Page 14 - Modul A+Matematik Tamb Tg 4_Pan Asia Publications

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7. (a) f(x) = x – 6kx + 10k + 1 4(11 + 3c) + 6b + 5c = –6 (x – 2)(x – 5) = 0
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= (x – 3k) – 9k + 10k + 1 17c + 6b = –50 … 5 x = 2 atau / or 5
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= (x – 3k) + k + 1 4 × 6 + 5 y = –1 atau / or 8
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Maka / Hence, k + 1 = r + 2k 47c + 138 = –50 7. 2y + x = 30
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r = k – 2k + 1 47c = –188 xy = 100
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r = (k – 1) 2 c = –4 y(30 – 2y) = 100
r = ±(k – 1) a = 11 + 3(–4) = –1 2y – 30y + 100 = 0
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(b) Apabila / When x = 3k = r – 1 7(–1) + 4 + 15 y – 15y + 50 = 0
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3k = (k – 1) – 1 b = 4 (y – 5)(y – 10) = 0
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3k = k – 2k = 3 y = 5 atau / or 10
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k – 5k = 0 x = 20 atau / or 10
k(k – 5) = 0 3. 2p – 4q + 5r = –33 … 1
k = 0 atau / or k = 5 4p – q = –5 … 2 8. x = 1 – 3y
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r = ±1, ±4 –2p + 2q – 3r = 19 … 3 x – xy + y = 21
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(1 – 3y) – y(1 – 3y) + y = 21
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8. (a) DPBR 1 + 3 1 – 6y + 9y – y + 3y + y = 21
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= y – xy – y(y – 2x) – –2q + 2r = –14 13y – 7y – 20 = 0
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1 (2x)(y – x) q – r = 7 … 4 (13y – 20)(y + 1) = 0
2 1 × 2 – 2 y = 20 atau / or –1
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1 2
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= y – xy – y + xy – xy + x 2 7q – 10r = 61 … 5 13
2 2 4 × 7 – 5 x = –3 8 atau / or 4
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2
1 2
= x – xy + y 3r = –12 13
2 2 r = –4, q = 3 x 2 21
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1 2
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(b) x − xy + y p = – 1 9. xy + π(x) + π   = 2 π … 1
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2 2 2
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1 2
= y – xy + x 2 4. (a) 1 550x + 750y + 1 250z = 33 300 y = (2πx) + 1.75π
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2 2
7
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= (y – xy) + x 2 31x + 15y + 25z = 666 … 1 = πx + π … 2
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2 x + y + z = 28 … 2
= 1  y – x  2 – x 2  + x 2 x + y – z = –2 … 3 Gantikan 2 ke dalam 1.
2 2 4 (b) 2 – 3 Substitute 2 into 1.
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
= 1  y – x  2 – x 2 + x 2 2z = 30 x  πx + π + πx 2 + πx 2 = 21 π
2 2 8 z = 15 2 4 4 16 2
7
7 2
= 1  y – x  2 + x x + y = 28 – 15 x 2 + x + x 2 + x 2 = 21
2 2 8 = 13 2 4 4 16 2
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Luas minimum / Minimum area Dari / From 1, 8x + 28x + 4x + x = 168
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7 2
= x unit 2 31x + 15y = 666 – 25(15) = 291 13x + 28x – 168 = 0
8 15x + 15y = 195 –28 ± ABBBBBBBBBBB
28 – 4(13)(–168)
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16x = 96 x = 2(13)
Bab 3 x = 6 = 2.68
y = 28 – 6 – 15
Kertas 2 / Paper 2 = 7
1. 2x + 5y + 2z = –38 … 1 Komputer / Computer: 6 Bab 5
–6x – 7z + y = –12 … 2 Komputer riba / Laptop: 7
4z + 3x – 2y = 17 … 3 Telefon pintar / Smart phone: 15 Kertas 2 / Paper 2
1 × 3 + 2 5. x + y + z = 16 … 1 1. (a) 3 200, 3 150, 3 100, …, 1 500.
16y – z = –126 … 4 x + y = z … 2 T n = 1 500 = 3 200 + (n – 1)(–50)
3 × 2 + 2 2y = 7 + x … 3 n = 35 hari / days
–3y + z = 22 … 5 Gantikan 2 ke dalam 1. (b) S 35 = 35 [2(3 200) – 50(34)]
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4 + 5 Substitute 2 into 1. S 35 = 82 250
13y = –104 z + z = 16
y = –8 2z = 16 Jumlah kos perbelanjaan
z = 22 – 24 z = 8 Total spending cost
= –2 Gantikan 3 ke dalam 1. = 82 250 × RM0.20
2x + 5(–8) + 2(–2) = –38 Substitute 3 into 1. = RM16 450
2x = 6 2y – 7 + y + 8 = 16 2. (a) 4, 6, 8, … n = 62
x = 3 3y = 15 (i) T 62 = 4 + 61(2)
2. 4a + 6b + 5c = –6 … 1 y = 5 = 126 cm

x = 3
7a – 4b – c = –15 … 2 \ Nombor itu ialah 358. 62
a = 11 + 3c … 3 \ The number is 358. (ii) S 62 = 2 [8 + 61(2)]
Gantikan 3 ke dalam 2. 6. 3x – y = 7 = 4 030 cm
Substitute 3 into 2. x + xy – y = 1 = 40.3 m
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7(11 + 3c) – 4b – c = –15 x + x(3x – 7) – (3x – 7) = 1 25 000
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5c – b = –23 … 4 x + 3x – 7x – (9x – 42x + 49) = 1 (b) 250 = 100 cm untuk lebar /
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Gantikan 3 ke dalam 1. –5x + 35x – 50 = 0 for width
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Substitute 3 into 1. x – 7x + 10 = 0 T n = 100 = 4 + (n – 1)(2)
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