Graphing Inequalities: Practice

The boundary passes through $(1, 0)$ and $(0, -2)$, so it is $y = 2x - 2$. It is solid (inclusive), and the shaded region is below and right of the line.

$\text{test } (0, 0):\ 0 \le 2(0) - 2 \;\Rightarrow\; 0 \le -2 \ \text{(false)}$

The origin is unshaded, so the solid line with the far side shaded is $y \le 2x - 2$.

The region is $y \ge 1$, $x \ge -1$, and $x + y \le 3$. Test the candidates.

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

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

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

$5 - 2(1) = 3, \quad \text{so the statement is}\ 3 \ge 3$

That is true. The point is on the boundary $x - 2y = 3$, and $\ge$ is inclusive, so the boundary is solid and the point is a solution.

Divide both sides by $-3$; the divisor is negative, so reverse the symbol.

$\frac{-3y}{-3} > \frac{6x - 9}{-3} \;\Rightarrow\; y > -2x + 3$

With $y$ greater than the line, shade above it.

Substitute $x = 0$ into the second boundary equation.

$2(0) + y = 6 \;\Rightarrow\; y = 6$

So the two boundaries meet at $(0, 6)$.

The first quadrant needs $x \ge 0$ and $y \ge 0$, and staying on the origin side of $x + y = 6$ needs $x + y \le 6$.

$\text{test } (1, 1):\ 1 \ge 0,\ 1 \ge 0,\ 2 \le 6 \ \text{(inside)}$

The combination $x \ge 0,\ y \ge 0,\ x + y \le 6$ gives exactly that triangle.

A solution would need $y \ge x + 2$ and $y \le x - 2$ at once, forcing $x + 2 \le y \le x - 2$.

$x + 2 \le x - 2 \;\Rightarrow\; 2 \le -2 \ \text{(false)}$

The boundaries are parallel and the shaded sides point apart, so there is no overlap.

The slope is $\frac{1 - (-2)}{1 - 0} = 3$ and the $y$-intercept is $-2$, so the line is $y = 3x - 2$. It is dashed (strict), and below the line means smaller $y$.

$\text{dashed } y = 3x - 2,\ \text{shaded below} \;\Rightarrow\; y < 3x - 2$

The strict symbol and the lower side give $y < 3x - 2$.

Read each line style from its symbol: strict gives dashed, inclusive gives solid.

$y > 2x - 1 \Rightarrow \text{dashed}, \qquad y \le -x + 4 \Rightarrow \text{solid}$

So the first boundary is dashed and the second is solid.

A false test result means the origin is not a solution.

$2(0) + 3(0) \le -6 \;\Rightarrow\; 0 \le -6 \ \text{(false)}$

So shade the side away from the origin; the shaded region does not contain it.

The boundary $x = 4$ is vertical, $\le$ is inclusive so the line is included, and smaller $x$ is to the left.

$x \le 4 \;\Rightarrow\; \text{on or left of } x = 4$

The region is the solid vertical line and everything to its left.

Set $y = 0$ in the second boundary and solve for $x$.

$2x + 0 = 8 \;\Rightarrow\; x = 4$

The two boundaries meet at $(4, 0)$.