1. put 9y=-3x-39 into standard form 2. put -2x+2y=-2 into standard form 3. solve the system by either using substitution or elimination{-2x=20, 3x-4y=-2} 4.

1. 9y = -3x - 39add 3x to both sides
    3x + 9y = -39 <== standard form

2. -2x + 2y = -2. ..multiply by -1
     2x - 2y = 2 <== standard form

3. -2x = 20
     x = -20/2
     x = -10now sub this back into the second equation

     3x - 4y = -2
    3(-10) - 4y = -2
    -30 - 4y = -2
     -4y = -2 + 30
     -4y = 28
      y = -28/4
      y = -7

solution is (-10,-7) <== solution

1. Put 9y=-3x-39 into standard form 
           3x + 9y = -39 ...this is standard form

2. Put -2x+2y=-2 into standard form
           2x - 2y = 2  ...this is standard form

3. Solve the system by either using substitution or elimination
{-2x=20, 3x-4y=-2}

-2x = 20
   x = -10

substitute x = -10 into  3x-4y=-2

3x-4y=-2
3(-10)-4y=-2
-30 - 4y = -2
4y = -28
  y = -7

answer
x = -10 and y = -7

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

The answer is something i don’t know