A) \[0\]
B) \[1\]
C) \[2\]
D) \[4\]
Correct Answer: C
Solution :
Given, \[x+y=3\] and \[y=3mx+1\] \[\therefore \] \[x+3mx+1=3\] \[\Rightarrow \] \[x(1+3m)=2\] \[\Rightarrow \] \[x=\frac{2}{1+3m}\] Since, x is an integer, therefore \[1+3m=\pm 1,\pm 2\] \[\therefore \] \[3m=\pm 1-1,\,\,3m=\pm 2-1\] \[\Rightarrow \] \[m=0,\,-\frac{2}{3},\,\,m=\frac{1}{3},-1\] Since, m is an integer. So, \[m=0\] and \[-1\] exist.You need to login to perform this action.
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