الأسبوع 6 · العمليات على الكسور الجبرية النسبية: الجمع والطرح والضرب والقسمة والمعكوسان وخواص العمليات · 3 دروس
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Adding and subtracting rational algebraic fractions: factorise, common denominator, simplest form
Mathematics Prep 3 Term 1, Unit 2, Lesson 5
GoalAdd and subtract rational fractions, write the domain first, and give the answer in simplest form.
New words
common denominator · a denominator that every fraction in the sum can be changed tomissing factor · the factor of the common denominator that a fraction does not yet havedomain of a sum · all real numbers except the zeros of every denominatorsimplest form · numerator and denominator have no common factor left
Explanation
Adding rational fractions works like adding number fractions. (1) Factorise every denominator and find the domain: remove the zeros of ALL denominators. (2) Simplify each fraction if you can. (3) Find a common denominator that contains each denominator’s factors once (or as many times as needed). (4) Multiply the top of each fraction by the factor its denominator is missing. (5) Add or subtract the tops over the common denominator; when subtracting, put the whole second top in brackets so its sign changes everywhere. (6) Factorise the new top and cancel. Never add the denominators: 1/a + 1/b is not 2/(a + b). The domain of the result is the domain found in step 1, even if a factor cancels at the end.
Step
Work for 2x ÷ (x² − 1) − 1 ÷ (x − 1)
1 factorise, domain
(x − 1)(x + 1); R − {1, −1}
2 common denominator
(x − 1)(x + 1)
3 missing factor
(x + 1)
4 combine
[2x − (x + 1)] ÷ ((x − 1)(x + 1))
5 simplest form
1 ÷ (x + 1)
Worked examples
Add x ÷ (x − 2) + 3 ÷ (x + 2). Domain: R − {2, −2}. Common denominator (x − 2)(x + 2). Multiply the first top by (x + 2) and the second by (x − 2): [x(x + 2) + 3(x − 2)] ÷ (x² − 4) = (x² + 2x + 3x − 6) ÷ (x² − 4) = (x² + 5x − 6) ÷ (x² − 4) = (x + 6)(x − 1) ÷ (x² − 4). No factor cancels. Check at x = 3: 3 ÷ 1 + 3 ÷ 5 = 18/5, and (9 + 15 − 6) ÷ 5 = 18/5 ✓.
Add (x − 3) ÷ (x² − 9) + 2 ÷ (x + 3). Domain: x² − 9 = (x − 3)(x + 3), so R − {3, −3}. Simplify the first fraction: 1 ÷ (x + 3). The denominators are now equal: 1 ÷ (x + 3) + 2 ÷ (x + 3) = 3 ÷ (x + 3). Check at x = 1: (−2) ÷ (−8) + 2 ÷ 4 = 1/4 + 1/2 = 3/4 and 3 ÷ 4 = 3/4 ✓.
Subtract 2x ÷ (x² − 1) − 1 ÷ (x − 1). Domain R − {1, −1}. The second fraction is missing (x + 1): [2x − (x + 1)] ÷ ((x − 1)(x + 1)) = (2x − x − 1) ÷ ((x − 1)(x + 1)) = (x − 1) ÷ ((x − 1)(x + 1)) = 1 ÷ (x + 1). The domain stays R − {1, −1}. Check at x = 2: 4/3 − 1 = 1/3 ✓.
Class activity
Bracket check. Subtract 3 ÷ (x + 1) − (x + 4) ÷ (x + 1) twice: once keeping the brackets, once forgetting them. Which answer is right at x = 1? Test with numbers.
