Cambridge International Mathematics

Discovery Linear inequalities of the form ax+by < d or ax+by > d

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Consider the regions on either side of the line with equation

3 x+2y=6.

What to do:

a Copy and complete:

Point 3 x+2y Point 3 x+2y

A(1,3) 3(1) + 2(3) = 9 H(¡ 2 ,4) 3(¡2) + 2(4) = 2

B(, ) I

C(, ) J

D(, ) K

E(, ) L

F(, ) M

G(, ) O(0,0)

b Usinga, what inequality defines the region under the line 3 x+2y=6?

What inequality defines the region above the line 3 x+2y=6?

You should have discovered that:

All points satisfying ax+by < d lie on one side of the line ax+by=d and all points

satisfying ax+by > d lie on the other side.

To find the region which corresponds to an inequality, we substitute into the inequality a point not on the

boundary line, usually O(0,0).

If a true statement results then this point lies in the region we want. If not, then the required region is the

other side of the line.

Example 4 Self Tutor

Graph 3 x¡ 4 y> 12.

y

O x

3

2

F

A

E

D

H

M

LI

J

C

B

K

y

O x

3 ¡-¡4 ¡=¡12xy

()0 -3,

()4 0,

the required region

R

The boundary line is 3 x¡ 4 y=12.

When x=0, ¡ 4 y=12

) y=¡ 3

When y=0, 3 x=12

) x=4

So,(0,¡3)and(4,0)lie on the boundary.

If we substitute(0,0)into 3 x¡ 4 y> 12

we obtain 0 > 12 which is false.

) (0,0)does not lie in the region.

So, in this case we want the region below the line 3 x¡ 4 y=12.

It is not included in the region.

642 Inequalities (Chapter 32)

IGCSE01

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Y:\HAESE\IGCSE01\IG01_32\642IGCSE01_32.CDR Monday, 27 October 2008 3:06:16 PM PETER