Question

a+6b=1
a+3b=-2

Answer 1

I assume you want to solve for a and b?

Answer 2

it has to be either no solution, infinite many solutions or an ordered pair

Answer 3

But yes I am solving for a and b

Answer 4

Ok, that's fairly easy. Let's use the fact that both equations have a 1 coeffecient in front of the a. So we can subtract the second equation from the first, like so \[a+6b-(a+3b)=1-(-2)\]\[3b=3\]At which point it becomes obvious that b=1. Then you can take that value of b you just found and plug it into either equation to get a. =)

Answer 5

So then it would look like this a+3(1)=-2

Answer 6

Right.

Answer 7

So then, I would add 3(1)=-2 whch ten would be 3=-2, would I divide or multiply?

Answer 8

not add but multiply

Answer 9

Well, you'd have \[a+3=-2\] so you would just subtract 3 from both sides.

Answer 10

oh okay, so then it would be 1

Answer 11

It wouldn't have solution then? right?

Answer 12

What is -2-3?

Answer 13

oh its -5

Answer 14

Right! :)

Answer 15

so the ordered pair would be (1,-5)

Answer 16

Yes, if you wrote b first.

Answer 17

oh, (-5,1)