Re: Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


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Posted by Subhotosh Khan on October 11, 2002 at 13:42:59:

In Reply to: Re: Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! posted by Subhotosh Khan on October 11, 2002 at 13:40:05:

: : OK, here's the problem. I got one answer, my book got a different answer, and my calculator says both are wrong. Here's the equation:
: Assuming the actual problem is
: 1/(8*x)-1/(7*x)=2, solve for x

: : 1/8x-1/7x=2, solve for x

: : I got:
: : 7/56x-8/56x=2
: : -1/56x=2
: : -56x^-1=2 this should be (1/56)*x^(-1)=2
: : -28x^-1=1
: : x=28

: : The book got:
: : [1/7x-1/8x=2]56
: : 7-8=112x this should be 8/x - 7/x = 112
: : -1=112x
: : x=-1/112

: : My calculator got:
: : 1/8x-1/7x=2, x=-112 (using q solver), when I substituted -112 for x, it came out equaling 2
: : (1/{8[-112]}-1/{7[-112]}=2).

: : I know the calculator is right, and I did find a way to get that answer, and I also see how I got my answer and the book got its answer, but I can't figure out what the book, or I, did wrong.
: : P.S. those are one-eighth and one-seventh x.
: : :-)

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