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Solve the following simultaneous equations
log x + 2 log y = 2
｛ 3x - 2y = 2
maths 緊急緊急 [10]
2008-02-18 9:12 am
回答 (2)
2008-02-18 1:42 pm
✔ 最佳答案
log x + 2 log y = 2=> x*y^2=100
=>x=100/y^2...(i)
Substituting (i) into 3x-2y=2
=>300/y^2-2y=2
=>300-2y^3=2y^2
=>150-y^3=y^2
=>y^3+y^2=150
=>y(y^2+1)=150
hmmm, then i am not sure...
2008-02-18 6:51 pm
3x-2y=2-------1
logx+2logy=2-------2
from 2,
logxy^2=2
xy^2=100------3
from1,
y=(3x-2)/2-----4
put 4 int to 3
9x^3-12x^2+4x=400
(x-4)(9x^2+24x+100)=0
thus,x=4
put x=4 onto 3
y=5 or y=-5(rejected)
therefore,x=4,y=5
logx+2logy=2-------2
from 2,
logxy^2=2
xy^2=100------3
from1,
y=(3x-2)/2-----4
put 4 int to 3
9x^3-12x^2+4x=400
(x-4)(9x^2+24x+100)=0
thus,x=4
put x=4 onto 3
y=5 or y=-5(rejected)
therefore,x=4,y=5
參考: meself
收錄日期: 2021-04-20 13:42:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080218000051KK00207