Cambridge International Mathematics

Formulae and simultaneous equations (Chapter 7) 163

WHICH VARIABLE TO ELIMINATE

There is always a choice whether to eliminatexory, so our choice depends on which variable is easier to

eliminate.

InExample 18, try to solve by multiplying (1) by 5 and (2) by¡ 4. This eliminatesyrather thanx. The

final solution should be the same.

EXERCISE 7E.3

1 What equation results when the following are added vertically?

a 3 x+4y=6

8 x¡ 4 y=5

b 2 x¡y=7

¡ 2 x+5y=5

c 7 x¡ 3 y=2

2 x+3y=7

d 6 x¡ 11 y=12

3 x+11y=¡ 6

e ¡ 7 x+2y=5

7 x¡ 3 y=6

f 2 x¡ 3 y=¡ 7

¡ 2 x¡ 8 y=¡ 4

2 Solve the following using the method of elimination:

a 5 x¡y=4

2 x+y=10

b 3 x¡ 2 y=7

3 x+2y=¡ 1

c ¡ 5 x¡ 3 y=14

5 x+8y=¡ 29

d 4 x+3y=¡ 11

¡ 4 x¡ 2 y=6

e 2 x¡ 5 y=14

4 x+5y=¡ 2

f ¡ 6 x¡y=17

6 x+5y=¡ 13

3 Give the equation that results when both sides of the equation:

a 2 x+5y=1 are multiplied by 5 b 3 x¡y=4 are multiplied by¡ 1

c x¡ 7 y=8 are multiplied by 3 d 5 x+4y=9 are multiplied by¡ 2

e ¡ 3 x¡ 2 y=2 are multiplied by 6 f 4 x¡ 2 y=3 are multiplied by¡ 4

4 Solve the following using the method of elimination:

a 2 x+y=8

x¡ 3 y=11

b 3 x+2y=7

x+3y=7

c 5 x¡ 2 y=17

3 x¡y=9

d 2 x+3y=13

3 x+2y=17

e 4 x¡ 3 y=1

2 x+5y=7

f 2 x+5y=14

5 x¡ 3 y+27=0

g 7 x¡ 2 y=20

4 x+3y=¡ 1

h 3 x¡ 2 y=5

5 x¡ 3 y=8

i 3 x¡ 4 y=¡ 15

2 x+3y=7

j 4 x+3y=14

5 x¡ 2 y=29

k 2 x¡ 3 y=1

5 x+7y=3

l 8 x¡ 5 y=¡ 9

3 x+4y=¡ 21

m 3 x+8y+34=0

5 x¡ 4 y=0

n 5 x¡ 6 y=8

6 x¡ 7 y¡9=0

o 2 x¡ 7 y¡18 = 0

3 x¡ 5 y¡5=0

5 Use the method of elimination to attempt to solve:

a 2 x¡y=3

4 x¡ 2 y=6

b 3 x+4y=6

6 x+8y=7

Comment on your results.

IGCSE01

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Y:\HAESE\IGCSE01\IG01_07\163IGCSE01_07.CDR Monday, 15 September 2008 4:47:07 PM PETER