Cambridge International Mathematics

160 Formulae and simultaneous equations (Chapter 7)

g y=3x+2

y=2x+3

h y=3x+1

y=3x+5

i y=5x¡ 2

y=10x¡ 4

2 Find the simultaneous solution to the following pairs of equations:

a y=x+4

y=5¡x

b y=x+1

y=7¡x

c y=2x¡ 5

y=3¡ 2 x

d y=x¡ 4

y=¡ 2 x¡ 4

e y=3x+2

y=¡ 2 x¡ 3

f y=4x+6

y=6¡ 2 x

SOLUTION BY SUBSTITUTION

The method ofsolution by substitutionis used when at least one equation is given with eitherxoryas

thesubjectof the formula, or if it is easy to makexorythe subject.

Example 14 Self Tutor

Solve simultaneously, by substitution: y=9¡x

2 x+3y=21

y=9¡x..... (1) 2 x+3y=21..... (2)

Since y=9¡x, then 2 x+ 3(9¡x)=21

) 2 x+27¡ 3 x=21

) 27 ¡x=21

) x=6

Substituting x=6 into (1) gives y=9¡6=3.

The solution is: x=6, y=3:

Check: (1) 3=9¡ 6 X (2) 2(6) + 3(3) = 12 + 9 = 21 X

Example 15 Self Tutor

Solve simultaneously, by substitution: 2 y¡x=2

x=1+8y

2 y¡x=2..... (1) x=1+8y..... (2)

Substituting (2) into (1) gives 2 y¡(1 + 8y)=2

) 2 y¡ 1 ¡ 8 y=2

) ¡ 6 y=3

) y=¡^12

Substituting y=¡^12 into (2) gives x=1+8£¡^12 =¡ 3

The solution is: x=¡ 3 , y=¡^12.

Check: (1) 2(¡^12 )¡(¡3) =¡1+3=2 X

(2) 1+8(¡^12 )=1¡4=¡ 3 X

We substitute

for in

the other equation.

9 ¡¡¡xy

x

yx

is the subject of

the second equation,

so we substitute

for in

the first equation.

1+8¡¡

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Y:\HAESE\IGCSE01\IG01_07\160IGCSE01_07.CDR Monday, 15 September 2008 4:36:36 PM PETER