Past papers › WAEC · 2011 · May/June · General Maths · Paper 2 › Question 2 Question WAEC General Maths 2011 Theory Expressions, formulae & change of subject Indices & standard form Linear & simultaneous equations Expressions, formulae & change of subject, Indices & standard form, Linear & simultaneous equations
(a) Make q q q t = p q r − r 2 q t = \sqrt{\dfrac{pq}{r} - r^2q} t = r pq − r 2 q
(b) If 9 ( 1 − x ) = 27 y 9^{(1 - x)} = 27^y 9 ( 1 − x ) = 2 7 y x − y = − 1 1 2 x - y = -1\frac12 x − y = − 1 2 1 x + y x + y x + y
Worked solution (try it first) (a) q q q is inside a square root, so square both sides:
t 2 = p q r − r 2 q t^2 = \frac{pq}{r} - r^2 q t 2 = r pq − r 2 q .
Multiply every term by
r r r to clear the fraction:
r t 2 = p q − r 3 q rt^2 = pq - r^3 q r t 2 = pq − r 3 q .
Both terms on the right contain
q q q , so take it out as a common factor:
r t 2 = q ( p − r 3 ) rt^2 = q(p - r^3) r t 2 = q ( p − r 3 ) .
Divide by the bracket:
q = r t 2 p − r 3 q = \frac{rt^2}{p - r^3} q = p − r 3 r t 2 .
(b) Write both sides as powers of 3:
9 = 3 2 9 = 3^2 9 = 3 2 and
27 = 3 3 27 = 3^3 27 = 3 3 .
So
3 2 ( 1 − x ) = 3 3 y 3^{2(1 - x)} = 3^{3y} 3 2 ( 1 − x ) = 3 3 y , and the powers must be equal:
2 − 2 x = 3 y 2 - 2x = 3y 2 − 2 x = 3 y , which is
2 x + 3 y = 2 2x + 3y = 2 2 x + 3 y = 2 (1).
The second equation is
x − y = − 3 2 x - y = -\frac32 x − y = − 2 3 (2).
Multiply (2) by 3:
3 x − 3 y = − 9 2 3x - 3y = -\frac92 3 x − 3 y = − 2 9 (3).
Add (1) and (3):
5 x = − 5 2 5x = -\frac52 5 x = − 2 5 , so
x = − 1 2 x = -\frac12 x = − 2 1 .
From (2),
y = x + 3 2 = 1 y = x + \frac32 = 1 y = x + 2 3 = 1 .
So
x + y = − 1 2 + 1 = 1 2 x + y = -\frac12 + 1 = \frac12 x + y = − 2 1 + 1 = 2 1 .
Watch out
When you square p q r − r 2 q \sqrt{\frac{pq}{r} - r^2q} r pq − r 2 q p q r − r 2 q \frac{pq}{r} - r^2q r pq − r 2 q q q q q q q In (b), you can only compare the powers once both sides are written with the same base. Report a problem with this question