Solve 6/4b-8+2b/3

$$\frac { 6 } { 4 b - 8 } + \frac { 2 b } { 3 }$$

Evaluate

$$\frac{4b^{2}-8b+9}{6\left(b-2\right)}$$

Short Solution Steps

Factor 4b-8.

$$\frac{6}{4\left(b-2\right)}+\frac{2b}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(b-2\right) and 3 is 12\left(b-2\right). Multiply \frac{6}{4\left(b-2\right)} times \frac{3}{3}. Multiply \frac{2b}{3} times \frac{4\left(b-2\right)}{4\left(b-2\right)}.

$$\frac{6\times 3}{12\left(b-2\right)}+\frac{2b\times 4\left(b-2\right)}{12\left(b-2\right)}$$

Since \frac{6\times 3}{12\left(b-2\right)} and \frac{2b\times 4\left(b-2\right)}{12\left(b-2\right)} have the same denominator, add them by adding their numerators.

$$\frac{6\times 3+2b\times 4\left(b-2\right)}{12\left(b-2\right)}$$

Do the multiplications in 6\times 3+2b\times 4\left(b-2\right).

$$\frac{18+8b^{2}-16b}{12\left(b-2\right)}$$

Factor the expressions that are not already factored in \frac{18+8b^{2}-16b}{12\left(b-2\right)}.

$$\frac{2\left(4b^{2}-8b+9\right)}{12\left(b-2\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4b^{2}-8b+9}{6\left(b-2\right)}$$

Expand 6\left(b-2\right).

$$\frac{4b^{2}-8b+9}{6b-12}$$

Factor

$$\frac{4b^{2}-8b+9}{6\left(b-2\right)}$$