Solve 15b/3-b+8b/b^2-9*10+21/4

$$\frac { 15 b } { 3 - b } + \frac { 8 b } { b ^ { 2 } - 9 } \cdot \frac { 10 + 21 } { 4 }$$

Evaluate

$$\frac{b\left(15b-17\right)}{9-b^{2}}$$

Short Solution Steps

Add 10 and 21 to get 31.

$$\frac{15b}{3-b}+\frac{8b}{b^{2}-9}\times \frac{31}{4}$$

Multiply \frac{8b}{b^{2}-9} times \frac{31}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{15b}{3-b}+\frac{8b\times 31}{\left(b^{2}-9\right)\times 4}$$

Cancel out 4 in both numerator and denominator.

$$\frac{15b}{3-b}+\frac{2\times 31b}{b^{2}-9}$$

Factor b^{2}-9.

$$\frac{15b}{3-b}+\frac{2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3-b and \left(b-3\right)\left(b+3\right) is \left(b-3\right)\left(b+3\right). Multiply \frac{15b}{3-b} times \frac{-\left(b+3\right)}{-\left(b+3\right)}.

$$\frac{15b\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)}+\frac{2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

Since \frac{15b\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)} and \frac{2\times 31b}{\left(b-3\right)\left(b+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{15b\left(-1\right)\left(b+3\right)+2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

Do the multiplications in 15b\left(-1\right)\left(b+3\right)+2\times 31b.

$$\frac{-15b^{2}-45b+62b}{\left(b-3\right)\left(b+3\right)}$$

Combine like terms in -15b^{2}-45b+62b.

$$\frac{-15b^{2}+17b}{\left(b-3\right)\left(b+3\right)}$$

Expand \left(b-3\right)\left(b+3\right).

$$\frac{-15b^{2}+17b}{b^{2}-9}$$

Expand

$$-\frac{15b^{2}-17b}{b^{2}-9}$$

Short Solution Steps

Add 10 and 21 to get 31.

$$\frac{15b}{3-b}+\frac{8b}{b^{2}-9}\times \frac{31}{4}$$

Multiply \frac{8b}{b^{2}-9} times \frac{31}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{15b}{3-b}+\frac{8b\times 31}{\left(b^{2}-9\right)\times 4}$$

Cancel out 4 in both numerator and denominator.

$$\frac{15b}{3-b}+\frac{2\times 31b}{b^{2}-9}$$

Factor b^{2}-9.

$$\frac{15b}{3-b}+\frac{2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3-b and \left(b-3\right)\left(b+3\right) is \left(b-3\right)\left(b+3\right). Multiply \frac{15b}{3-b} times \frac{-\left(b+3\right)}{-\left(b+3\right)}.

$$\frac{15b\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)}+\frac{2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

Since \frac{15b\left(-1\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)} and \frac{2\times 31b}{\left(b-3\right)\left(b+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{15b\left(-1\right)\left(b+3\right)+2\times 31b}{\left(b-3\right)\left(b+3\right)}$$

Do the multiplications in 15b\left(-1\right)\left(b+3\right)+2\times 31b.

$$\frac{-15b^{2}-45b+62b}{\left(b-3\right)\left(b+3\right)}$$

Combine like terms in -15b^{2}-45b+62b.

$$\frac{-15b^{2}+17b}{\left(b-3\right)\left(b+3\right)}$$

Expand \left(b-3\right)\left(b+3\right).

$$\frac{-15b^{2}+17b}{b^{2}-9}$$