Page 52 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan
3. Selesaikan setiap sistem persamaan linear berikut dengan menggunakan kaedah penghapusan.
Solve each of the following systems of linear equations by using the elimination method. SP 3.1.2 TP2

(a) 4x − 4y + 7z = −1 .............. 
−6x − 4y − 2z = −2 ................... −5x + 2y − z = −10 ........... b
−3x − 5y + 7z = −61 ................b −9x + 4y − 5z = −10 ......... c
8x − y + 8z = −20 ....................c Pilih  dan b
untuk
b × 2: −6x − 10y + 14z = −122 .......d menghapuskan x. b × 2: −10x + 4y − 2z = −20 ...... d
Choose  and b to
 + d: −6x + 5z = −21 ............... e
 − d: 6y − 16z = 120 ......................e eliminate x.
 + c: −5x + 2z = −11 ............... f
b × 8: −24x − 40y + 56z = −488 .....f Pilih b dan c
c × 3: 24x − 3y + 24z = −60 .............g untuk e × 5: −30x + 25z = −105 .......... g
f + g: −43y + 80z = −548 ..............h menghapuskan x. f × 6: −30x + 12z = −66 ............ h
Choose b and c to
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eliminate x. g − h: 13z = −39
e × 5: 30y − 80z = 600 ......................i Pilih e dan h z = −3
h + i: −13y = 52 untuk Gantikan z = –3 ke dalam e / Substitute z = –3 into e
y = −4 menghapuskan z. −6x + 5(−3) = −21
Choose e and h to
eliminate z. −6x − 15 = −21
Gantikan y = –4 ke dalam e / Substitute y = –4 into e x = 1
6(−4) − 16z = 120 Gantikan x = 1 dan z = –3 ke dalam 
−24 − 16z = 120 Substitute x = 1 and z = –3 into 
z = −9 4(1) − 4y + 7(−3) = −1
Gantikan y = –4 dan z = –9 ke dalam c −4y − 17 = −1
Substitute y = –4 and z = –9 into c y = −4
8x − (−4) + 8(−9) = −20 ∴ x = 1, y = −4, z = −3
8x − 68 = −20
x = 6
∴ x = 6, y = −4, z = −9

(b) 7x + 8y − 4z = 129............. (c) 3x − 2y + 5z = −3 ...............
8x − 11y + 5z = 81.............b 5x + 4y − 2z = 8 ..................b
2x − 3y + z = 15 .................c 7x − 6y + 3z = −41 .............c

c × −4: −8x + 12y − 4z = −60.......d  × 2: 6x − 4y + 10z = −6............d
 − d: 15x − 4y = 189....................e d + b: 11x + 8z = 2 .....................e
c × 5: 10x − 15y + 5z = 75.............f  × 3: 9x − 6y + 15z = −9............f
f − b: 2x − 4y = −6........................g f − c: 2x + 12z = 32 ...................g
e − g: 13x = 195 e × 2: 22x + 16z = 4 .....................h
x = 15 g × 11: 22x + 132z = 352 ...........i
i − h: 116z = 348
Gantikan x = 15 ke dalam g / Substitute x = 15 into g z = 3
2(15) − 4y = −6
30 − 4y = −6 Gantikan z = 3 ke dalam g / Substitute z = 3 into g
y = 9 2x + 12(3) = 32
x = −2
Gantikan x = 15 dan y = 9 ke dalam c
Substitute x = 15 and y = 9 into c Gantikan x = –2 dan z = 3 ke dalam b
2(15) − 3(9) + z = 15 Substitute x = –2 and z = 3 into b
3 + z = 15 5(−2) + 4y − 2(3) = 8
z = 12 4y − 16 = 8

∴ x = 15, y = 9, z = 12 y = 6
∴ x = −2, y = 6, z = 3

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03 Hybrid PBD Mate Tambahan Tg4.indd 34 29/09/2021 3:30 PM