Let's solve this by substitution. We're told:
3x + y = 9
6x = 4 - 2y
Let's start with the second equation, and solve for y:
6x = 4 - 2y
3x = 2 - y
2 - 3x = y
Now we can substitute it into the first equation, replacing all occurrences of y with 2 - 3x:
3x + (2 - 3x) = 9
2 = 9
Ah! That of course makes no sense, which tells us that the lines do not intersect at all, but are in fact parallel to each other. Let's confirm that by rearranging each of them into standard form:
3x + y = 9
y = 9 - 3x
y = -3x + 9
6x = 4 - 2y
y = 2 - 3x
y = -3x + 7
So there you have it. Those expressions describe two lines that have a slope of 3 and a y position that differs by 2.
At the point (2, 4).
It works out that they intersect at: (4, -7)
x = 3 and y = 2 so the lines intersect at the point (3, 2)
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
Those two statements are linear equations, not lines. If the equations are graphed, each one produces a straight line. The lines intersect at the point (-1, -2).
At the point (2, 4).
It works out that they intersect at: (4, -7)
By a process of elimination and substitution the lines intersect at: (4, -7)
By a process of elimination and substitution the lines intersect at: (1/4, 0)
x = 1 and y = 6 so the lines intersect at the point (1, 6)
x = 3 and y = 2 so the lines intersect at the point (3, 2)
x = 2 and y = 1 The lines intersect at the point (2, 1)
Solving the simultaneous equations works out as x = -2 and y = -2 So the lines intersect at: (-2, -2)
The lines are perpendicular, and intersect at the point (1.35, 3.55) .
Those two statements are linear equations, not lines. If the equations are graphed, each one produces a straight line. The lines intersect at the point (-1, -2).
Two straight lines that intersect.
None. When these two equations are graphed, the two lines are parallel. Since they never intersect, there is no point that satisfies both equations.