Page 302 - Buku Teks Digital Mate KSSM T5

7. (a) 3. y = 4x – 5 (3, 7) 4.
F y = –3x + 4 (1, 1)
D
y > 4x – 5 (2, 4), (–2, 0) y > –3x + 4 (–1, 8), (–0.5, 7)
A C
E
y < 4x – 5 (0, –6), (4, 5) y < –3x + 4 (–2, 3), (0, 1)
B
(b) 3.08 km
8. (a) X Praktis Kendiri 6.1c
●P ●G
●D ●J ●A Y 1. (a) y (b) y
Z ●B ●E ●L ●T 1
3
●C ●R ●N ●F y = —x + 3
●H ●M
●K ●I O x 3
1
(b) (i) {C, R, K, I} –2 y = –2 y < —x + 3
3
(ii) {P, G, D, C, R, K, I} y < –2 –3 O 3 x
(iii){E}
9. (b) (i) RM1 080 (ii) 40 y
10. (a) ketiga (b) 484 (c) 13 068 (c) x = 2 (d) y y = x + 2
y > x + 2 2
BAB 6 Ketaksamaan Linear Dalam Dua
Pemboleh Ubah O 2 x –2 O x
Praktis Kendiri 6.1a x ≤ 2
1. (a) 25x + 45y ≤ 250 atau 5x + 9y ≤ 50
(b) 2x + 1.5y ≤ 500 atau 4x + 3y ≤ 1 000 (e) y (f) y
(c) 0.3x + 0.4y ≤ 50 atau 3x + 4y ≤ 500 3 y = x
1
(d) 1.5x + 3.5y ≥ 120 atau 3x + 7y ≥ 240 y ≥ – —x – 2 y ≥ x
2
x x
Praktis Kendiri 6.1b – 4 –2 O –3 O 3

1
–2 y = – —x – 2
1. Rantau y > —x – 2 2 –3
2
y 3 y = —x – 2
2
3
1 (3, 1)
2. (a) y (x = 0) (b) y
2
1
x y = –x – 2 (1.5, –1) y = —x
1
O 1 2 3 3 y > —x 2
(1, –1) 2 x ≤ 0 2
–1 (1.5, –1) y > –x – 2 (3, 1), (1, –1) 1
3 x x
(2, –2) 2 O O 2
–2 (3, –2) y < –x – 2 (2, –2), (3, –2)
3
2
Rantau y < —x – 2
3
(c) y (d) y
2. Rantau y > – –x + 2 2y = x + 4
1
2
y x + y ≥ –3 2
6 O x O x

(–3, 5) (4, 5) –3 – 4
4 2y < x + 4
–3
2 (2, 1) x + y = –3
1
(–3, 1) y = – –x + 2
x 2 (e) y (f) y
–4 –2 O 2 4 6 2y + x = 2
1
Rantau y < – –x + 2
–2 (1, –2) 2 2 2y + x ≥ 2
1
x x
1
y = – –x + 2 (2, 1) O 2 O 2
2
1
y > – –x + 2 (–3, 5),(4, 5) y ≤ –x + 2 y = –x + 2
2 Saiz sebenar
1
y < – –x + 2 (–3, 1), (1, –2)
2
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