Solve 2a/2a-1-1/2a-4a^2=

$$\frac { 2 a } { 2 a - 1 } - \frac { 1 } { 2 a - 4 a ^ { 2 } } =$$

Evaluate

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

Short Solution Steps

Factor 2a-4a^{2}.

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

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

$$\frac{2a\times 2a}{2a\left(2a-1\right)}-\frac{-1}{2a\left(2a-1\right)}$$

Since \frac{2a\times 2a}{2a\left(2a-1\right)} and \frac{-1}{2a\left(2a-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2a\times 2a-\left(-1\right)}{2a\left(2a-1\right)}$$

Do the multiplications in 2a\times 2a-\left(-1\right).

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

Expand 2a\left(2a-1\right).

$$\frac{4a^{2}+1}{4a^{2}-2a}$$

Factor

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