Solve 4b/4b^2-1+2b+1/3-6b+2b-1/4b+2=

$$\frac { 4 b } { 4 b ^ { 2 } - 1 } + \frac { 2 b + 1 } { 3 - 6 b } + \frac { 2 b - 1 } { 4 b + 2 } =$$

Evaluate

$$\frac{2b+1}{6\left(2b-1\right)}$$

Short Solution Steps

Factor 4b^{2}-1. Factor 3-6b.

$$\frac{4b}{\left(2b-1\right)\left(2b+1\right)}+\frac{2b+1}{3\left(-2b+1\right)}+\frac{2b-1}{4b+2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2b-1\right)\left(2b+1\right) and 3\left(-2b+1\right) is 3\left(2b-1\right)\left(2b+1\right). Multiply \frac{4b}{\left(2b-1\right)\left(2b+1\right)} times \frac{3}{3}. Multiply \frac{2b+1}{3\left(-2b+1\right)} times \frac{-\left(2b+1\right)}{-\left(2b+1\right)}.

$$\frac{3\times 4b}{3\left(2b-1\right)\left(2b+1\right)}+\frac{\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Since \frac{3\times 4b}{3\left(2b-1\right)\left(2b+1\right)} and \frac{\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\times 4b+\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Do the multiplications in 3\times 4b+\left(2b+1\right)\left(-1\right)\left(2b+1\right).

$$\frac{12b-4b^{2}-2b-2b-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Combine like terms in 12b-4b^{2}-2b-2b-1.

$$\frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Factor 4b+2.

$$\frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{2\left(2b+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(2b-1\right)\left(2b+1\right) and 2\left(2b+1\right) is 6\left(2b-1\right)\left(2b+1\right). Multiply \frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)} times \frac{2}{2}. Multiply \frac{2b-1}{2\left(2b+1\right)} times \frac{3\left(2b-1\right)}{3\left(2b-1\right)}.

$$\frac{2\left(8b-4b^{2}-1\right)}{6\left(2b-1\right)\left(2b+1\right)}+\frac{\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)}$$

Since \frac{2\left(8b-4b^{2}-1\right)}{6\left(2b-1\right)\left(2b+1\right)} and \frac{\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(8b-4b^{2}-1\right)+\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)}$$

Do the multiplications in 2\left(8b-4b^{2}-1\right)+\left(2b-1\right)\times 3\left(2b-1\right).

$$\frac{16b-8b^{2}-2+12b^{2}-6b+3-6b}{6\left(2b-1\right)\left(2b+1\right)}$$

Combine like terms in 16b-8b^{2}-2+12b^{2}-6b+3-6b.

$$\frac{4b+4b^{2}+1}{6\left(2b-1\right)\left(2b+1\right)}$$

Factor the expressions that are not already factored in \frac{4b+4b^{2}+1}{6\left(2b-1\right)\left(2b+1\right)}.

$$\frac{\left(2b+1\right)^{2}}{6\left(2b-1\right)\left(2b+1\right)}$$

Cancel out 2b+1 in both numerator and denominator.

$$\frac{2b+1}{6\left(2b-1\right)}$$

Expand 6\left(2b-1\right).

$$\frac{2b+1}{12b-6}$$

Expand

$$\frac{2b+1}{6\left(2b-1\right)}$$

Short Solution Steps

Factor 4b^{2}-1. Factor 3-6b.

$$\frac{4b}{\left(2b-1\right)\left(2b+1\right)}+\frac{2b+1}{3\left(-2b+1\right)}+\frac{2b-1}{4b+2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2b-1\right)\left(2b+1\right) and 3\left(-2b+1\right) is 3\left(2b-1\right)\left(2b+1\right). Multiply \frac{4b}{\left(2b-1\right)\left(2b+1\right)} times \frac{3}{3}. Multiply \frac{2b+1}{3\left(-2b+1\right)} times \frac{-\left(2b+1\right)}{-\left(2b+1\right)}.

$$\frac{3\times 4b}{3\left(2b-1\right)\left(2b+1\right)}+\frac{\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Since \frac{3\times 4b}{3\left(2b-1\right)\left(2b+1\right)} and \frac{\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{3\times 4b+\left(2b+1\right)\left(-1\right)\left(2b+1\right)}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Do the multiplications in 3\times 4b+\left(2b+1\right)\left(-1\right)\left(2b+1\right).

$$\frac{12b-4b^{2}-2b-2b-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Combine like terms in 12b-4b^{2}-2b-2b-1.

$$\frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{4b+2}$$

Factor 4b+2.

$$\frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)}+\frac{2b-1}{2\left(2b+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\left(2b-1\right)\left(2b+1\right) and 2\left(2b+1\right) is 6\left(2b-1\right)\left(2b+1\right). Multiply \frac{8b-4b^{2}-1}{3\left(2b-1\right)\left(2b+1\right)} times \frac{2}{2}. Multiply \frac{2b-1}{2\left(2b+1\right)} times \frac{3\left(2b-1\right)}{3\left(2b-1\right)}.

$$\frac{2\left(8b-4b^{2}-1\right)}{6\left(2b-1\right)\left(2b+1\right)}+\frac{\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)}$$

Since \frac{2\left(8b-4b^{2}-1\right)}{6\left(2b-1\right)\left(2b+1\right)} and \frac{\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(8b-4b^{2}-1\right)+\left(2b-1\right)\times 3\left(2b-1\right)}{6\left(2b-1\right)\left(2b+1\right)}$$

Do the multiplications in 2\left(8b-4b^{2}-1\right)+\left(2b-1\right)\times 3\left(2b-1\right).

$$\frac{16b-8b^{2}-2+12b^{2}-6b+3-6b}{6\left(2b-1\right)\left(2b+1\right)}$$

Combine like terms in 16b-8b^{2}-2+12b^{2}-6b+3-6b.

$$\frac{4b+4b^{2}+1}{6\left(2b-1\right)\left(2b+1\right)}$$

Factor the expressions that are not already factored in \frac{4b+4b^{2}+1}{6\left(2b-1\right)\left(2b+1\right)}.

$$\frac{\left(2b+1\right)^{2}}{6\left(2b-1\right)\left(2b+1\right)}$$

Cancel out 2b+1 in both numerator and denominator.

$$\frac{2b+1}{6\left(2b-1\right)}$$

Expand 6\left(2b-1\right).

$$\frac{2b+1}{12b-6}$$