Elimination

Elimination
以下數條,請幫忙
1.{a/2 +b/3=13
{a/3 -b/4=3

We have

a/2 + b/3 = 13 ... (1)
a/3 - b/4 = 3 ... (2)

Consider 4×(1) - 6×(2):

4(a/2 + b/3) - 6(a/3 - b/4) = 4(13) - 6(3)

2a + 4b/3 - 2a + 3b/2 = 34

17b/6 = 34

b = 12

Consider 9×(1) + 12×(2):

9(a/2 + b/3) + 12(a/3 - b/4) = 9(13) - 12(3)

9a/2 + 3b + 4a - 3b = 153

17a/2 = 153

a = 18

So a = 18, b = 12.

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2.{4x- 3/2y=6
{2x+ 3/2y=6

We have

4x - 3/2y = 6 ... (1)
2x + 3/2y = 6 ... (2)

Consider (1) + (2):

(4x - 3/2y) + (2x + 3/2y) = 6 + 6

6x = 12

x = 2

Consider (1) - 2×(2):

4x - 3/2 y - 2(2x + 3/2 y) = 6 - 2(6)

4x - 3/2 y - 4x - 3y = -6

-9y/2 = -6

y = 4/3

So x = 2, y = 4/3.

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3.{x/5 y/4=1
{5-y=x-1

你的問題似乎漏打了一個 "+" 或 "-" 號。

如果是 "+"：

We have

x/5 + y/4 = 1 ... (1)
5 - y = x - 1 ... (2)

Rearrange (2) becomes

x + y = 6 ... (3)

Consider 4×(1) - (3)

4(x/5 + y/4) - (x + y) = 4(1) - 6

4x/5 + y - x - y = -2

-x/5 = -2

x = 10

Consider 5×(1) - (3)

5(x/5 + y/4) - (x + y) = 5(1) - 1

x + 5y/4 - x - y = 4

y/4 = 4

y = 16

So the solution is x = 10, y = 16.

如果是 "-"：

We have

x/5 - y/4 = 1 ... (1)
5 - y = x - 1 ... (2)

Rearrange (2) becomes

x + y = 6 ... (3)

Consider 4×(1) + (3)

4(x/5 - y/4) + (x + y) = 4(1) + 6

4x/5 - y + x + y = 10

9x/5 = 10

x = 50/9

Consider 5×(1) - (3)

5(x/5 - y/4) - (x + y) = 5(1) - 1

x - 5y/4 - x - y = 4

-9y/4 = 4

y = -16/9

So the solution is x = 50/9, y = -16/9

我認為 "+" 的情況比較可能，因為答案較美觀 ^^

1. first equation x 1/3, second equation x 1/2
first equation - secondary equation:
(a/2 x 1/3 + b/3 x 1/3) - (a/3 x 1/2 - b/4 x 1/2) = 13/3 - 3/2
a/6+b/9-a/6+b/8=(13x2-3x3)/6
b/9+b/8=17/6
8b+9b/72=17/6
17b/72=17/6
b=72/6
b=12
put b=12 into second equation: a/3-12/4=3
a/3=6
a=18

2. second equation x 2
then second equation - first equation
[2(2x)+2(3/2)y]-(4x-3/2y)=2(6)-6
4x+3y-4x+3/2y=6
9/2y=6
y=4/3
put y=4/3 into first equation
4x- 3/2(4/3)=6
4x-2=6
4x=8
x=2

第三條, 是否打錯題目????