Thank you for the opportunity to help you with your question!
I had this question explained in my previous answer that expired even though the time froze, which I apologize for. Therefore if you need an explanation please ask me in the follow up question, I will be happy to explain it then.
The answer is Graph #3 from the left because the x and y points from the equation are x =2 and y = 4 and with <= 12 the shaded area will be below the line and the points will have solid circles.Please let me know if you need any clarification. I'm always happy to answer your questions.
How I got x =2 and y =4 are as follows:
Took the original equation; 6x +3y <= 12 and simplified first
*moved a common multiplier out of 6x+3y which was 3, which can go into 6 and 3
3(2x+y) <= 12
*Next, I divided by 3 to get x and y on one side
2x+y <= 4
Solved for the x-equation first
2x <= 4 -y
*divide by 2 to get x by itself
x<= 2-2y (EQ #1)
Solved for y-equation now
2x+y <= 4
*subtract 2x from both sides to get y by itself
y <= 4 -2x (EQ #2)
Now to solve EQ #1 the x-intercepts at y = 0
x<= 2 -2y
x<= 2 - 2(0)
x<= 2 therefore the point on the graph would be (2,0)
Now to solve EQ #2 the x-intercepts at x = 0
y<= 4 - 2x
y<= 4 -2(0)
y<= 4 therefore the other point on the graph would be (0,4)
**Hope this explains it in better detail and glad to have you as a student.
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