Page 56 - Hybrid PBD 2022 Form 4 Additional Mathematics

Matematik Tambahan Tingkatan 4 Bab 3 Sistem Persamaan

3.2 Persamaan Serentak yang Melibatkan Satu Persamaan Linear dan Satu
Persamaan Tak Linear Buku Teks
PBD
PBD
PBD Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation ms. 79 – 84
6. Selesaikan setiap persamaan serentak yang berikut. SP 3.2.1 TP3
Solve each of the following simultaneous equations. E-pop Quiz

(i) −x + 2y = 5 ................. (ii) x + y = 5 ...................
2
2
x + y = 25 .................b xy − y = 3 .................b
Menggunakan kaedah penggantian: Menggunakan kaedah penghapusan:
Using the substitution method: Using the elimination method:
2
Dari  / From : x = 2y − 5 ............c  × y: xy + y = 5y ................c
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Gantikan c ke dalam b / Substitute c into b c − b: y − y = 5y − 3
2
2
2
(2y − 5) + y = 25 y − 6y + 3 = 0
2
4y − 20y + 25 + y = 25 −(−6) ±  − 4(1)(3)
2
2
(−6)
2
5y − 20y = 0 y = 2(1)
2
5y(y − 4) = 0 = 5.4495 atau / or 0.5505
y = 0 , y = 4
Gantikan y = 5.4495 ke dalam 
Gantikan y = 0 ke dalam c Substitute y = 5.4495 into 
Substitute y = 0 into c x + 5.4495 = 5
x = 2(0) − 5 x = −0.4495
= −5
Gantikan y = 0.5505 ke dalam 
Gantikan y = 4 ke dalam c Substitute y = 0.5505 into 
Substitute y = 4 into c x + 0.5505 = 5
x = 2(4) − 5 x = 4.4495
= 3
Maka, x = –0.4495, y = 5.4495 dan x = 4.4495,
Maka, x = –5, y = 0 dan x = 3, y = 4. y = 0.5505.
Thus, x = –5, y = 0 and x = 3, y = 4.
Thus, x = –0.4495, y = 5.4495 and x = 4.4495, y = 0.5505.
(a) 2x + y = 7 ............................. (b) 2x − y = 1 ..............................
2
2
4y − 3x − xy = 0 .................b 9x − 2y + 9 = 0 .................b
Dari  / From : y = 7 − 2x .................c Dari  / From : y = 2x − 1 .................c
Gantikan c ke dalam b / Substitute c into b Gantikan c ke dalam b / Substitute c into b
4(7 − 2x) − 3x − x(7 − 2x) = 0 9x − 2(2x − 1) + 9 = 0
2
2
28 − 8x − 3x − 7x + 2x = 0 9x − 2(4x − 4x + 1) + 9 = 0
2
2
2
2
2
2x − 18x + 28 = 0 9x − 8x + 8x − 2 + 9 = 0
2
x − 9x + 14 = 0 x + 8x + 7 = 0
2
2
(x − 7)(x − 2) = 0 (x + 1)(x + 7) = 0
x = 7, x = 2 x = −1, x = −7
Gantikan x = 7 ke dalam c Gantikan x = −1 ke dalam c
Substitute x = 7 into c Substitute x = −1 into c
y = 7 − 2(7) y = 2(−1) − 1
= −7 = −3
Gantikan x = 2 ke dalam c Gantikan x = −7 ke dalam c
Substitute x = 2 into c Substitute x = −7 into c
y = 7 − 2(2) y = 2(−7) − 1
= 3 = −15
Maka, x = 7, y = –7 dan x = 2, y = 3.
Thus, x = 7, y = –7 and x = 2, y = 3. Maka, x = −1, y = −3 dan x = −7, y = −15.
Thus, x = −1, y = −3 and x = −7, y = −15.

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