數學方程式問題

Let x be the cost of one apple and y be the cost of one orange.

The cost of an apple is a half of the cost of an orange:
x = y/2

Mrs. Chu buys 3 apples and 4 oranges at $44:
3x + 4y = $44

3x + 4y = $44
3(y/2) + 4y = $44 (substitute y/2 for x)
2(3y/2 + 4y) = 2($44)
3y + 8y = $88
11y = $88
y = $8

x = y/2
x = $8/2 (substitute $8 for y)
x = $4

∴ an apple costs $4 and an orange costs $8.
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"元“ 的數量取決於不同的未知數的數量。
(例：這兩條數式都各有兩個不同的未知數，因此它們是 ”二元“ 的。)

”次“ 的數量取決於未知數的最高冪 (次方數) 的數量
(例：這兩條數式的未知數最高冪是 1 ，即是最高次方數是 1 ，因此它們是 “一次” 的。)

因此，這兩條數式是 “二元一次聯立方程式” 。

Cost of Apple = a , cost of orange = 2a

Then
3a + 4 x (2a) = 44
3a + 8a = 44
11a = 44
a = 4

i.e Apple is $4, Orange is $8