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February 1, 2015

February 1, 2015

Posted by **Muiz** on Friday, January 11, 2013 at 12:49am.

- math -
**Reiny**, Friday, January 11, 2013 at 8:02amI will assume:

2x^-1 - 3y^-1 = 4 ---> 2/x - 3/y = 4

and

4x^-1 + y^-1 = 1 ----> 4/x + 1/y = 1

double the first : 4/x - 6/y = 8

keep the 2nd : 4/x + 1/y = 1

subtract them:

-7/y = 7

7y = -7

y = -1

back into the 2nd:

4/x + 1/-1 = 1

4/x = 2

2x = 4

x = 2

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