Solve, graph, and write the solution in interval notation. 1) -3<x+3 <= 3

Step 1
1. $-<!--\; -\; -->3<x+3\le 3$. On adding -3 to all the 3 sides, we get $-<!--\; -\; -->6<x\le 0$. Thus x belongs to (-6,0] . The graph is the part of 2nd and 3rd quadrants between the line $x=-<!--\; -\; -->6$ ( excluding this line) and the Y-Axis, i.e. the line $x=0$ ( including this line).
2. $5\le <!--\; \le \; -->2x-<!--\; -\; -->1\le <!--\; \le \; -->9$. On adding 1 to all the 3 sides, we get $6\le <!--\; \le \; -->2x\le <!--\; \le \; -->10$. Now, on dividing all the 3 sides by 2, we get $3\le <!--\; \le \; -->x\le <!--\; \le \; -->5$. Thus, x belongs to [3,5]. The graph is the part of 1st and 4th quadrants between the line $x=3$ the line $x=5$ (including both these lines).
3. $-<!--\; -\; -->22\le <!--\; \le \; -->4y-<!--\; -\; -->2<-<!--\; -\; -->10$. On adding 2 to all the 3 sides, we get $-<!--\; -\; -->20\le <!--\; \le \; -->4y<-<!--\; -\; -->8$. Now, on dividing all the 3 sides by 4, we get $-<!--\; -\; -->5\le <!--\; \le \; -->y<-<!--\; -\; -->2$.Thus, y belongs to [-5,-2). The graph is the part of 3rd and 4th quadrants between the line $y=-<!--\; -\; -->5$( including this line) and the line $y=-<!--\; -\; -->2$ ( excluding this line).
4. $2x<8$ and $x+11>6$. On dividing both the sides of the 1st inequality by 2, we get $x<4$. Further, on adding -11 to both the sides of the 2nd inequality, we get $x>-<!--\; -\; -->5$. Thus, x belongs to (-5, 4). The graph is the part of 1st and 2nd quadrants between the line $x=-<!--\; -\; -->5$ and the line $x=4$(excluding both these lines).
5. $x\ge <!--\; \ge \; -->-<!--\; -\; -->9\text{and}x+2<2$. On adding -2 to both the sides of the 2nd inequality, we get $x<0$. Thus, x belongs to [-9,0). The graph is the part of 1st and 2nd quadrants between the line $x=-<!--\; -\; -->9$ and the line $x=0$ (including the 1st line but excluding the 2nd line).
Step 2
6. $-<!--\; -\; -->3(x+2)\ge <!--\; \ge \; -->0$ and $x+9\le <!--\; \le \; -->1$. On dividing both the sides of the 1st inequality by -3, we get $x+2\ge <!--\; \ge \; -->0$. Now, on adding -2 to both the sides , we get $x\ge <!--\; \ge \; -->-<!--\; -\; -->2$. Now, on adding -9 to both the sides of the 2nd inequality, we get $x\le <!--\; \le \; -->-<!--\; -\; -->8$. The conditions $x\ge <!--\; \ge \; -->-<!--\; -\; -->2$ and $x\le <!--\; \le \; -->-<!--\; -\; -->8$ cannot be satisfied together. Hence there is no solution.
7. $x<5$ and $x-<!--\; -\; -->4\ge <!--\; \ge \; -->3$. On adding 4 to both the sides of the 2nd inequality, we get $x\ge <!--\; \ge \; -->7$.The conditions $x<5$ and $x\ge <!--\; \ge \; -->7$ cannot be satisfied together. Hence there is no solution.
8. $x+1\ge <!--\; \ge \; -->4$ and $x-<!--\; -\; -->6\ge <!--\; \ge \; -->2$. On adding -1 to both the sides of the 1st inequality, we get $x\ge <!--\; \ge \; -->3$. On adding 6 to both the sides of the 2nd inequality, we get $x\ge <!--\; \ge \; -->8$. Hence $x\ge <!--\; \ge \; -->8$. Thus, x belongs to $[8,\mathrm{\infty <!--\; \infty \; -->})$. The graph is the part of the 1st and 4th quadrants to the right of the line $x=8$( including this line).
9. $x+8\ge <!--\; \ge \; -->15\text{or}5x\le <!--\; \le \; -->30$. On adding -8 to both the sides of the 1st inequality, we get $x\ge <!--\; \ge \; -->7$. On dividing both the sides of the 2nd inequality by 5, we get $x\le <!--\; \le \; -->6$. Thus, x belongs to $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},6]\cup <!--\; \cup \; -->[7,\mathrm{\infty <!--\; \infty \; -->})$. The graph is the part of xy plane to the left of the line $x=6$ ( including this line) and to the right of the line $x=7$ ( including this line).
10. $x-<!--\; -\; -->3\le <!--\; \le \; -->-<!--\; -\; -->9\text{or}x+4\ge <!--\; \ge \; -->3$. On adding 3 to both the sides of the 1st inequality, we get $x\le <!--\; \le \; -->-<!--\; -\; -->6$. On adding -4 to both the sides of the 2nd inequality, we get $x\ge <!--\; \ge \; -->-<!--\; -\; -->1$. Thus, x belongs to $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}-<!--\; -\; -->6]\cup <!--\; \cup \; -->(-<!--\; -\; -->1,\mathrm{\infty <!--\; \infty \; -->})$. The graph is the part of xy plane to the left of the line $x=-<!--\; -\; -->6$ ( including this line) and to the right of the line $x=-<!--\; -\; -->1$ ( including this line).