Page 46 - Steady Mate Tg1 - Book B

Matematik Tingkatan 1 Bab 6 Persamaan Linear

L Selesaikan setiap persamaan linear serentak yang berikut dengan menggunakan kaedah penghapusan. TP 3 ARAS S
Solve each of the following simultaneous linear equations by using elimination method.

Contoh Kenal pasti pemboleh ubah Contoh
dengan pekali yang sama.
x + y = 5 ............................ a Identify the variable with the 2x + 3y = 5 ........................ a Darabkan a dan b dengan
2x – y = 4 ........................... b same coefficient. 3x + 4y = 6 ........................ b suatu nombor yang sesuai
untuk menyamakan pekali x.
a + b : 3x + 0 = 9 Hapuskan pemboleh ubah y. a × 3: 6x + 9y = 15 ......... c Multiply a and b by a
suitable number to equate
3x = 9 Eliminate the variable of y. b × 2: 6x + 8y = 12 ......... d the coefficient of x.
x = 3
c – d: 0 + y = 3
Hapuskan pemboleh ubah x.
y = 3 Eliminate the variable of x.
Gantikan x = 3 ke dalam a.
Substitute x = 3 into a. Gantikan y = 3 ke dalam a.
3 + y = 5 Substitute y = 3 into a.
y = 5 – 3 2x + 3(3) = 5
= 2 2x + 9 = 5
2x = 5 – 9
Maka / Thus, x = 3 dan / and y = 2. 2x = –4
x = –2

Maka / Thus, x = –2 dan / and y = 3.

1. x – y = 6 ........................................... a 3. 5x – 2y = 6 ....................................... a
2x + y = 3 ........................................ b 2x + 3y = 10 .................................... b
a × 3: 15x – 6y = 18 ..................... c
a + b: 3x + 0 = 9 b × 2: 4x + 6y = 20 ..................... d
3x = 9 c + d: 19x + 0 = 38
x = 3
x = 2
Gantikan x = 3 ke dalam b. Gantikan x = 2 ke dalam a.
Substitute x = 3 into b. Substitute x = 2 into a.
2(3) + y = 3 5(2) – 2y = 6
6 + y = 3 10 – 2y = 6
y = 3 – 6 –2y = 6 – 10
y = –3 –2y = – 4
y = 2
Maka / Thus, x = 3 dan / and y = –3.
Maka / Thus, x = 2 dan / and y = 2.
2. 5x – 2y = 5 ........................... a 4. 4x – 5y = 13 ......................... a
3x – 2y = –1 ........................ b 2x – 3y = 7 ......................... b
a – b: 2x + 0 = 6 b × 2: 4x – 6y = 14 .......... c
2x = 6 a – c: 0 + y = –1
x = 3 y = –1
Gantikan x = 3 ke dalam b. Gantikan y = –1 ke dalam a.
Substitute x = 3 into b. Substitute y = –1 into a.
3(3) – 2y = –1 4x – 5(–1) = 13
9 – 2y = –1 4x + 5 = 13
–2y = –1 – 9 4x = 13 – 5
–2y = –10 4x = 8
y = 5 x = 2

Maka / Thus, x = 3 dan / and y = 5. Maka / Thus, x = 2 dan / and y = –1.

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