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Question
Solve each of the following systems of equations by the method of cross-multiplication :
`2/x + 3/y = 13`
`5/4 - 4/y = -2`
where `x != 0 and y != 0`
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Solution
The given system of equation is
``2/x + 3/y = 13`
`5/x - 4/y = -2 " where " x!= 0 and y != 0`
Let `1/x = u` and `1/y = v`, Then the given system of equation becomes
2u + 3v = 13
5u - 4v = -2
here
`a_1 = 2, b_1 = 3, c_1 = -13`
`a_2 = 5, b_2 = -4, c_2 = 2`
By cross multiplication, we have
`=> u/(3xx2-(-13)xx(-4)) = (-v)/(2xx2-(-13)xx5) = 1/(2xx (-4)-3 xx 5)`
`=> u/(6 - 52) = (-v)/(4 + 65) = 1/(-8 - 15)`
`=> u/(-46) = (-v)/69 = 1/(-23)`
Now
`u/(-46) = 1/(-23)`
`=> u = (-46)/(-23) = 2`
And
`(-v)/69 =1/(-23)`
`=> v = (-69)/(-23) = 3`
Now
`x = 1/u = 1/2`
And
`y = 1/v = 1/3``
Hence `x = 1/2, y = 1/3` is the solution of the given system of equations.
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