multiply equation 2 with 2=2x1+2x2+6x3=2-------(3)
x1+2x2+6x3=2-------(1)
subtract 3 from 1= x1=0
substitute value of x1 in equation 1,2
2x2+6x3=2
x2+3x3=1
both equations are same so there are infinte number of solution
let x2=t
x2+3x3=1
t+3x3=1
x3=(1-t)/3
(x1,x2,x3)=(0,t,(1-t)/3)
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