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Question
Solve the following systems of equations:
`x/3 + y/4 =11`
`(5x)/6 - y/3 = -7`
Solution
The given equations are:
`x/3 + y/4 =11`...(i)
`(5x)/6 - y/3 = -7` .....(ii)
From (i) we get
`(4x + 3y)/12 = 11`
=> 4x + 3y = 132 ....(iii)
From (ii), we get
`(5x + 2y)/6 = -7`
=> 5x - 2y = -42 ....(iv)
Let us eliminate y from the given equations. The coefficients of y in the equations(iii) and (iv) are 3 and 2 respectively. The L.C.M of 3 and 2 is 6. So, we make the coefficient of y equal to 6 in the two equations.
Multiplying (iii) by 2 and (iv) by 3, we get
8x + 6y = 264 ...(v)
15x - 6x = -126 ...(vi)
Adding (v) and (vi), we get
8x + 15x = 264 - 126
=> 23x = 138
`=> x = 138/23 = 6`
Substituting x = 6 in (iii), we get
4 x 6 + 3y = 132
=> 3y = 132 - 24
3y = 108
`=> y = 108/3 = 36`
Hence, the solution of the given system of equations is x = 6, y = 36.
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