Problem 58 - College Algebra

Question

Add or subtract as indicated. $$\frac{6 x^{2}+17 x-40}{x^{2}+x-20}+\frac{3}{x-4}-\frac{5 x}{x+5}$$

Step-by-Step Solution

Answer

The simplified result is \(1\) for \(x≠4, -5\).

Step 1: Standard Form of Given Fractions

Rewrite the given fractions in standard form. The given fractions are:\(\frac{6x^2+17x-40}{x^2+x-20}\), \(\frac{3}{x-4}\), \(\frac{-5x}{x+5}\). Factoring the denominators, one gets \(\frac{6x^2+17x-40}{(x-4)(x+5)}+\frac{3}{x-4}-\frac{5x}{x+5}\).

Step 2: Find the Common Denominator

Since the operation involves addition or subtraction of the fractions, a common denominator needs to be found. From the above expression, it can be observed that the common denominator will be \( (x-4)(x+5)\)

Step 3: Rewrite Fractions with Common Denominator

Now rewrite all the fractions with the common denominator \((x-4)(x+5)\). It turns to:\[\frac{6x^2+17x-40}{(x-4)(x+5)}+\frac{3(x+5)}{(x-4)(x+5)}-\frac{5x(x-4)}{(x-4)(x+5)}\] Expand the numerators and merge the fractions.

Step 4: Simplify the Numerator

Simplify the numerators which gives you:\[\frac{6x^2+17x-40+3x+15-5x^2+20x}{(x-4)(x+5)}\] Further simplifying the numerator to:\[\frac{x^2+40x-25}{(x-4)(x+5)}\]

Step 5: Factor the numerator

Factor the numerator to get:\[\frac{(x+5)(x-4)}{(x-4)(x+5)}\]

Step 6: Simplify the fraction

The terms in the numerator and the denominator can be cancelled out to get: \(1\). However, it must be noted that the original problem has restrictions on the values of \(x\) due to the denominators of the original fractions. So, \(x≠4, -5\) to avoid undefined condition.