The correct option is
A 2By solving both equations for x coordinate, we get
mx+1=9−3x4
⇒4mx+4=9−3x
Or, 3x+4mx+4=9
Or, x=5/(3+4m)
Since x has to be an integer, possible values of x are 5,−5,1,−1 for which,
the denominator on R.H.S. has to be 1,−1,5,−5 respectively.
Now, 3+4m=1,implies m=−0.5
3+4m=−1, implies m=−1
3+4m=5, implies m=0.5
3+4m=−5, implies m=−2.
Hence, two integral values of m are possible.