# Math

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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

• Math -

I will do #3, you try the others

3.

2x - 6y = -2
x = 3y-1

sub into the 1st:
2(3y-1) - 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

• Math -

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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