Cambridge International AS and A Level Mathematics Pure Mathematics 1

Algebra

30

P1^

1

Linear simultaneous equations

EXAMPLE 1.34 At a poultry farm, six hens and one duck cost $40, while four hens and three

ducks cost $36. What is the cost of each type of bird?

SOLUTION

Let the cost of one hen be $h and the cost of one duck be $d.

Then the information given can be written as:

6 h + d = (^40) ^1
4 h + 3 d = 36. ^2
There are several methods of solving this pair of equations.
Method 1: Elimination
Multiplying equation ^1 by 3 ⇒ 18 h + 3 d = 120
Leaving equation ^2 ⇒ 4 h + 3 d = 36
Subtracting ⇒ 14 h = 84
Dividing both sides by 14 ⇒ h = 6
Substituting h = 6 in equation ^1 gives 36 + d = 40
⇒ d = 4
Therefore a hen costs $6 and a duck $4.
Note
1 The first step was to multiply equation  1 by 3 so that there would be a term 3d
in both equations. This meant that when equation  2 was subtracted, the variable
d was eliminated and so it was possible to find the value of h.
2 The value h = 6 was substituted in equation ^1 but it could equally well have
been substituted in the other equation. Check for yourself that this too gives the
answer d = 4.
Before looking at other methods for solving this pair of equations, here is another
example.
EXAMPLE 1.35 Solve 3 x + 5 y = (^12) ^1
2 x − 6 y = − (^20) ^2
SOLUTION
(^) ^1 × 6 ⇒ 18 x + 30 y = 72
(^) ^2 × 5 ⇒ 10 x − 30 y = − 100
Adding ⇒ 28 x = − 28
Giving x = − 1
Substituting x = −1 in equation ^1 ⇒ − 3 + 5 y = 12
Adding 3 to each side ⇒ 5 y = 15
Dividing by 5 ⇒ y = 3
Therefore x = −1, y = 3.