Solve 9/2x^2+x-28-3/2x^2-9x+7

$$\frac { 9 } { 2 x ^ { 2 } + x - 28 } - \frac { 3 } { 2 x ^ { 2 } - 9 x + 7 }$$

Evaluate

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

Short Solution Steps

Factor 2x^{2}+x-28. Factor 2x^{2}-9x+7.

$$\frac{9}{\left(2x-7\right)\left(x+4\right)}-\frac{3}{\left(x-1\right)\left(2x-7\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2x-7\right)\left(x+4\right) and \left(x-1\right)\left(2x-7\right) is \left(x-1\right)\left(2x-7\right)\left(x+4\right). Multiply \frac{9}{\left(2x-7\right)\left(x+4\right)} times \frac{x-1}{x-1}. Multiply \frac{3}{\left(x-1\right)\left(2x-7\right)} times \frac{x+4}{x+4}.

$$\frac{9\left(x-1\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}-\frac{3\left(x+4\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}$$

Since \frac{9\left(x-1\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)} and \frac{3\left(x+4\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(x-1\right)-3\left(x+4\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}$$

Do the multiplications in 9\left(x-1\right)-3\left(x+4\right).

$$\frac{9x-9-3x-12}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}$$

Combine like terms in 9x-9-3x-12.

$$\frac{6x-21}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}$$

Factor the expressions that are not already factored in \frac{6x-21}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}.

$$\frac{3\left(2x-7\right)}{\left(x-1\right)\left(2x-7\right)\left(x+4\right)}$$

Cancel out 2x-7 in both numerator and denominator.

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

Expand \left(x-1\right)\left(x+4\right).

$$\frac{3}{x^{2}+3x-4}$$

Factor

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