PreCalculus
posted by Caroline .
Solve by using the substitution method
1.x=3y+4 and 3x+9y=12
3(3y+4)+9y=12
9y+12+9y12
0=0
x=3(0)+4
x=4
Would there be infinite solutions or would it be (0,4)?
2. x=3y1 and 2x6y=2
2x6y=2
2(3y1)6y=2
6y26y=2
0=0
x=3y1
x=3(0)1
x=1
Would there be infinite solutions or would it be (1,0)?

for your first one, the variable dropped out, but you ended up with a true statement, 0=0
If that happens you will have an infinite number of solutions. In effect , the two equations are really the same equation.
look at the 2nd
3x + 9y = 12 , divide by 3
x + 3y = 4
or
x = 3y + 4 , which was your first equation.
If your variable drops out, but you end up with a false statement, such as 4 = 9
then there is no solution.
In that case, what you would be given would be two lines that are parallel, thus they never meet, thus no solution
The same result was true in your second question
notice by dividing each term of the 2nd by 2, then rearranging you get your first equation
I also noticed that in both cases you substituted x=0
but nowhere did it say that x = 0 , it said 0 = 0
no mention of x
So if you have a true statement, > infinite number of solutions
If you have a false statement > no solution at all 
Thank you so much.

welcome