Solve 1/1-9x^2-x+2/3x^2-16x+5

$$\frac { 1 } { 1 - 9 x ^ { 2 } } - \frac { x + 2 } { 3 x ^ { 2 } - 16 x + 5 }$$

Evaluate

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

Short Solution Steps

Factor 1-9x^{2}. Factor 3x^{2}-16x+5.

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

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

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

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

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

Do the multiplications in x-5-\left(x+2\right)\left(-3x-1\right).

$$\frac{x-5+3x^{2}+x+6x+2}{\left(x-5\right)\left(-3x-1\right)\left(3x-1\right)}$$

Combine like terms in x-5+3x^{2}+x+6x+2.

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

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

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

Cancel out 3x-1 in both numerator and denominator.

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

Expand \left(x-5\right)\left(-3x-1\right).

$$\frac{x+3}{-3x^{2}+14x+5}$$

Expand

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

Short Solution Steps

Factor 1-9x^{2}. Factor 3x^{2}-16x+5.

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

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

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

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

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

Do the multiplications in x-5-\left(x+2\right)\left(-3x-1\right).

$$\frac{x-5+3x^{2}+x+6x+2}{\left(x-5\right)\left(-3x-1\right)\left(3x-1\right)}$$

Combine like terms in x-5+3x^{2}+x+6x+2.

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

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

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

Cancel out 3x-1 in both numerator and denominator.

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

Expand \left(x-5\right)\left(-3x-1\right).

$$\frac{x+3}{-3x^{2}+14x+5}$$