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 [tex]y=5x-1[/tex], then when [tex]x=0[/tex], you get [tex]y=-1[/tex]; when [tex]x=1[/tex], you get [tex]y=4[/tex]. So plot the points (0, -1) and (1, 4), then strike a line through.
1. Notice that dividing both sides of [tex]2y=10x-2[/tex] by 2 returns [tex]y=5x-1[/tex], 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 [tex]x[/tex] or [tex]y[/tex], you can always find a corresponding value for the other variable.)
2. See attached image. [tex]3x-y=2[/tex] 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 [tex]x+3y=9[/tex], so there will be an infinite number of solutions.
4. (by substitution) Since [tex]y=-7x+3[/tex], we can replace [tex]y[/tex] in the second equation:
[tex]-7x+3+7x=10\implies3=10[/tex]
but this is false, so there are no solutions to this system.
5. (by substitution) Since [tex]x=2y+2[/tex], in the first equation we have
[tex]-5(2y+2)+3y=-10y-10+3y=-7y-10=11\implies-7y=21\implies y=-3[/tex]
Then back in the second equation we find
[tex]x=2(-3)+2=-6+2=-4[/tex]
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.
