1條F.2 maths 題20分,超急

A(x-3) + B(2x+5) ≡ 29x + 1
Ax - 3A + 2Bx + 5B ≡ 29x + 1
(A+2B)x + (5B-3A) ≡ 29x + 1

A + 2B = 29 --- (1)
5B - 3A = 1 --- (2)

(1) × 3 :

3A + 6B = 87 --- (3)

(2) + (3) :

11B = 88
B = 88/11
B = 8 --- (4)

Sub. (4) into (1)

A + 2(8) = 29
A + 16 = 29
A = 29 - 16
A = 13

Answer : A = 13 , B = 8 .

kevinhtk1998 您好！我是 lop，高興能解答您的問題。

If the question is like this ↓ ↓ ↓ ↓ ↓

A(x-3) + B(2x+5) ≡ 29 + 1
Ax - 3A + 2Bx + 5B ≡ 30
(A+3B)x + (5B-3A) ≡ 30

A + 3B = 0 --- (1)
5B - 3A = 30 --- (2)

(1) × 3 :

3A + 9B = 0 --- (3)

(2) + (3) :

14B = 30
B = 30/14
B = 15/7 --- (4)

Sub. (4) into (1)

A + 3(15/7) = 0
A + 45/7 = 0
A = -45/7

Answer : A = -45/7 , B = 15/7 .

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If the question is this ↓ ↓ ↓ ↓ ↓

A(x-3) + B(2x+5) ≡ 29x + 1
Ax - 3A + 2Bx + 5B ≡ 29x + 1
(A+2B)x + (5B-3A) ≡ 29x + 1

A + 2B = 29 --- (1)
5B - 3A = 1 --- (2)

(1) × 3 :

3A + 6B = 87 --- (3)

(2) + (3) :

11B = 88
B = 88/11
B = 8 --- (4)

Sub. (4) into (1)

A + 2(8) = 29
A + 16 = 29
A = 29 - 16
A = 13

Answer : A = 13 , B = 8 .

Think the question should read :

If A(x-3)+B(2x+5)=29x+1 ,find the constants A and B.

A(x-3)+B(2x+5)=29x+1
Ax - 3A + 2Bx + 5B = 29x + 1
(A+2B)x + (-3A + 5B) = 29x + 1

therefore, we got :
A + 2B = 29 ........ (1)
-3A + 5B = 1 ........(2)

3 x (1) + (2) : 11B = 29x3 + 1
11B = 88
B = 8

thence (1) : A + 16 = 29
A = 13