Graphing Inequalities: Practice

Substitute $x = -2$, $y = 3$ into the right side $-x + 1$.

$3 \ge -(-2) + 1 \;\Rightarrow\; 3 \ge 3$

The point lies on the boundary, and because $\ge$ is inclusive the boundary is solid, so it is a solution.

The boundary passes through $(0, 1)$ and $(-1, 0)$, so it is $y = x + 1$. It is solid (inclusive) and the shaded region is above it.

$\text{solid line } y = x + 1,\ \text{shaded above} \;\Rightarrow\; y \ge x + 1$

The origin is unshaded, and $0 \ge 1$ is false, which matches.

The boundary passes through $(0, 2)$ and $(2, 0)$, so it is $y = -x + 2$. It is dashed (strict) and the shaded region, containing the origin, is below it.

$\text{dashed line } y = -x + 2,\ \text{shaded below} \;\Rightarrow\; y < -x + 2$

The origin is shaded, and $0 < 2$ is true, which matches.

Test the point in each inequality; a system needs both to hold.

$2 > 0 \ \text{(true)}, \qquad 2 < 1 + 3 = 4 \ \text{(true)}$

Both are true, so the point is in the feasible region.

Test each point against both conditions.

$(2, 1):\ 2 + 1 = 3 \le 5 \ \text{and}\ 2 \ge 1 \ \text{(both true)}$

The point $(0, 2)$ fails $x \ge 1$, while $(4, 3)$ and $(1, 6)$ fail $x + y \le 5$. Only $(2, 1)$ works.

The boundary $y = 2$ is horizontal, $\ge$ is inclusive so the line is solid, and larger $y$ is upward.

$y \ge 2 \;\Rightarrow\; \text{solid line } y = 2,\ \text{shade above}$

Inclusive and greater-than together give a solid line shaded upward.

Substitute into the left side $2x - y$.

$2(-3) - (-1) = -6 + 1 = -5, \quad \text{so}\ -5 \le -4$

That is true, so the point is a solution.

A false result means the origin is not a solution, so the solutions lie across the boundary.

$0 > 12 \ \text{(false)} \;\Rightarrow\; \text{shade the other side}$

One test point settles the whole graph; there is no need to keep testing.

The line $y = 3x$ passes through the origin, so the origin cannot decide the side. The point $(2, 6)$ also lies on the line since $6 = 3(2)$.

$(1, 0):\ 0 \le 3(1) = 3 \ \text{(off the line)}$

Only $(1, 0)$ is off the boundary, so it is the usable test point.

Because the origin fails the test, the solution region is on the far side of the boundary.

$0 - 2(0) > 4 \;\Rightarrow\; 0 > 4 \ \text{(false)}$

So shade the half-plane on the opposite side from the origin.

A point must satisfy both $y \ge x$ and $y \le x$ at once, which forces equality.

$y \ge x \ \text{and}\ y \le x \;\Rightarrow\; y = x$

The overlap is exactly the line $y = x$.

Check the point against all three inequalities.

$1 \ge 0, \quad 1 \ge 0, \quad 1 + 1 = 2 \le 3 \ \text{(all true)}$

Every condition holds, so the point is in the feasible region.