maths! maths! maths!

1.x + y = 4 … (1)
x – y = 2 … (2)
(1)–(2) => 2y = 2 => y = 1 (c)
2.2x – y = 5 ...(1)
x + 3y = 6 …(2)
3x(1)+(2) => 6x + x = 15 + 6
7x = 21 => x = 3 (c)
3.2x – 3y = 2 … (1)
3x – 5y = 2 … (2)
3x(1) – 2x(2) =>
–9y + 10y = 6 – 4
y = 2
Sub into (1)
2x – 6 = 2 = > 2x = 8 => x = 4 (b)
4.
Let price of apple and orange be $x and $y each.
2x + 3y = 13 … (1)
5x + 4y = 22 … (2)
5x(1) – 2x(2) =>
15y – 8y = 65 – 44
7y = 21 => y = 3
Sub into (1), 2x + 9 = 13 => x = 2
Let Peter buys u apples, so 11 – u oranges
Payment = 2u + 3(11 – u) = 29
–u + 33 = 29 => u = 4 (a)

1. c
x+y-(x-y)=4-2
x+y-x+y=2
2y=2
y=1

2. c

3. c
compare 2x-3y and 3x-5y, then you can see the difference between
the two =)

4. d
Let $a be the price of apples and $b be the price of oranges.
Mary: 2a+3b=13------------>10a+15b=65
John: 5a+4b=22------------->10a+8b=44
10a+15b-10a-8b = 65-44
7b = 21
b = 3
2a+3b = 13
2a+3*3 = 13
2a+9 = 13
2a = 4
a = 2
Peter paid $7 more than John, so he bought 1 orange and 2 apples
more.

1. x + y =4 and x - y = 2
兩式相減以除去 x 項,得 2y = 2
y = 1
2. 2x -y = 5 and x + 3y = 6
第一式乘3, 得6x - 3y = 15, 然後加第二式以除去 y項,
得 7x = 21
x = 3
3.
2x - 3y = 2 ... (1)
3x - 5y = 2 ... (2)
(1)乘以 3得 6x - 9y = 6 ... (3)
(2)乘以 2得 6x - 10y = 4 ... (4)
(3) - (4) 得 y = 2
所以答案係 b
4. 設 蘋果每個值 a元,橙每個值b元,
2a + 3b = 13 ... (1)
5a + 4b = 22 ... (2)
(1)乘以 5 得 10a + 15b = 65 ... (3)
(2)乘以 2 得 10a + 8b = 44 ... (4)
(3) - (4) 得 7b = 21
b = 3
代 b = 3 入 (1)得 a = 2
Peter bought 11 fruits ( including apples and oranges only ) and paid $ 29.
如果 4a + 7b = 29
所以答案係 a

1. C
x+y=4_______________________(1)
x-y=2________________________(2)
(1)-(2):
2y=2
y=1

2.C
2x -y = 5______________________(1)
x + 3y = 6______________________(2)

From(1),we have
y=2x-5_________________________(3)

Subst.(3) into (2)
x+3(2x-5)=6
x+6x-15=6
7x=21
x=3



3.B
2x - 3y =2_________________________(1)
3x - 5y = 2_________________________(2)

From(1),we have
3y=2x-2
y=(2x-2)/3__________________________(3)

Subst.(3) into (2)
3x – 5[(2x-2)/3]= 2
9x-5(2x-2)=6
9x-10x+10=6
-x=-4
x=4

Subst. x=4 into (1)
2(4)-3y=2
8-3y=2
3y=6
y=2


4A
Let the price of apples be x,then the price of oranges be y.

2x+3y=13___________________________(1)

5x+4y=22____________________________(2)

Solve (1) and(2),we have:
x=2,y=3

Let the number of apples Peter bought be z,then number of oranges Peter bought be 11-z.
2z+3(11-z)=29
2z+33-3z=29
-z=-4
z=4
∴Peter bought 4 apples.

1 . x - y = 2
x = 2 + y
x + y = 4
2 + y+ y = 4
2 y = 2
y = 1
c

2 . c 3

3 . 2x - 3y = 2
3x - 5y = 2

b {x = 4 y= 2


4 . a 4
第4條要用2次聯立方程