Solve x-1/x^2+x-3/3x^2+7x+4-x-5/x^2-2x/frac{31x+28{xleft(x+1right)left(3x+4right)left(x-2right)}}

$$\frac{ \frac{ x-1 }{ { x }^{ 2 } +x } - \frac{ 3 }{ 3 { x }^{ 2 } +7x+4 } - \frac{ x-5 }{ { x }^{ 2 } -2x } }{ \frac{ 31x+28 }{ x \left( x+1 \right) \left( 3x+4 \right) \left( x-2 \right) } }$$

Evaluate

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Short Solution Steps

Factor x^{2}+x. Factor 3x^{2}+7x+4.

$$\frac{\frac{x-1}{x\left(x+1\right)}-\frac{3}{\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x^{2}-2x}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

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

$$\frac{\frac{\left(x-1\right)\left(3x+4\right)}{x\left(x+1\right)\left(3x+4\right)}-\frac{3x}{x\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x^{2}-2x}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

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

$$\frac{\frac{\left(x-1\right)\left(3x+4\right)-3x}{x\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x^{2}-2x}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

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

$$\frac{\frac{3x^{2}+4x-3x-4-3x}{x\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x^{2}-2x}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

Combine like terms in 3x^{2}+4x-3x-4-3x.

$$\frac{\frac{3x^{2}-2x-4}{x\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x^{2}-2x}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

Factor x^{2}-2x.

$$\frac{\frac{3x^{2}-2x-4}{x\left(x+1\right)\left(3x+4\right)}-\frac{x-5}{x\left(x-2\right)}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

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

$$\frac{\frac{\left(3x^{2}-2x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)}-\frac{\left(x-5\right)\left(x+1\right)\left(3x+4\right)}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

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

$$\frac{\frac{\left(3x^{2}-2x-4\right)\left(x-2\right)-\left(x-5\right)\left(x+1\right)\left(3x+4\right)}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

Do the multiplications in \left(3x^{2}-2x-4\right)\left(x-2\right)-\left(x-5\right)\left(x+1\right)\left(3x+4\right).

$$\frac{\frac{3x^{3}-6x^{2}-2x^{2}+4x-4x+8-3x^{3}-7x^{2}-4x+15x^{2}+35x+20}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

Combine like terms in 3x^{3}-6x^{2}-2x^{2}+4x-4x+8-3x^{3}-7x^{2}-4x+15x^{2}+35x+20.

$$\frac{\frac{31x+28}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)}}{\frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}}$$

Divide \frac{31x+28}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)} by \frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)} by multiplying \frac{31x+28}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)} by the reciprocal of \frac{31x+28}{x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}.

$$\frac{\left(31x+28\right)x\left(x+1\right)\left(3x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+1\right)\left(3x+4\right)\left(31x+28\right)}$$

Cancel out x\left(x-2\right)\left(x+1\right)\left(3x+4\right)\left(31x+28\right) in both numerator and denominator.

$$1$$

Factor

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