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Rational numbers · 4 marks

1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?
2. Verify the distributive law for rational numbers.

  1. 1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?
  2. 2. Verify the distributive law for rational numbers.
Answer: 1. Rewrite each fraction as an equivalent fraction whose denominator is a common multiple of both denominators (best: the LCM). E.g. 14 + 16 = 312 + 212 = 512. 2. With p = 12, q = 23, r = −14: both p(q + r) and pq + pr equal 524, so the law holds.

Step-by-step solution

Idea: Multiplying the top and bottom of a fraction by the same non-zero number does not change its value. That lets us give two fractions the same denominator. For the distributive law, work out both sides separately and compare.

1. While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

  1. Find a common multiple of the two denominators; the smallest one, the LCM, keeps the numbers small. (Multiplying the two denominators always works too: b × d.)½ mark
  2. Change each fraction to an equivalent fraction with that denominator by multiplying its numerator and denominator by the same number: ab = adbd and cd = bcbd.½ mark
  3. Now add or subtract the numerators and keep the common denominator. Example: 14 + 16: LCM of 4 and 6 is 12, so 14 = 312, 16 = 212 and the sum is 512.1 mark
Rewrite both as equivalent fractions with a common denominator (the LCM), then add or subtract the numerators.

2. Verify the distributive law for rational numbers.

  1. Distributive law: p(q + r) = pq + pr. Take p = 12, q = 23, r = −14.½ mark
  2. LHS: q + r = 23 − 14 = 812 − 312 = 512. So p(q + r) = 12 × 512 = 524.½ mark
  3. RHS: pq = 12 × 23 = 26 = 13 and pr = 12 × −14 = −18. So pq + pr = 824 − 324 = 524.½ mark
  4. LHS = RHS = 524, so the distributive law holds for these rational numbers. (In general, ab(cd + ef) = ab × cf + dedf = acf + adebdf = acbd + aebf, because integers obey the distributive law.)½ mark
12(23 + −14) = 524 = 12 × 23 + 12 × −14 ✓
1. Convert both fractions to equivalent fractions over a common denominator (the LCM), then add or subtract numerators. 2. For p = 1/2, q = 2/3, r = −1/4, both p(q + r) and pq + pr equal 5/24, verifying the distributive law.

Check: 512 ≈ 0.4167 and 14 + 16 = 0.25 + 0.1667 = 0.4167 ✓. 524 ≈ 0.2083 = 0.3333 − 0.125 ✓.

Answer to write in the exam

1.

Take the LCM of the denominators as the common denominator.

ab = adbd, cd = bcbd (multiply top and bottom by the same number)

e.g. 14 + 16 = 312 + 212 = 512

∴ Convert to equivalent fractions with a common denominator, then add/subtract numerators.

2.

p = 12, q = 23, r = −14

LHS = p(q + r) = 12 × (812 − 312) = 12 × 512 = 524

RHS = pq + pr = 13 + −18 = 824 − 324 = 524

∴ LHS = RHS; the distributive law is verified.

Common mistakes that cost marks

  • Adding tops and bottoms: 14 + 16 = 210. Only the numerators are added, after the denominators are made equal.
  • Changing the denominator without changing the numerator, e.g. writing 14 = 112.
  • In the distributive law, multiplying p by q only and forgetting pr.

How this can come in the exam

MCQ (1 mark)

56 − 38 =

  1. 22
  2. 1124
  3. 224
  4. 2924
Show answer

(B) 1124
LCM 24: 2024 − 924 = 1124.

Short answer (2 marks)

Verify p(q + r) = pq + pr for p = −25, q = 13, r = 16.

Show answerLHS: 13 + 16 = 12, and −25 × 12 = −15 (1 mark). RHS: −215 + −230 = −430 + −230 = −630 = −15. LHS = RHS ✓ (1 mark).

Try one yourself

Find 310 + 415 by first making the denominators equal.

Show answer

LCM 30: 930 + 830 = 1730.

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