}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

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

$$\frac{\frac{-2x^{2}}{y}+\frac{5xy}{y}}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Since \frac{-2x^{2}}{y} and \frac{5xy}{y} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express \frac{1}{y}x as a single fraction.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \left(\frac{x}{y}\right)^{2}}$$

To raise \frac{x}{y} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \frac{x^{2}}{y^{2}}}$$

Express -4\times \frac{x^{2}}{y^{2}} as a single fraction.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 25 times \frac{y^{2}}{y^{2}}.

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

Since \frac{25y^{2}}{y^{2}} and \frac{-4x^{2}}{y^{2}} have the same denominator, add them by adding their numerators.

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

Divide \frac{-2x^{2}+5xy}{y} by \frac{25y^{2}-4x^{2}}{y^{2}} by multiplying \frac{-2x^{2}+5xy}{y} by the reciprocal of \frac{25y^{2}-4x^{2}}{y^{2}}.

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

Cancel out y in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{xy\left(-2x+5y\right)}{\left(-2x-5y\right)\left(2x-5y\right)}$$

Extract the negative sign in -2x+5y.

$$\frac{-xy\left(2x-5y\right)}{\left(-2x-5y\right)\left(2x-5y\right)}$$

Cancel out 2x-5y in both numerator and denominator.

$$\frac{-xy}{-2x-5y}$$

Expand

$$\frac{xy}{2x+5y}$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{\left(-2\times \frac{1}{y}x+5\right)\times \frac{1}{x}}{\left(-4y^{-2}x^{2}+25\right)x^{-2}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{\left(-2\times \frac{1}{y}x+5\right)x^{1}}{-4y^{-2}x^{2}+25}$$

Expand the expression.

$$\frac{-2\times \frac{1}{y}x^{2}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express -2\times \frac{1}{y} as a single fraction.

$$\frac{\frac{-2}{y}x^{2}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express \frac{-2}{y}x^{2} as a single fraction.

$$\frac{\frac{-2x^{2}}{y}+5x}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

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

$$\frac{\frac{-2x^{2}}{y}+\frac{5xy}{y}}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Since \frac{-2x^{2}}{y} and \frac{5xy}{y} have the same denominator, add them by adding their numerators.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \left(\frac{1}{y}x\right)^{2}}$$

Express \frac{1}{y}x as a single fraction.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \left(\frac{x}{y}\right)^{2}}$$

To raise \frac{x}{y} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\frac{-2x^{2}+5xy}{y}}{25-4\times \frac{x^{2}}{y^{2}}}$$

Express -4\times \frac{x^{2}}{y^{2}} as a single fraction.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 25 times \frac{y^{2}}{y^{2}}.

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

Since \frac{25y^{2}}{y^{2}} and \frac{-4x^{2}}{y^{2}} have the same denominator, add them by adding their numerators.

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

Divide \frac{-2x^{2}+5xy}{y} by \frac{25y^{2}-4x^{2}}{y^{2}} by multiplying \frac{-2x^{2}+5xy}{y} by the reciprocal of \frac{25y^{2}-4x^{2}}{y^{2}}.

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

Cancel out y in both numerator and denominator.

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

Factor the expressions that are not already factored.

$$\frac{xy\left(-2x+5y\right)}{\left(-2x-5y\right)\left(2x-5y\right)}$$

Extract the negative sign in -2x+5y.

$$\frac{-xy\left(2x-5y\right)}{\left(-2x-5y\right)\left(2x-5y\right)}$$

Cancel out 2x-5y in both numerator and denominator.

$$\frac{-xy}{-2x-5y}$$