Problem 72 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{x}{x^{2}-10 x+25}-\frac{x-4}{2 x-10}$$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-x^2+21x-80}{2(x-5)^2}\)

Step 1: Determine the least common denominator (LCD)

The denominators of the given fractions are \(x^{2}-10x+25\) and \(2x-10\). The factorization of the first fraction gives \((x-5)^2\), and the factorization of the second fraction gives \(2(x-5)\). Thus, the LCD of these two fractions is \(2(x-5)^2\) because it's the smallest exponent of the common factor \((x-5)\), which appears in the two fractions.

Step 2: Transform the fractions into equivalent fractions with the common denominator

The first fraction's denominator is missing a factor of 2 from the LCD. Therefore, multiply the numerator and the denominator of the first fraction by 2. The first fraction becomes \(\frac{2x}{2(x-5)^2}\). The second fraction's denominator is missing a factor of \((x-5)\) from the LCD. Therefore, multiply the numerator and the denominator of the second fraction by \((x-5)\). The second fraction becomes \(\frac{(x-4)(x-5)}{2(x-5)^2}\). The new fraction arrangement appears as \[ \frac{2x}{2(x-5)^2} - \frac{(x-4)(x-5)}{2(x-5)^2}\]

Step 3: Simplify the fractions

Since the fractions now have the same denominator, subtract the numerators \[ \frac{2x-(x-4)(x-5)}{2(x-5)^2}\]. Now expand \((x-4)(x-5)\) in the numerator and simplify to get: \[\frac{2x-x^2+20x-80}{2(x-5)^2} = \frac{21x-80-x^2}{2(x-5)^2}\]. The simplified expression is \[\frac{-x^2+21x-80}{2(x-5)^2}\].