For 1 and 2, you plot both lines, and wherever they intersect is the solution to the system. Given the equation of a line, I think the easiest way to plot it is to find two points on the line, then draw a line through them. For example, if
, then when
, you get
; when
, you get
. So plot the points (0, -1) and (1, 4), then strike a line through.
1. Notice that dividing both sides of
by 2 returns
, same as the first equation. So the system of equations reduces to one equation, which can have an infinite number of solutions. (This is because for any choice of
or
, you can always find a corresponding value for the other variable.)
2. See attached image.
is given by the purple line.
For 3-6, you have several options. The two simplest methods of solving them are by substitution or elimination.
3. Like with (1), notice that dividing both sides of the first equation by 2 gives
, so there will be an infinite number of solutions.
4. (by substitution) Since
, we can replace
in the second equation:

but this is false, so there are no solutions to this system.
5. (by substitution) Since
, in the first equation we have

Then back in the second equation we find

So (-4, -3) is the only solution here.
6. (by substitution) Notice that the left hand sides of both equations are the same, so we end up with 7 = 12, but this is false, so no solution exists.