Solve -6/x^2-4x+3-3/3-x-4/x-1

$$\frac{ -6 }{ { x }^{ 2 } -4x+3 } - \frac{ 3 }{ 3-x } - \frac{ 4 }{ x-1 }$$

Evaluate

$$-\frac{1}{x-1}$$

Short Solution Steps

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

$$\frac{-6}{\left(x-3\right)\left(x-1\right)}-\frac{3}{3-x}-\frac{4}{x-1}$$

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

$$\frac{-6}{\left(x-3\right)\left(x-1\right)}-\frac{3\left(-1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{4}{x-1}$$

Since \frac{-6}{\left(x-3\right)\left(x-1\right)} and \frac{3\left(-1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-6-3\left(-1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{4}{x-1}$$

Do the multiplications in -6-3\left(-1\right)\left(x-1\right).

$$\frac{-6+3x-3}{\left(x-3\right)\left(x-1\right)}-\frac{4}{x-1}$$

Combine like terms in -6+3x-3.

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

Factor the expressions that are not already factored in \frac{-9+3x}{\left(x-3\right)\left(x-1\right)}.

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

Cancel out x-3 in both numerator and denominator.

$$\frac{3}{x-1}-\frac{4}{x-1}$$

Since \frac{3}{x-1} and \frac{4}{x-1} have the same denominator, subtract them by subtracting their numerators. Subtract 4 from 3 to get -1.

$$\frac{-1}{x-1}$$

Differentiate w.r.t. x

$$\frac{1}{\left(x-1\right)^{2}}$$