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प्रश्न
Solve the following systems of equations:
4u + 3y = 8
`6u - 4y = -5`
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उत्तर
Taking `1/x = u` then given equations become
4u + 3y = 8 ......(i)
6u - 4y = -5 ..........(ii)
From (i), we get
4u = 8 -3y
`=> u = (8 - 3y)/4`
Substituting u = `(8 - 3y)/4` in (ii) we get
From (ii), we get
`6((8-3y)/4) - 4y = -5`
`=> (3(8 - 3y))/4 - 4y = -5`
`=> (24 - 9y)/2 - 4y= -5`
`=> (24 - 9y - 8y)/2 = -5`
=> 24 - 17y = -10
=> -17y = -10 -24
=> -17y = -34
`=> y = (-34)/(-17) = 2`
Putting y = 2 in u =`(8-3y)/4` we get
`u = (8 - 3xx 2)/4 = (8-6)/4 = 2/4 = 1/2`
Hence x = 1/u = 2
So, the solution of the given system of equation is x 2, y
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