Problem 43 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-3}{x(x+1)}\).

Step 1: Identifying the common denominator

The common denominator needs to be a multiple of both \(x+1\) and \(x\). Since these two expressions are relatively prime (i.e., they don't share factors), their product, \(x(x+1)\), is a common denominator.

Step 2: Adjusting the fractions to have a common denominator

You can rewrite \(\frac{3}{x+1}\) as \(\frac{3x}{x(x+1)}\) and \(\frac{3}{x}\) as \(\frac{3(x+1)}{x(x+1)}\), each time multiplying the numerator and denominator of the fraction by the missing expression from the common denominator. So \(\frac{3}{x+1}-\frac{3}{x}\) becomes \(\frac{3x}{x(x+1)} - \frac{3(x+1)}{x(x+1)}\).

Step 3: Subtraction

Subtract the new expressions. \[\frac{3x}{x(x+1)} - \frac{3(x+1)}{x(x+1)} = \frac{3x - 3(x+1)}{x(x+1)}\] This simplifies to \(\frac{3x - 3x - 3}{x(x+1)}\) or \(\frac{-3}{x(x+1)}\).