Solve the following simultaneous equations by the method of elimination.
1) 5r+13s = 1-2r+4s = 4
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1) Mr Law is 24 years older than his daughter. 10 tears ago, his daughter was half his age. Find the present ages of Mr Law and his daughter respectively.
2) The sum of the digits in a two-digit number is 14.If the digits are reversed, the value of the number decreases by 18. Find the original number.
3) Suppose the cost $C of printing n magazines is calculated by the following formula :
C = a+bn
where a and b are non-zero constants.
Given that the costs of printing 500 magazines and 2000 magazines are $4500 and $12 000 respectively, find
a) the values of a and b,
b) the cost of printing 6000 magazines
MATHS - elimination + 文字題
2013-02-19 5:48 am
回答 (2)
2013-02-19 7:14 am
✔ 最佳答案
1)5r + 13s= 4...... [1]
1 - 2r + 4s= 4...... [2]
[1] x 2 :
10r + 26s = 8 ...... [3]
[2] x 5 :
5 - 10r + 20s = 20
- 10r + 20s = 15 ...... [4]
[3] + [4] :
46s = 23
s = 0.5
Put s = 0.5 into [1] :
5r + 13(0.5) = 4
5r = -2.5
r = -0.5
Hence, r = -0.5and s = 0.5
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1)
Let m and d be the present age of Mr. Law and that of his daughter respectively.
m - d = 24 ...... [1]
(d - 10) = (m - 10)/2 ...... [2]
From [2] :
2d - 20 = m - 10
-m + 2d = 10 ...... [3]
[1] + [3] :
d = 34
Put d = 34 into [1] :
m - (34) = 24
m = 58
The present age of Mr. Law = 58
Thepresent age of Mr. Law daughter = 34
2)
Let t and u be the tens digits and the units digit respectively.
t + u = 14 ...... [1]
(10t + u) - (10u + t) = 18 ...... [2]
From [2] :
10t + u - 10u - t = 18
9t - 9u = 18
t - u = 2 ...... [3]
[1] + [3] :
2t = 16
t = 8
[1] - [3] :
2u = 12
u = 6
The original number = 86
3)
a)
4500 = a + 500b ...... [1]
12000 = a + 2000b ...... [2]
[2] - [1] :
1500b = 7500
b = 5
Put b = 5 into [1] :
a + 500(5) = 4500
a = 2000
Hence, a = 2000and b = 5
b)
C = 2000 + 5n
When n = 6000 :
C = 2000 + 5(6000)
C = 32000
The required cost of printing = $32000
參考: 賣女孩的火柴
2013-02-22 3:23 am
Elimination:
1) 5r+13s = 1-2r+4s = 4
5r+13s = 4 ...(1)
1-2r+4s = 4 ... (2)
From (2),
1-2r+4s=4
-2r+4s=4-1
-2r+4s=3 ... (3)
(1)x2 :
2(5r+13s)=2x4
10r+26s=8 ... (4)
(3)x5:
5(-2r+4s)=5x3
-10r+20s=15 ... (5)
(4)+(5):
10r+26s+(-10r+20s)=8+15
46s = 23
s=0.5
Sub s=0.5 into (1)
5r+13s=4
5r+7.5=4
5r=-2.5
r=-0.5
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1.Let x be the present age of Mr Law.
x-24-10=(x-10)/2
2(x-34)=[(x-10)/2]x2
2x-68=x-10
2x-x=-10+68
x=58
When x=58,
her daughter present age=58-24=34
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2.Let x and y be the ten digit and one digit respectively.
x+y=14 ...(1)
(10x+y)-(10y+x)=18 ...(2)
From (2),
(10x+y)-(10y+x)=18
10x+y-10y-x=18
9x-9y=18 ...(3)
(1)x9:
9x+9y=126 ...(4)
(3)+(4):
9x-9y+(9x+9y)=18+126
18x=144
x=8
Sub x=8 into (1),
x+y=14
8+y=14
y=6
So the original no. is 86.
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3.a) 4500 =a+500b ...(1)
12000 =a+2000b ...(2)
(1)x4:
4500x4=4a+2000b
18000=4a+2000b ...(3)
(3)-(2):
18000-12000=4a+2000b-(a+2000b)
6000=3a
a=2000
Sub a=2000 into (1)
4500 = a+500b
4500 = 2000+500b
2500 = 500b
b=5
b) C=a+6000b
C=2000+6000(5)
C=32000
So the cost is $32000.
1) 5r+13s = 1-2r+4s = 4
5r+13s = 4 ...(1)
1-2r+4s = 4 ... (2)
From (2),
1-2r+4s=4
-2r+4s=4-1
-2r+4s=3 ... (3)
(1)x2 :
2(5r+13s)=2x4
10r+26s=8 ... (4)
(3)x5:
5(-2r+4s)=5x3
-10r+20s=15 ... (5)
(4)+(5):
10r+26s+(-10r+20s)=8+15
46s = 23
s=0.5
Sub s=0.5 into (1)
5r+13s=4
5r+7.5=4
5r=-2.5
r=-0.5
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1.Let x be the present age of Mr Law.
x-24-10=(x-10)/2
2(x-34)=[(x-10)/2]x2
2x-68=x-10
2x-x=-10+68
x=58
When x=58,
her daughter present age=58-24=34
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2.Let x and y be the ten digit and one digit respectively.
x+y=14 ...(1)
(10x+y)-(10y+x)=18 ...(2)
From (2),
(10x+y)-(10y+x)=18
10x+y-10y-x=18
9x-9y=18 ...(3)
(1)x9:
9x+9y=126 ...(4)
(3)+(4):
9x-9y+(9x+9y)=18+126
18x=144
x=8
Sub x=8 into (1),
x+y=14
8+y=14
y=6
So the original no. is 86.
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3.a) 4500 =a+500b ...(1)
12000 =a+2000b ...(2)
(1)x4:
4500x4=4a+2000b
18000=4a+2000b ...(3)
(3)-(2):
18000-12000=4a+2000b-(a+2000b)
6000=3a
a=2000
Sub a=2000 into (1)
4500 = a+500b
4500 = 2000+500b
2500 = 500b
b=5
b) C=a+6000b
C=2000+6000(5)
C=32000
So the cost is $32000.
參考: me
收錄日期: 2021-04-13 19:18:09
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