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

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

Evaluate

$$0$$

Short Solution Steps

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

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

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

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

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

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

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

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

Combine like terms in -4-x+1.

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

Extract the negative sign in -3-x.

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

Cancel out x+3 in both numerator and denominator.

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

Since \frac{-1}{x-1} and \frac{1}{x-1} have the same denominator, add them by adding their numerators. Add -1 and 1 to get 0.

$$\frac{0}{x-1}$$

Zero divided by any non-zero term gives zero.

$$0$$

Factor

$$0$$