maths~~~F.2

x-2y-8=0 --- (1)
5x=6-7y---- (2)

In 1
x = 2y + 8

put 1 into 2
5(2y+8) = 6 - 7y
10y + 40 = 6- 7y
17y = -34
y=-2 ----- (3)

put 3 into 2
5x=6-7(-2)
x=4


2.Use the method of substitutation and the method of elimination.
2x-3y=0 ----- (1)
3x-4y=1 ----- (2)

In 1
x=3y / 2 ---- (3)

put 3 into 2

3(3y/2)- 4y =1
(9y/2) - 4y = 1
9y-8y = 2
y=2 ---- (4)

put 4 into 1
2x-3(2)=0
x=3

3.Solve the simultaneous equation
72-2m+n=6m+7n ---- (1)
6m+7n=3m+8n-25 ----(2)

in 1
72 - 8m = 6n
36 - 4m = 3n
n = (36-4m)/3 ---- (3)

in 2
6m+7n=3m+8n-25
3m =n -25 ---- (4)

put 3 into 4
3m =((36-4m)/3) -25
3m + 25 = (36-4m)/3
9m +75 = 36 -4m
13m = -39
m=-3 ---- (5)

put 5 into 3
n = (36-4(-3))/3
n=8

Common salt consists of sodium and chlorine in the ratio 23 : 35.5 by weight. How many grams of sodium are there in 117 g of salt?

grams of sodium
= 117 (23/(23+35.5))
= 46 g

2007-01-03 23:31:51 補充：
sorry my mistke In Q3put 5 into 3n = (36-4(-3))/3n =( 36+12) /3n=16

1) x-2y-8=0........(1)
5x=6-7y.........(2)
From (1), x=2y+8 so 5x=5(2y+8)=10y+40.....(3)
so 10y+40=6-7y, and therefore y=2. Put y=2 into (1), x=12
2) 2x=3y........(1)
3x-4y=1........(2)
from (1), 6x=9y and from (2) 6x-8y=2 by simple mutiplication
so 9y=8y+2 and therefore y=2
3) Divide it into two parts,
72-2m+n=6m+7n..........(1)
and 6m+7n=3m+8n-25...........(2)
from (1), 8m+6n=72, so 4m+3n=36......(3)
from (2), 3m-n+25=0........(4)
from (4), n=3m+25.........(5)
put (5) into (3), so 4m+3(3m+25)=36, so 13m=-39 and therefore m=-3, so n=16
4) grams of sodium = 117*23/(23+35.5)=46