(a-b)x-(a+b)y=0 ((turn subject to a))
&
c= 1/6a - 1/2b((turn subject to b))
how to calculate
maths
2006-10-29 12:28 am
回答 (2)
2006-10-29 1:01 am
( a-b)x-(a+b)y=0 ((turn subject to a))
ax-bx-ay-by=0
a(x-y)=bx+by
a=(bx+by)/(x-y)
c= 1/6a - 1/2b((turn subject to b))
c+1/(6a)=-1/(2b)
6a+1/c=-2b
b=-(6ac+1)/(2c)
ax-bx-ay-by=0
a(x-y)=bx+by
a=(bx+by)/(x-y)
c= 1/6a - 1/2b((turn subject to b))
c+1/(6a)=-1/(2b)
6a+1/c=-2b
b=-(6ac+1)/(2c)
2006-10-29 12:43 am
(a-b)x-(a+b)y=0
ax-bx-ay-by=0
a(x-y)-b(x+y)=0
a(x-y)=b(x+y)
a=b(x+y)/(x-y)
c=1/6a-1/2b
1/2b=1/6a-c
b=1/3a-2c
ax-bx-ay-by=0
a(x-y)-b(x+y)=0
a(x-y)=b(x+y)
a=b(x+y)/(x-y)
c=1/6a-1/2b
1/2b=1/6a-c
b=1/3a-2c
收錄日期: 2021-04-23 00:09:54
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https://hk.answers.yahoo.com/question/index?qid=20061028000051KK03017