A binary operation takes two numbers and gives one answer, by a rule the question states. The symbol is usually ∗ * ∗ ∘ \circ ∘ ⊕ \oplus ⊕ Δ \Delta Δ a ∗ b = 2 a + b a * b = 2a + b a ∗ b = 2 a + b 5 ∗ 3 5 * 3 5 ∗ 3 first number in place of a a a second in place of b b b 5 ∗ 3 = 2 ( 5 ) + 3 = 13 5 * 3 = 2(5) + 3 = 13 5 ∗ 3 = 2 ( 5 ) + 3 = 13
a * b = 2a + b
5 * 3 = 2(5 ) + 3 = 13
Using the rule The first number replaces a, the second replaces b Try it
Working out an operation Pick a rule, then change the numbers
a * b = 2a + b
3 * 2 = 2(3 ) + 2 = 8
2 * 3 = 2(2 ) + 3 = 7
8 a * b 7 b * a no same?
2a + b ab − a a² − b a + b − 3 Swap Brackets a = 3 b = 2
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 a a a b b b a ∗ b a * b a ∗ b b ∗ a b * a b ∗ a
Common mistake
Putting the numbers in the wrong places. In a ∗ b a * b a ∗ b a a a a ∗ b = 3 a − b a * b = 3a - b a ∗ b = 3 a − b 4 ∗ 5 = 7 4 * 5 = 7 4 ∗ 5 = 7 5 ∗ 4 = 3 ( 5 ) − 4 = 11 5 * 4 = 3(5) - 4 = 11 5 ∗ 4 = 3 ( 5 ) − 4 = 11
Also put negative numbers in brackets when you substitute: a 2 a^2 a 2 a = − 3 a = -3 a = − 3 ( − 3 ) 2 = 9 (-3)^2 = 9 ( − 3 ) 2 = 9 − 9 -9 − 9
Brackets first
In ( a ∗ b ) ∗ c (a * b) * c ( a ∗ b ) ∗ c first number of the next step. In a ∗ ( b ∗ c ) a * (b * c) a ∗ ( b ∗ c ) b ∗ c b * c b ∗ c second number.
a b c a * b (a * b) * c 1st 2nd Brackets first Work 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.
Check yourself
Two operations are defined by a ∘ b = 2 a − b a \circ b = 2a - b a ∘ b = 2 a − b a Δ b = a b + 1 a \,\Delta\, b = ab + 1 a Δ b = ab + 1 2 Δ ( 3 ∘ 1 ) 2 \,\Delta\, (3 \circ 1) 2 Δ ( 3 ∘ 1 )
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 ∗ * ∗ x ∗ y = x + y + 2 x y x * y = x + y + 2xy x ∗ y = x + y + 2 x y x , y ∈ R x, y \in \mathbb R x , y ∈ R ( 2 ∗ 3 ) ∗ 5 (2 * 3) * 5 ( 2 ∗ 3 ) ∗ 5 ( x ∗ 7 ) = ( x ∗ 5 ) ∗ 2 (x * 7) = (x * 5) * 2 ( x ∗ 7 ) = ( x ∗ 5 ) ∗ 2
(i) The bracket first
2 ∗ 3 = 2 + 3 + 2 ( 2 ) ( 3 ) {2 * 3 = 2 + 3 + 2(2)(3)} 2 ∗ 3 = 2 + 3 + 2 ( 2 ) ( 3 ) = 5 + 12 = 17 {= 5 + 12 = 17} = 5 + 12 = 17 Think first. Work out 2 * 3 with x = 2 and y = 3.
Then with 5
17 ∗ 5 = 17 + 5 + 2 ( 17 ) ( 5 ) {17 * 5 = 17 + 5 + 2(17)(5)} 17 ∗ 5 = 17 + 5 + 2 ( 17 ) ( 5 ) = 22 + 170 = 192 {= 22 + 170 = 192} = 22 + 170 = 192 Think first. Now x = 17 and y = 5.
(ii) Write out each side
x ∗ 7 = x + 7 + 14 x = 15 x + 7 {x * 7 = x + 7 + 14x = 15x + 7} x ∗ 7 = x + 7 + 14 x = 15 x + 7 x ∗ 5 = x + 5 + 10 x = 11 x + 5 {x * 5 = x + 5 + 10x = 11x + 5} x ∗ 5 = x + 5 + 10 x = 11 x + 5 Think first. What is x * 7? And x * 5?
The right-hand side
( 11 x + 5 ) ∗ 2 {(11x + 5) * 2} ( 11 x + 5 ) ∗ 2 11 x + 5 {11x + 5} 11 x + 5 = ( 11 x + 5 ) + 2 + 2 ( 11 x + 5 ) ( 2 ) {= (11x + 5) + 2 + 2(11x + 5)(2)} = ( 11 x + 5 ) + 2 + 2 ( 11 x + 5 ) ( 2 ) = 11 x + 7 + 44 x + 20 {= 11x + 7 + 44x + 20} = 11 x + 7 + 44 x + 20 = 55 x + 27 {= 55x + 27} = 55 x + 27 Think first. (11x + 5) * 2: the first number is 11x + 5.
Solve
Take 55 x + 7 {55x + 7} 55 x + 7 − 40 x = 20 {-40x = 20} − 40 x = 20
So x = − 1 2 {x = -\frac12} x = − 2 1 { − 1 2 } {\left\{-\frac12\right\}} { − 2 1 }
Think first. 15x + 7 = 55x + 27.
← Back Next step → Show all steps Start again
Your turn
(c) Given that w ∗ u = w + u + 1 w * u = w + u + 1 w ∗ u = w + u + 1 ( y ∗ 4 ) ∗ y = 12 (y * 4) * y = 12 ( y ∗ 4 ) ∗ y = 12 y y y
Worked solution (try it first) (c) Work out the bracket first:
y ∗ 4 = y + 4 + 1 = y + 5 y * 4 = y + 4 + 1 = y + 5 y ∗ 4 = y + 4 + 1 = y + 5 .
Then
( y + 5 ) ∗ y = ( y + 5 ) + y + 1 = 2 y + 6 (y + 5) * y = (y + 5) + y + 1 = 2y + 6 ( y + 5 ) ∗ y = ( y + 5 ) + y + 1 = 2 y + 6 .
So
2 y + 6 = 12 2y + 6 = 12 2 y + 6 = 12 , which gives
y = 3 y = 3 y = 3 .
Watch out
In (a), R S → \overrightarrow{RS} R S O S → − O R → \overrightarrow{OS} - \overrightarrow{OR} O S − O R O R → \overrightarrow{OR} O R In (c), the answer to the bracket, y + 5 y + 5 y + 5 first number of the next step. Report a problem with this question
More past questions like this
WAEC 2017 · Paper 2 · Q13 The operation ( ∗ ) (*) R \mathbb R x ∗ y = x + y 2 x * y = \dfrac{x + y}{2} x , y ∈ R x, y \in \mathbb R JAMB 1995 · UME · Q22 If a ∗ b = + a b a * b = +\sqrt{ab} 2 ∗ ( 12 ∗ 27 ) 2 * (12 * 27) JAMB 1997 · UME · Q21 Two binary operations ∗ * ⊕ \oplus m ∗ n = m n − n − 1 m * n = mn - n - 1 m ⊕ n = m n + n − 2 m \oplus n = mn + n - 2 JAMB 1997 · UME · Q22 If x ∗ y = x + y − x y x * y = x + y - xy x x ( x ∗ 2 ) + ( x ∗ 3 ) = 68 (x * 2) + (x * 3) = 68 JAMB 1998 · UME · Q22 The binary operation ∗ * x ∗ y = x y − y − x x * y = xy - y - x x x y y x ∗ 3 = 2 ∗ x x * 3 = 2 * x JAMB 1999 · UME · Q14 If m ∗ n = m n − n m m * n = \frac mn - \frac nm m m n n − 3 ∗ 4 -3 * 4 JAMB 2000 · UME · Q12 A binary operation ∗ * a ∗ b = a b a * b = a^b a ∗ 2 = 2 − a a * 2 = 2 - a a a JAMB 2002 · UME · Q34 The binary operation ∗ * p ∗ q = p q + p + q p * q = pq + p + q 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) JAMB 2004 · UME · Q21 If the operation ∗ * p ∗ q = p q p * q = \sqrt{pq} 4 ∗ ( 8 ∗ 32 ) 4 * (8 * 32) JAMB 2010 · UTME · Q23 If x ∗ y = x + y 2 x * y = x + y^2 ( 2 ∗ 3 ) ∗ 5 (2 * 3) * 5 JAMB 2011 · UTME · Q23 A binary operation ⊕ \oplus x ⊕ y = x y + x + y x \oplus y = xy + x + y 3 ⊕ ( − 2 3 ) 3 \oplus \left(-\frac23\right) JAMB 2012 · UTME · Q21 The binary operation ∗ * p ∗ q = p q + p − q p * q = pq + p - q 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) JAMB 2013 · UTME · Q21 If a binary operation ∗ * x ∗ y = x + 2 y x * y = x + 2y 2 ∗ ( 3 ∗ 4 ) 2 * (3 * 4) JAMB 2014 · UTME · Q21 A binary operation ∗ * x ∗ y = x y x * y = xy x ∗ x = 12 − x x * x = 12 - x x x JAMB 2017 · UTME (set 2) · Q30 The operation ∗ * R \mathbb{R} x ∗ y = 3 x + 2 y − 1 x * y = 3x + 2y - 1 3 ∗ ( − 1 ) 3 * (-1) JAMB 1993 · UME · Q24 A binary operation ∗ * x ∗ y = x y x * y = x^y x x y y x ∗ 2 = x x * 2 = x