Solve y-3/4y-8div(y+2-5/y-2)

$$\frac { y - 3 } { 4 y - 8 } \div ( y + 2 - \frac { 5 } { y - 2 } )$$

Evaluate

$$\frac{1}{4\left(y+3\right)}$$

Short Solution Steps

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

$$\frac{\frac{y-3}{4y-8}}{\frac{\left(y+2\right)\left(y-2\right)}{y-2}-\frac{5}{y-2}}$$

Since \frac{\left(y+2\right)\left(y-2\right)}{y-2} and \frac{5}{y-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{y-3}{4y-8}}{\frac{\left(y+2\right)\left(y-2\right)-5}{y-2}}$$

Do the multiplications in \left(y+2\right)\left(y-2\right)-5.

$$\frac{\frac{y-3}{4y-8}}{\frac{y^{2}-2y+2y-4-5}{y-2}}$$

Combine like terms in y^{2}-2y+2y-4-5.

$$\frac{\frac{y-3}{4y-8}}{\frac{y^{2}-9}{y-2}}$$

Divide \frac{y-3}{4y-8} by \frac{y^{2}-9}{y-2} by multiplying \frac{y-3}{4y-8} by the reciprocal of \frac{y^{2}-9}{y-2}.

$$\frac{\left(y-3\right)\left(y-2\right)}{\left(4y-8\right)\left(y^{2}-9\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(y-3\right)\left(y-2\right)}{4\left(y-3\right)\left(y-2\right)\left(y+3\right)}$$

Cancel out \left(y-3\right)\left(y-2\right) in both numerator and denominator.

$$\frac{1}{4\left(y+3\right)}$$

Expand the expression.

$$\frac{1}{4y+12}$$

Expand

$$\frac{1}{4\left(y+3\right)}$$

Short Solution Steps

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

$$\frac{\frac{y-3}{4y-8}}{\frac{\left(y+2\right)\left(y-2\right)}{y-2}-\frac{5}{y-2}}$$

Since \frac{\left(y+2\right)\left(y-2\right)}{y-2} and \frac{5}{y-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{y-3}{4y-8}}{\frac{\left(y+2\right)\left(y-2\right)-5}{y-2}}$$

Do the multiplications in \left(y+2\right)\left(y-2\right)-5.

$$\frac{\frac{y-3}{4y-8}}{\frac{y^{2}-2y+2y-4-5}{y-2}}$$

Combine like terms in y^{2}-2y+2y-4-5.

$$\frac{\frac{y-3}{4y-8}}{\frac{y^{2}-9}{y-2}}$$

Divide \frac{y-3}{4y-8} by \frac{y^{2}-9}{y-2} by multiplying \frac{y-3}{4y-8} by the reciprocal of \frac{y^{2}-9}{y-2}.

$$\frac{\left(y-3\right)\left(y-2\right)}{\left(4y-8\right)\left(y^{2}-9\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(y-3\right)\left(y-2\right)}{4\left(y-3\right)\left(y-2\right)\left(y+3\right)}$$

Cancel out \left(y-3\right)\left(y-2\right) in both numerator and denominator.

$$\frac{1}{4\left(y+3\right)}$$

Expand the expression.

$$\frac{1}{4y+12}$$