Solve 5h+4/h/h-frac{15{h}}=

$$\frac { 5 h + \frac { 4 } { h } } { h - \frac { 15 } { h } } =$$

Evaluate

$$\frac{5h^{2}+4}{h^{2}-15}$$

Short Solution Steps

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

$$\frac{\frac{5hh}{h}+\frac{4}{h}}{h-\frac{15}{h}}$$

Since \frac{5hh}{h} and \frac{4}{h} have the same denominator, add them by adding their numerators.

$$\frac{\frac{5hh+4}{h}}{h-\frac{15}{h}}$$

Do the multiplications in 5hh+4.

$$\frac{\frac{5h^{2}+4}{h}}{h-\frac{15}{h}}$$

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

$$\frac{\frac{5h^{2}+4}{h}}{\frac{hh}{h}-\frac{15}{h}}$$

Since \frac{hh}{h} and \frac{15}{h} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{5h^{2}+4}{h}}{\frac{hh-15}{h}}$$

Do the multiplications in hh-15.

$$\frac{\frac{5h^{2}+4}{h}}{\frac{h^{2}-15}{h}}$$

Divide \frac{5h^{2}+4}{h} by \frac{h^{2}-15}{h} by multiplying \frac{5h^{2}+4}{h} by the reciprocal of \frac{h^{2}-15}{h}.

$$\frac{\left(5h^{2}+4\right)h}{h\left(h^{2}-15\right)}$$

Cancel out h in both numerator and denominator.

$$\frac{5h^{2}+4}{h^{2}-15}$$

Expand

$$\frac{5h^{2}+4}{h^{2}-15}$$

Short Solution Steps

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

$$\frac{\frac{5hh}{h}+\frac{4}{h}}{h-\frac{15}{h}}$$

Since \frac{5hh}{h} and \frac{4}{h} have the same denominator, add them by adding their numerators.

$$\frac{\frac{5hh+4}{h}}{h-\frac{15}{h}}$$

Do the multiplications in 5hh+4.

$$\frac{\frac{5h^{2}+4}{h}}{h-\frac{15}{h}}$$

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

$$\frac{\frac{5h^{2}+4}{h}}{\frac{hh}{h}-\frac{15}{h}}$$

Since \frac{hh}{h} and \frac{15}{h} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{5h^{2}+4}{h}}{\frac{hh-15}{h}}$$

Do the multiplications in hh-15.

$$\frac{\frac{5h^{2}+4}{h}}{\frac{h^{2}-15}{h}}$$

Divide \frac{5h^{2}+4}{h} by \frac{h^{2}-15}{h} by multiplying \frac{5h^{2}+4}{h} by the reciprocal of \frac{h^{2}-15}{h}.

$$\frac{\left(5h^{2}+4\right)h}{h\left(h^{2}-15\right)}$$

Cancel out h in both numerator and denominator.

$$\frac{5h^{2}+4}{h^{2}-15}$$