Problem 46 - College Algebra

Question

Add or subtract as indicated. $$\frac{3 x}{x-3}-\frac{x+4}{x+2}$$

Step-by-Step Solution

Answer

The result of the operation \(\frac{3x}{x-3} - \frac{x+4}{x+2}\) is \(\frac{(2x-3)(x+4)}{(x-3)(x+2)}\).

Step 1: Identify the denominators

The denominators of the fractions are \(x-3\) and \(x+2\). In order to add or subtract fractions, they must have a common denominator.

Step 2: Find the common denominator

To find a common denominator, the denominators can be multiplied together to get \((x-3) * (x+2)\) as the common denominator.

Step 3: Rewrite the fractions with the common denominator

Rewrite the fractions such that they both have the common denominator. This gives \(\frac{3x * (x+2)}{(x-3)*(x+2)} - \frac{(x+4) * (x-3)}{(x-3)*(x+2)}\).

Step 4: Expand the numerators

Expand the numerators: \(\frac{3x*x+6x}{(x-3)*(x+2)} - \frac{x*x -3x+12}{(x-3)*(x+2)}\) which simlifies to \(\frac{3x^2+6x}{x^2 -x -6} - \frac{x^2 -3x +12}{x^2 -x -6}\).

Step 5: Subtract the fractions

Now the fractions can be subtracted since they have the same denominator. This gives \(\frac{3x^2+6x - (x^2 -3x +12)}{x^2 -x -6}\) which simplifies to \(\frac{2x^2+9x -12}{x^2 -x -6}\).

Step 6: Factorize the numerator and denominator, if possible

In this case, the numerator and denominator can be factorized, giving \(\frac{(2x-3)(x+4)}{(x-3)(x+2)}\).