Intermediate Algebra
Simplify x5x2\(\frac{x^5}{x^2}\), assuming x≠0x ≠ 0.
Simplify x+52x+10\(\frac{x+5}{2x+10}\) by factoring out the greatest common factor, and state any excluded value.
Simplify x3−x2−6xx2−4\(\frac{x^3-x^2-6x}{x^2-4}\) and state any domain exclusions.
Which of the following denominators is already fully factored and is irreducible over the integer values?
Factor the numerical denominators 8484 and 3636 completely into prime factors, then find the LCD of 184\(\frac{1}{84}\) and 136\(\frac{1}{36}\).
Rewrite the pair of rational expressions 130x\(\frac{1}{30x}\) and 120x2\(\frac{1}{20x^2}\) so both have the LCD 60x260x^2.
Simplify 3x−6x−2\(\frac{3x-6}{x-2}\).
Given 5x+3−2x−1\(\frac{5}{x+3}\)-\(\frac{2}{x-1}\), form a single rational expression, expand and combine terms, and present the simplified numerator.
Compute 12x+3x+1\(\frac{1}{2x}\)+\(\frac{3}{x+1}\).
Solve the rational equation 16+1x=14\(\displaystyle\]\frac\)16+\(\frac{1}{x}\)=\(\frac\)14 for xx.
Solve the following rational equation for xx:
1x2−5x+6+1x−2=1x+3\(\displaystyle\]\frac{1}{x^2 - 5x + 6}\)+\(\frac{1}{x - 2}\)=\(\frac{1}{x + 3}\)
Which justification best explains why multiplying both sides of a rational equation by the LCD is a valid step when solving rational equations (provided you track restrictions)?