Solve the system of equations:
y=2x+2 ................(1)
y=-2x+a+1............(2)
for x and y
(1)-(2)
4x-a+1=0 => x=(a-1)/4
(1)+(2)
2y=a+3 => y=(a+3)/2
x>0 if (a-1)/4>0 => a>1
y>0 if (a+3)/2>0 => a>-3
For both x & y >0, we have the condition a>1.
Note that if a=1, then the intersection poin lies ON the x-axis, and therefore NOT in quadrant 1.
