Given the equations:

a + 3b = -3 ..............(1)

2b = a -12

==> -a+2b = -12 ............(2)

We have a system of two equations and two variables.

Then, we will use the elimination method to find a and b.

==> We will add (1) + (2).

==> 5b= -15

Now we will divide by 5.

==> b = -15/5 = -3

==> b= -3

Now we will substitute into (1) to find a.

==> a + 3b = -3

==> a + 3(-3) = -3

==> a -9 = -3

==> a = 9-3 = 6

==> a = 6

Then, the answer is:

**a = 6 and b = -3**

We have the following equations to solve for a and b:

a+3b=-3...(1)

2b=a-12...(2)

Adding (1) and (2)

=> a + 3b + 2b = -3 + a - 12

=> 5b =-15

=> b = -3

substituting this in (2)

=> -6 = a - 12

=> a = 12 - 6

=> a = 6

**Therefore a = 6 and b = -3.**

To find a and b:

a+3b = -3....(1)

2b = a-12. Or a-12 = 2b. Or a-2b =12...(2).

(1) - (2) = (a+3b)- (a-2b) = -3 -12 = 15.

3b+2b = -15,

5b = -15

5b/5 = -15/5 = -3.

Therefore b = -3. We substitute b= -3 in a+3b = -3. So a+3(-3) = -3.

So a = -3+3(3) = 6.

So a= 6 and b = -3.