Graphing Inequalities: Additional Practice (Set 1)

The boundary passes through $(0, 1)$ and $(2, 2)$, a slope of $\tfrac{1}{2}$, so it is $y = \tfrac{1}{2}x + 1$. It is dashed (strict) and the region below it, containing the origin, is shaded.

$\text{test } (0, 0):\ 0 < \tfrac{1}{2}(0) + 1 \;\Rightarrow\; 0 < 1 \ \text{(true)}$

The dashed line with the lower side gives $y < \tfrac{1}{2}x + 1$.

Test each point against both inequalities.

$(3, 1):\ 1 > -2 \ \text{and}\ 1 < 2(3) = 6 \ \text{(both true)}$

The point $(0, 0)$ fails $y < 2x$ since $0 < 0$ is false, $(1, 3)$ fails $y < 2x$ since $3 < 2$ is false, and $(-2, -3)$ fails $y > -2$. Only $(3, 1)$ works.

Substitute $x = 2$ into $y = x + 1$.

$y = 2 + 1 = 3$

So the boundaries intersect at $(2, 3)$.

The line through $(0, 3)$ and $(3, 0)$ is $x + y = 3$. It is solid (inclusive), and the origin is shaded.

$\text{test } (0, 0):\ 0 + 0 = 0 \le 3 \ \text{(true)}$

Solid and the origin side give $x + y \le 3$.

Substitute into the left side $2x + 5y$.

$2(4) + 5(-1) = 8 - 5 = 3, \quad \text{so the statement is}\ 3 < 3$

That is false. The point is on the boundary, and the strict $<$ excludes it.

Both conditions constrain $x$ only, trapping it between two vertical lines.

$-2 < x < 3$

The overlap is the vertical strip between $x = -2$ and $x = 3$.

Test each point against the three inequalities.

$(2, 1):\ 1 \ge 0,\ 2 \ge 0,\ 2 + 2(1) = 4 \le 4 \ \text{(all true)}$

The point $(0, 3)$ gives $0 + 6 = 6 \le 4$ (false), $(4, 1)$ gives $4 + 2 = 6 \le 4$ (false), and $(-1, 1)$ fails $x \ge 0$. Only $(2, 1)$ works.

On or above $y = 1$ is $y \ge 1$, and on or below $y = 4$ is $y \le 4$; both are inclusive.

$1 \le y \le 4$

The system $y \ge 1$ and $y \le 4$ gives that horizontal band.

The test result decides membership.

$0 \le 3(0) - 2 \;\Rightarrow\; 0 \le -2 \ \text{(false)}$

Since the origin is not a solution, shade the side that does not contain it.

The feasible region is the overlap of all three half-planes, so membership requires every inequality to hold.

$\text{feasible} \;\Leftrightarrow\; \text{all three inequalities true}$

Missing even one condition puts the point outside the region.

The boundary comes from the matching equation, and $\le$ is inclusive, so the line is solid.

$6x + y \le 6 \;\Rightarrow\; \text{solid line } 6x + y = 6$

Replace the symbol with $=$ and draw it solid.

A point must satisfy both, so it must clear the stricter lower bound.

$y \ge 2 \ \text{and}\ y \ge 5 \;\Rightarrow\; y \ge 5$

Every $y \ge 5$ automatically satisfies $y \ge 2$, so $y \ge 5$ is the overlap.