Graphing Inequalities: Additional Practice (Set 2)

The region is below the dashed line $y = x$ and above the dashed line $y = -2$, so it needs $y < x$ and $y > -2$.

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

The others fail: $(0, 1)$ has $1 < 0$ false, $(-1, -3)$ has $-3 > -2$ false, and $(0, -2)$ lies on a dashed edge.

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

$\text{test } (0, 0):\ 0 \ge -0 - 1 \;\Rightarrow\; 0 \ge -1 \ \text{(true)}$

The origin is shaded, so the solid line with the upper side gives $y \ge -x - 1$.

Test each candidate in $3x - 5y < 15$.

$(0, 5):\ 3(0) - 5(5) = -25 < 15 \ \text{(true)}$

The point $(5, 0)$ gives $15 < 15$ (false, on the boundary), $(6, 0)$ gives $18 < 15$ (false), and $(5, -5)$ gives $40 < 15$ (false). Only $(0, 5)$ is a solution.

The region is $x \ge 0$, $y \ge 0$, and $2x + y \le 4$. Test the candidates.

$(1, 1):\ 1 \ge 0,\ 1 \ge 0,\ 2(1) + 1 = 3 \le 4 \ \text{(all true)}$

The point $(2, 2)$ gives $2(2) + 2 = 6 \le 4$ (false), $(-1, 1)$ fails $x \ge 0$, and $(1, 3)$ gives $2 + 3 = 5 \le 4$ (false). Only $(1, 1)$ works.

The three boundaries meet in pairs at the corners. Lines $x = 1$ and $y = 1$ meet at $(1, 1)$; $y = 1$ and $x + y = 5$ meet at $(4, 1)$; $x = 1$ and $x + y = 5$ meet at $(1, 4)$.

$\text{corners: } (1, 1),\ (4, 1),\ (1, 4)$

The bounded overlap is the triangle with those corners.

Evaluate the right side at $x = -1$.

$-3(-1) = 3, \quad \text{so the statement is}\ 4 > 3$

That is true, so $(-1, 4)$ is a solution.

Evaluate the right side at $x = 2$.

$\tfrac{3}{2}(2) = 3, \quad \text{so the statement is}\ 3 > 3$

That is false, and the point lies on the line $y = \tfrac{3}{2}x$, which the strict $>$ excludes.

Substitute into the left side $3x - y$.

$3(2) - 2 = 4, \quad \text{so the statement is}\ 4 \ge 4$

That is true. The point is on the boundary, and $\ge$ is inclusive, so it is a solution.

Test each point against all three inequalities.

$(2, 1):\ 1 \le 2(2) = 4,\ 1 \ge -1,\ 2 \le 3 \ \text{(all true)}$

The point $(0, 2)$ fails $y \le 2x$, $(4, 0)$ fails $x \le 3$, and $(1, -3)$ fails $y \ge -1$. Only $(2, 1)$ works.

Divide every term by $-2$; the divisor is negative, so reverse the symbol.

$\frac{-2y}{-2} \le \frac{6x - 4}{-2} \;\Rightarrow\; y \le -3x + 2$

The flip turns $\ge$ into $\le$.

The inclusive $\ge$ makes the boundary solid, and the failed origin test sends the shading to the far side.

$-0 + 4(0) \ge 8 \;\Rightarrow\; 0 \ge 8 \ \text{(false)}$

Shade the half-plane on the opposite side from the origin.

A solid line marks an inclusive inequality. Among the choices, only $\ge$ is inclusive.

$3x - y \ge 7:\ \ge\ \text{is inclusive} \;\Rightarrow\; \text{solid}$

The others use strict symbols and are drawn dashed.