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Chapter 0: Problem 55

Add or subtract as indicated. $$\frac{x+3}{x^{2}-1}-\frac{x+2}{x-1}$$

Short Answer

Expert verified
The solution to the problem is \(\frac{1-x}{x-1}\) for \(x鈮�-1\) and \(x鈮�1\).

Step by step solution

01

Rewrite the denominator

Rewrite the denominator since \(x^2-1 = (x-1)(x+1)\). Now we have \(\frac{x+3}{(x-1)(x+1)} - \frac{x+2}{x-1}\)
02

Find the least common denominator (LCD)

Now, to add or subtract fractions, the denominators must be the same. The LCD for these fractions is \((x-1)(x+1)\)
03

Rewrite the second fraction

Express \(\frac{x+2}{x-1}\) to have the common denominator. This yields \(\frac{(x+2)(x+1)}{(x-1)(x+1)}\).
04

Rewrite the expression

Now rewrite the original expression using equal denominators: \(\frac{x+3}{(x-1)(x+1)} - \frac{(x+2)(x+1)}{(x-1)(x+1)}\).
05

Subtract the fractions

Now that the denominators are the same, you subtract the numerators to yield: \(\frac{x+3 - [(x+2)(x+1)]}{(x-1)(x+1)}\).
06

Simplify the expression

Expand and simplify the numerator to obtain: \(\frac{x+3 - x^2-2x-2}{(x-1)(x+1)}\) which simplifies to \(\frac{-x^2 + x + 1}{(x-1)(x+1)}\)
07

Factor the simplified expression

Finally, the simplified expression can be factored to get \(\frac{(1-x)(x+1)}{(x-1)(x+1)}\) which simplifies to \(\frac{1-x}{x-1}\). However, this is only valid if \(x鈮�-1\) and \(x鈮�1\).

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