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Question

The number of solutions of the system of equations 2x+yz=7, x3y+2z=1 and x+4y3z=5 is

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is A 0
Consider the given system of equations
2x+yz=7 ....(i)
x3y+2z=1 .....(ii)
and x+4y3z=5 .....(iii)
From equations (i) and (ii), we get
5xy=15 ....(iv)
and from equations (i) and (iii), we get
5xy=16 ....(v)
Since, equations (iv) and (v) shows that they are parallel and so solution does not exist.

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