Answer:
3x+y=8 and 3x+y=9
Step-by-step explanation:
Assume these two equations:
a1x+b1y+c1=0
a2x+b2y+c2=0
If (a1/a2) = (b1/b2) ≠ (c1/c2) than those two would have no solution.
In these case, take a look at the last option:
[tex]3x + y = 8 \\ 3x + y = 9[/tex]
Re-write 'em as a standard form:
[tex]3x + y - 8 = 0 \\ 3x + y - 9 = 0[/tex]
So (a1/a2) = (b1/b2) ≠ (c1/c2) is true here:
[tex] \frac{3}{3} = \frac{1}{1} ≠ \frac{ - 8}{ - 9} [/tex]
And they do not have any collision point
