Math
posted by Neil .
Solve by using the subatitution method. (Please keep fractions as they are, don't convert to decimal.)
1. y= 3x  1
2x  3y= 8
2. 1 + 3y= 10
5x + 2y= 6
3. 2x  6y= 2
x= 3y  1
4. y= 1/7x + 3
x  7y= 4

I will do #3, you try the others
3.
2x  6y = 2
x = 3y1
sub into the 1st:
2(3y1)  6y = 2
6y  2  6y = 2
0 = 0
Ahhh, since the variables dropped out and we ended up with a TRUE statement, the two equations are really one and the same equation.
(if you simplify the first, you get the second ...
2x  6y = 2
divide by 2
x  3y = 1
x = 3y  1, which was the second equation )
So any ordered pair which satisfies the 1st will obviously also satisfy the second.
Had we ended up with a false statement, such as 3 = 0, there would have been no solution at all 
Hm... so, 0=0 is the only answer for #3? You don't have to do anything else? Like plugging in for the second equation or something.
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