Can someone help me with this Algebra problem? I'm SO CONFUSED!?

Question

Solve the system of equations using substitution:
x=2y + 1
-6y = -3x + 3

A. Infinitely many solutions
B. no solution
C. (0,0)
D. (0,0) and (0,2)

Answer 1

put eq. 1 in 2
-6y=-3(2y+1) + 3
-6y=-6y-3+3
0=0
Since these two are equations for the same line there are infinite solutions. Choice A.

Answer 2

since x=2y+1 substitute into -6y = -3x + 3, ie.

-6y = -3 (2y + 1) + 3
-6y = -6y -3 + 3
0 = 0

this means there is no solution. Observe that if you rearrange
x = 2y + 1 you get

-2y = -x + 1

now multiply the equation by 3 and you get
-6y = -3x + 3 which is the second equation

What this means is that the lines parallel each other exactly and will never cross each other, hence there is no solution.

Answer 3

the 1st equation x=2y + 1 can be substituted into the second equation

-6y = -3x + 3
-6y = -3(2y + 1) + 3
-6y = -6y - 3 + 3
0=0
no solution

Answer 4

You can put in the x=(2y+1) where the x is in the other problem.
So
-6y = -3(2y+1) + 3. (You're just substituting).
Then figure the rest out.
-6y = -6y -3 + 3.
Which, when it it continued, ends up being 0=0.
Which means there are infinte solutions, because any number you stick in there will work.

Answer 5

x = 2y + 1
-6y = -3x + 3
Substitute #1 for x in #2:
-6y = -3(2y + 1) + 3
-6y = -6y -3 + 3
y is indeterminate
no solution

Answer 6

All you have you to do is plug x=2y+1 into -6y=-3x+3.
So the equation is -6y=-3(2y+1)+3.
-6y=-6y+0
So y=1.
Then plug y back into one of the original equations and solve for x.
The correct answer is (3,1), even though that is not one of the listed options.

Answer 7

-6y=-3(2y+1)+3
-6y=-6y-3+3
-6y+6y=0
0=0: therefore
NO SOLUTION

Answer 8

x = 2y + 1.
-6y = -3x + 3
The above equations can be rewritten as,
x - 2y = 1....................(1)
3x - 6y = 3..................(2)
(1) * 3 = 3x -6y = 3.....(3)
As eqn (3) &(2) are the same it has infinite solutions, because both the solutions represent the same line.

Answer 9

no solution

Answer 10

b)