Past papers › WAEC · 2014 · Nov/Dec · General Maths · Paper 2 › Question 1 Question WAEC General Maths 2014 Theory Expressions, formulae & change of subject Indices & standard form Linear & simultaneous equations Expressions, formulae & change of subject, Indices & standard form, Linear & simultaneous equations
(a) Factorize completely: m 2 − 2 m n + n 2 − 9 r 2 m^2 - 2mn + n^2 - 9r^2 m 2 − 2 mn + n 2 − 9 r 2
Show the answer ( m − n + 3 r ) ( m − n − 3 r ) (m - n + 3r)(m - n - 3r) ( m − n + 3 r ) ( m − n − 3 r )
(b) Solve simultaneously the equations 5 x − 4 y = 6 5x - 4y = 6 5 x − 4 y = 6 3 3 ( y − x ) = 1 27 3^{3(y - x)} = \frac{1}{27} 3 3 ( y − x ) = 27 1
Worked solution (try it first) (a) The first three terms are a perfect square:
m 2 − 2 m n + n 2 = ( m − n ) 2 m^2 - 2mn + n^2 = (m - n)^2 m 2 − 2 mn + n 2 = ( m − n ) 2 .
So the expression is
( m − n ) 2 − ( 3 r ) 2 (m - n)^2 - (3r)^2 ( m − n ) 2 − ( 3 r ) 2 , a difference of two squares.
Using
a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a - b)(a + b) a 2 − b 2 = ( a − b ) ( a + b ) with
a = m − n a = m - n a = m − n and
b = 3 r b = 3r b = 3 r :
( m − n − 3 r ) ( m − n + 3 r ) (m - n - 3r)(m - n + 3r) ( m − n − 3 r ) ( m − n + 3 r ) .
(b) Write
1 27 \frac1{27} 27 1 as a power of 3:
1 27 = 3 − 3 \frac1{27} = 3^{-3} 27 1 = 3 − 3 .
So
3 3 ( y − x ) = 3 − 3 3^{3(y - x)} = 3^{-3} 3 3 ( y − x ) = 3 − 3 , and the powers are equal:
3 ( y − x ) = − 3 3(y - x) = -3 3 ( y − x ) = − 3 , so
y − x = − 1 y - x = -1 y − x = − 1 and
y = x − 1 y = x - 1 y = x − 1 .
Substitute into
5 x − 4 y = 6 5x - 4y = 6 5 x − 4 y = 6 :
5 x − 4 ( x − 1 ) = 6 5x - 4(x - 1) = 6 5 x − 4 ( x − 1 ) = 6 .
So
5 x − 4 x + 4 = 6 5x - 4x + 4 = 6 5 x − 4 x + 4 = 6 and
x = 2 x = 2 x = 2 .
Then
y = 2 − 1 = 1 y = 2 - 1 = 1 y = 2 − 1 = 1 .
Watch out
Spot that m 2 − 2 m n + n 2 m^2 - 2mn + n^2 m 2 − 2 mn + n 2 ( m − n ) 2 (m - n)^2 ( m − n ) 2 Report a problem with this question