Page 10 - Focus KSSM SPM 2021 TG4.5 - Maths DLP

Mathematics SPM Chapter 6 Linear Inequalities in Two Variables
(b) Line x + y = 2 is a dashed line and the shaded Try This! 6.1
region lies below the line. Therefore, the
inequality is x + y  2. 1. Represent the following situations in the form of
linear inequalities.
Try Questions 5 – 7 in Try This! 6.1 (a) A patient is at risk of having a heart problem if
the blood pressure in the ankle, k mm Hg, is
6 less than 90% of the blood pressure in the arm,
Draw and shade the region that satisfy the following p mm Hg.
linear inequalities. (b) Aisyah’s father and Haida’s father gives pocket
(a) y  3x + 1 (b) y  –2x + 3 money several times each month respectively.
1 Each time, Aisyah will get RM5 while Haida
(c) y  x + 2 gets RM8. Let’s say Aisyah’s father gave her
2 pocket money for a times while Haida’s father
Solution gave her pocket money for h times last month. Form 4
(a) y = 3x + 1 • Convert the given The total amount of cash they received last
month was more than RM50.
Form 4
x –1 1 linear inequality to (c) Ananda has a 2.4 m of wood to make a
linear equation form
y –2 4 to draw the straight rectangular photo frame for the project of the
line. Living Skills subject.
y
6 2. Determine whether each of the following point lies
• Check the
4 inequality symbol on the straight line, in the region above or in the
y 3x + 1 and draw the region below the straight line y = –2x + 7.
2 straight line. (a) (1, 5)
x • Shade the region (b) (2, 4)
–4 –2 O 2 4 that satisfies the (c) (0, –3)
–2 inequality.
3. Determine whether each of the following points
satisfies y = 4 – 5x, y  4 – 5x or y  4 – 5x.
(b) y = –2x + 3 (a) (–1, 7)
x –1 2 (b) (2, – 6)
y 5 –1 (c) (3, 0)
4. Determine whether each of the following point is the
y
solution for the linear inequalities given.
4 (a) (3, – 5); y  –3x + 8
y = –2x 3 (b) (– 6, 10); y  x + 5
2
(c) (–1, –3); x – 2y  4
x (d) (2, 2); 4 + x  3y
–2 0 2 4
–2
5. Shade the region that satisfies the given inequalities.
(a) y  3x + 2
1
(c) y = x + 2
2 y
x –2 2 y = 3x 2

y 1 3 0 x
y

4
1
2 y = x + 2 (b) y  –1
2 y
x
–2 0 2 4
–2 x
O
y –1
Try Question 8 in Try This! 6.1

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06 Focus SPM Maths F4.indd 90 17/02/2021 5:24 PM