Page 37 - Hybrid PBD Matematik TG3_Edisi Guru

Matematik Tingkatan 3 Bab 1

1. Hitung nilai-nilai x yang mungkin bagi persamaan 2 × 2 = 2 1 10 .
x 2
7x
x 2
Calculate the possible values of x for the equation 2 × 2 = 1 .
7x
2 10 KBAT Mengaplikasi
1 • Menyamakan indeksnya dan
x 2
2 × 2 =
7x
2 10 selesaikan dengan kaedah
x 2
2 × 2 = 2 pemfaktoran.
7x
–10
Equate the indices and solve using
2 x 2 + 7x = 2 –10 factorisation method.

Secara perbandingan / By comparison,
x + 7x = –10
2
x + 7x + 10 = 0
2
(x + 5)(x + 2) = 0
x + 5 = 0
x = –5
atau/ or
x + 2 = 0
x = –2

Maka, nilai-nilai x yang mungkin ialah –5 dan –2.
Hence, the possible values of x are –5 and –2.

2. Selesaikan persamaan serentak berikut. KBAT Mengaplikasi
Solve the following simultaneous equations.
1
64 × 2 = 2 dan/ and 3 × = 27
2
2y
x
x
3 y
• Menyamakan indeksnya dan
1
x
64 × 2 = 2 2 3 × = 27 selesaikan dengan kaedah
2y
x
2 × 2 = 2 2 3 y penggantian atau penghapusan.
2y
6(x)
3
–y
x
2 6x + 2y = 2 2 3 × 3 = 3 Equate the indices and solve using
3
6x + 2y = 2 …… 1 3 x – y = 3 substitution or elimination method.
x – y = 3 …… 2
Daripada / From 2, y = x – 3 …… 3
Gantikan 3 ke dalam 1, Gantikan x = 1 ke dalam 3,
Substitute 3 into 1, Substitute x = 1 into 3,
6x + 2(x – 3) = 2 y = x – 3
6x + 2x – 6 = 2 = 1 – 3
8x = 8 = –2
x = 1
Maka / Hence, x = 1 dan / and y = –2

Kuiz 1

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