Binary operations · Lesson 1 of 2

Working out an operation

What a binary operation is, putting numbers into its rule in the right order, brackets first, two operations in one question, and solving equations that use an operation.

14 minYou should already know: Expressions, formulae & change of subject
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A binary operation takes two numbers and gives one answer by a stated rule. The symbol is usually ∗*, ∘\circ, ⊕\oplus or Δ\Delta, and the question tells you what it means. For example, if a∗b=2a+ba * b = 2a + b, then to work out 5∗35 * 3 you put the first number in place of aa and the second in place of bb.

a * b = 2a + b
5 * 3 = 2(5) + 3 = 13
Using the ruleThe first number replaces a, the second replaces b

Try it

Working out an operationPick a rule, then change the numbers
a * b = 2a + b
3 * 2 = 2(3) + 2 = 8
2 * 3 = 2(2) + 3 = 7
8a * b7b * anosame?
Put the first number in place of a and the second in place of b. Swapping the numbers changes the answer: the order matters for this rule.

Pick a rule and change aa and bb. “Swap” works out a∗ba * b and b∗ab * a side by side: for most rules, the order of the two numbers matters. “Brackets” is the next idea.

Brackets first

In (a∗b)∗c(a * b) * c, work out the bracket first. Its answer becomes the first number of the next step. In a∗(b∗c)a * (b * c), the bracket b∗cb * c comes first, and its answer becomes the second number.

abca * b(a * b) * c1st2nd
Brackets firstWork out a * b, then combine that answer with c

Some questions use two different operations. Keep each rule for its own symbol, and still start with the bracket.

Solving equations with an operation

When an unknown is inside an operation, write the operation out with the rule first. That turns it into an ordinary equation, often linear, sometimes quadratic.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q13 (a)

An operation ∗* is defined by x∗y=x+y+2xyx * y = x + y + 2xy, x,y∈Rx, y \in \mathbb R. (i) Calculate (2∗3)∗5(2 * 3) * 5. (ii) Find the truth set of (x∗7)=(x∗5)∗2(x * 7) = (x * 5) * 2.

  1. (i) The bracket first

    2∗3=2+3+2(2)(3)=5+12=172 * 3 = 2 + 3 + 2(2)(3) = 5 + 12 = 17.

    Think first. Work out 2 * 3 with x = 2 and y = 3.

  2. Then with 5

    17∗5=17+5+2(17)(5)=22+170=19217 * 5 = 17 + 5 + 2(17)(5) = 22 + 170 = 192.

    Think first. Now x = 17 and y = 5.

  3. (ii) Write out each side

    x∗7=x+7+14x=15x+7x * 7 = x + 7 + 14x = 15x + 7, and x∗5=x+5+10x=11x+5x * 5 = x + 5 + 10x = 11x + 5.

    Think first. What is x * 7? And x * 5?

  4. The right-hand side

    (11x+5)∗2=(11x+5)+2+2(11x+5)(2)(11x + 5) * 2 = (11x + 5) + 2 + 2(11x + 5)(2). That is 11x+7+44x+20=55x+2711x + 7 + 44x + 20 = 55x + 27.

    Think first. (11x + 5) * 2: the first number is 11x + 5.

  5. Solve

    −40x=20-40x = 20, so x=−12x = -\frac12. The truth set is {−12}\left\{-\frac12\right\}.

    Think first. 15x + 7 = 55x + 27.

Your turn

WAEC 2018 · Paper 2 · Q13 (c)

  1. (c)

    Given that w∗u=w+u+1w * u = w + u + 1, if (y∗4)∗y=12(y * 4) * y = 12, find the value of yy.

Worked solution (try it first)

(c)

  1. Work out the bracket first: y∗4=y+4+1=y+5y * 4 = y + 4 + 1 = y + 5.
  2. Then (y+5)∗y=(y+5)+y+1=2y+6(y + 5) * y = (y + 5) + y + 1 = 2y + 6.
  3. So 2y+6=122y + 6 = 12, which gives y=3y = 3.

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