Solve 4/8x^2-18x-5-2/12x^2-5x-2

$$\frac { 4 } { 8 x ^ { 2 } - 18 x - 5 } - \frac { 2 } { 12 x ^ { 2 } - 5 x - 2 }$$

Evaluate

$$\frac{2}{\left(2x-5\right)\left(3x-2\right)}$$

Short Solution Steps

Factor 8x^{2}-18x-5. Factor 12x^{2}-5x-2.

$$\frac{4}{\left(2x-5\right)\left(4x+1\right)}-\frac{2}{\left(3x-2\right)\left(4x+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2x-5\right)\left(4x+1\right) and \left(3x-2\right)\left(4x+1\right) is \left(2x-5\right)\left(3x-2\right)\left(4x+1\right). Multiply \frac{4}{\left(2x-5\right)\left(4x+1\right)} times \frac{3x-2}{3x-2}. Multiply \frac{2}{\left(3x-2\right)\left(4x+1\right)} times \frac{2x-5}{2x-5}.

$$\frac{4\left(3x-2\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}-\frac{2\left(2x-5\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}$$

Since \frac{4\left(3x-2\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)} and \frac{2\left(2x-5\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\left(3x-2\right)-2\left(2x-5\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}$$

Do the multiplications in 4\left(3x-2\right)-2\left(2x-5\right).

$$\frac{12x-8-4x+10}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}$$

Combine like terms in 12x-8-4x+10.

$$\frac{8x+2}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}$$

Factor the expressions that are not already factored in \frac{8x+2}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}.

$$\frac{2\left(4x+1\right)}{\left(2x-5\right)\left(3x-2\right)\left(4x+1\right)}$$

Cancel out 4x+1 in both numerator and denominator.

$$\frac{2}{\left(2x-5\right)\left(3x-2\right)}$$

Expand \left(2x-5\right)\left(3x-2\right).

$$\frac{2}{6x^{2}-19x+10}$$

Factor

$$\frac{2}{\left(2x-5\right)\left(3x-2\right)}$$