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Question
Solve the following systems of equations:
`x + y/2 = 4`
`x/3 + 2y = 5`
Solution
The given system of equation is
`x + y/2 = 4` .......(i)
`x/3 + 2y = 5` ....(ii)
From (i), we get
`(2x + y)/2 = 4`
2x + y = 8
y = 8 - 2x
From (ii), we get
x + 6y = 15 ....(iii)
Substituting y = 8 - 2x in (iii), we get
x + 6(8 - 2x) = 15
`=> x + 48 - 12x = 15`
=> -11x = 15 - 48
=> -11x = -33
`=> x = (-33)/(-11) = 3`
Putting x = 3 in y = 8 - 2x we get
y = 8 - 2 x 3
= 8 - 6
= 2
y = 2
Hence, solution of the given system of equation is x= 3, y = 2
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