Use whole-factor denominators for sums, products and quotients.
Extended · Syllabus E2.3
Notes and key methods
A sum of fractions requires a common denominator and all corresponding numerator factors. Multiplication uses the product of the numerators over the product of the denominators. For division, invert the entire second fraction.
Keep exclusions from each original denominator, and exclude any input that makes a divisor fraction zero. Simplify only after correct whole-factor operations; cancellation across addition is invalid.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Operations with algebraic fractions
Let \(\displaystyle A=\)\(\displaystyle \frac{6}{x - 4}\) and \(\displaystyle B=\)\(\displaystyle \frac{3}{x - 6}\). Every operation uses the common original domain \(\displaystyle x\ne 4, 6\).
Find A+B as one fraction and state its original exclusions.
Simplify \(\displaystyle \frac{A}{B}\) and state its original exclusions.
Evaluate A+B at \(\displaystyle x=7\).
Write \(\displaystyle \frac{A}{B}= \frac{K(x- 6)}{x- 4}\). Find K.
Show the reasoning
(a)
Multiply each numerator by its missing denominator factor. Their sum is \(9 x - 48\). Neither original denominator may vanish.
Multiply \(\displaystyle \frac{6}{x - 4}\) by \(\displaystyle \frac{x}{3} - 2\). The nonzero constant numerator of B gives no further zeros, but B must itself remain defined.
Answer: \(\displaystyle \frac{2 x - 12}{x - 4}\), \(x\ne 4,6\)
(c)
At this permitted input, direct substitution gives \(2\) + \(3\) = \(5\).
Answer: \(5\)
(d)
Use a common denominator for addition, multiply whole factors for a product, and invert the full divisor for a quotient.
\(2\)
Answer: \(2\).
Example 2 · Operations with algebraic fractions
Let \(\displaystyle A=\)\(\displaystyle \frac{7}{x - 5}\) and \(\displaystyle B=\)\(\displaystyle \frac{4}{x - 8}\). Every operation uses the common original domain \(\displaystyle x\ne 5, 8\).
Find A+B as one fraction and state its original exclusions.
Simplify \(\displaystyle \frac{A}{B}\) and state its original exclusions.
Evaluate A+B at \(\displaystyle x=9\).
Write \(\displaystyle \frac{A}{B}= \frac{K(x- 8)}{x- 5}\). Find K.
Show the reasoning
(a)
Multiply each numerator by its missing denominator factor. Their sum is \(11 x - 76\). Neither original denominator may vanish.
Multiply \(\displaystyle \frac{7}{x - 5}\) by \(\displaystyle \frac{x}{4} - 2\). The nonzero constant numerator of B gives no further zeros, but B must itself remain defined.
At this permitted input, direct substitution gives \(\displaystyle \frac{7}{4}\) + \(4\) = \(\displaystyle \frac{23}{4}\).
Answer: \(\displaystyle \frac{23}{4}\)
(d)
Use a common denominator for addition, multiply whole factors for a product, and invert the full divisor for a quotient.
\(\displaystyle \frac{7}{4}\)
Answer: \(\displaystyle \frac{7}{4}\).
Example 3 · Operations with algebraic fractions
Let \(\displaystyle A=\)\(\displaystyle \frac{2}{x - 1}\) and \(\displaystyle B=\)\(\displaystyle \frac{5}{x - 5}\). Every operation uses the common original domain \(\displaystyle x\ne 1, 5\).
Find A+B as one fraction and state its original exclusions.
Simplify \(\displaystyle \frac{A}{B}\) and state its original exclusions.
Evaluate A+B at \(\displaystyle x=6\).
Write \(\displaystyle \frac{A}{B}= \frac{K(x- 5)}{x- 1}\). Find K.
Show the reasoning
(a)
Multiply each numerator by its missing denominator factor. Their sum is \(7 x - 15\). Neither original denominator may vanish.
Multiply \(\displaystyle \frac{2}{x - 1}\) by \(\displaystyle \frac{x}{5} - 1\). The nonzero constant numerator of B gives no further zeros, but B must itself remain defined.