Problem 58 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-7}{x+4}+\frac{x+4}{x-7}$$

Step-by-Step Solution

Answer

The simplified result is \(\frac{2x^2-6x+65}{(x+4)*(x-7)}\), where x != -4 or 7.

Step 1: Write out the fractions and the operation

The fractions are \(\frac{x-7}{x+4}\) and \(\frac{x+4}{x-7}\). The operation is addition.

Step 2: Cross-multiply

Cross-multiply the fractions which results in two new fractions that can be added together. This is done by multiplying the numerator of the first fraction by the denominator of the second fraction and the numerator of the second fraction by the denominator of the first fraction. This gives you \(\frac{(x-7)*(x-7)}{(x+4)*(x-7)}+ \frac{(x+4)*(x+4)}{(x-7)*(x+4)}\).

Step 3: Simplify the fractions

Now we simplify fractions and these expressions become \(\frac{(x-7)^2}{(x+4)*(x-7)}+ \frac{(x+4)^2}{(x-7)*(x+4)}\).

Step 4: Add the fractions

Now that the fractions have the same denominator, they can be added together. This results in \(\frac{(x-7)^2 + (x+4)^2}{(x+4)*(x-7)}\).

Step 5: Final Simplification

Expand the squared binomials in the numerator, and simplify the entire expression: \(\frac{(x^2-14x+49) + (x^2+8x+16)}{(x+4)*(x-7)}\) which simplifies to \(\frac{2x^2-6x+65}{(x+4)*(x-7)}\). This is the simplified result. Keep in mind that x cannot equal -4 or 7, as these values would make the denominator equal to zero.