Problem 44 - College Algebra

Question

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

Step-by-Step Solution

Answer

The answer after subtracting the fractions is \(\frac{x^2 + 12x}{x^2 + 3x}\)

Step 1 - Identify the Least Common Denominator (LCD)

In this case, the LCD is not simply the product of \(x\) and \(x+3\). Looking closely, \(x\) is common to both denominators; hence, we only need to add \(3\) to match \(x+3\). So, the LCD is \(x(x+3) = x^2 + 3x\).

Step 2 - Adjust Each Fraction

Next, each fraction is adjusted to have the same denominator. For every fraction, divide the LCD by the original denominator and multiply the result by the numerator. This gives: \[4\left(\frac{x^2+3x}{x}\right) - 3\left(\frac{x^2+3x}{x+3}\right) = \frac{4x^2+12x-3x^2}{x^2+3x}\]

Step 3 - Simplify the Fraction

Now, simplify the fraction by performing the basic subtraction in the numerator, and rearranging it: \[\frac{4x^2+12x-3x^2}{x^2+3x} = \frac{x^2 + 12x}{x^2 + 3x}\]This fraction cannot be simplified further as no common factor is present in the numerator and denominator.