math homework, help..? Options

[missmissy]

Ok, I was sick for a while and I talked to the teacher about it. He explained it to me but he is a horrible explainer so I never really got what he said. So he told me to look in the text book and I still don't get it.. Do you think you can help?

graph each system of equations using the coordinate plane provided. Then determine whether the system has one solution, no solution, or infinitely man solutions. If the solution has one, name it.

x+y=3
x-y=3

2x-y=6
4x-2y=12

x+y=0
x+y=2


Do you think you could do two of those and make a graph of it on photoshop for me? You know, like make a grid and plot the lines? I just need examples so I can figure out the rest. There is a whole sheet of the things I need to do..

[runforfun529]

x+y=3
x-y=3

2x-y=6
4x-2y=12

x+y=0
x+y=2


What you would have to do for these is use the Elimination method for solving equations. The first one is already set up to use this method. To do this you would do the following:

1) Add the variables horizonally so that one variable crosses out (y's in this case)

x+y=3
x-y=3
2x=6

2) Solve that equation.

x=3

3) Then you would put x back in one of the equations to find y. So you would end up getting 3+y=3, therefore y=0. And your point of intersection would be (3,0).

The next one is a little harder because the equations are not set up so that 1 of the variables will cancel out. So you would have to do this:

1) Multiply or Divide 1 or both equations by a certain number to get one of the variables to cross out when added. For this one I'm going to multiply the 1st fraction by -2.

2x-y=6
4x-2y=12

-becomes-

-4x+2y=-12
4x-2y=12

2) Add

-4x+2y=-12
4x-2y=12
0=0

3) Everything cancels out so therefore, there is no point that they intersect at and they are therefore parallel.

The last one is just like the second.