Kiersten, this isn't as bad as it looks, I promise.
We want to turn each one into a graphable equation. The first is already done. Start at our y-intercept of 3 and go up 3 and right 2. We can also go down 3 and left 2. Since we have "or equal to" we use a solid line. Now, shade everything below your line. I would use diagonal lines that are perpendicular to your line.
The second equation can be turned into y = -2x + 10. We start at 10 and go down 2 and right one, as well as up 2 and left 1. Since we have just <, we use a dotted line. Shade everything underneath it in the same manner as the first.
What about the third? This gives us a horizontal line at -1. Anywhere on this line, y = -1. Since we have just >, we use a dotted line again. Shade everything above the line in the same manner again.
Where all three shadings overlap is our solution. It should be the triangle formed by all three lines.
Want to check? Pick a point in that area and make sure it works in all three equations. Let's try (3,1)
1 < 3/2(3) + 3
1 < 2 + 3
1 < 5. True
2(3) + 1 < 10
6 + 1 < 10
7 < 10. True
1 > -1. True.
Hope this helps!
