Cambridge International Mathematics

162 Formulae and simultaneous equations (Chapter 7)

In problems where the coefficients ofx(ory) arenotthesame sizeoropposite in sign, we may first have

tomultiplyeach equation by a number.

Example 17 Self Tutor

Solve simultaneously, by elimination: 3 x+2y=7

2 x¡ 5 y=11

3 x+2y=7...... (1) 2 x¡ 5 y=11...... (2)

We can eliminateyby multiplying (1) by 5 and (2) by 2.

) 15 x+10y=35

+4x¡ 10 y=22

19 x =57 fadding the equationsg

) x=3 fdividing both sides by 19 g

Substituting x=3 into equation (1) gives

3(3) + 2y=7

) 9+2y=7

) 2 y=¡ 2

) y=¡ 1

So, the solution is: x=3, y=¡ 1.

Check: 3(3) + 2(¡1) = 9¡2=7 X 2(3)¡5(¡1)=6+5=11 X

Example 18 Self Tutor

Solve by elimination: 3 x+4y=14

4 x+5y=17

3 x+4y=14 ::::::(1)

4 x+5y=17 ::::::(2)

To eliminatex, multiply both sides of

(1) by 4 : 12 x+16y=56 ::::::(3)

(2) by¡ 3 : ¡ 12 x¡ 15 y=¡ 51 ::::::(4)

y=5 fadding (3) and (4)g

Substituting y=5 into (2) gives

4 x+ 5(5) = 17

) 4 x+25=17

) 4 x=¡ 8

) x=¡ 2

Thus x=¡ 2 and y=5:

Check:

(1) 3(¡2) + 4(5) = (¡6) + 20 = 14 X

(2) 4(¡2) + 5(5) = (¡8) + 25 = 17 X

IGCSE01

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Y:\HAESE\IGCSE01\IG01_07\162IGCSE01_07.CDR Monday, 15 September 2008 4:44:38 PM PETER