Regions in Two Variables

Draw the line $y = x$ roughly. Now pick the point $(3,1)$ and the point $(1,3)$, one on each side. For each, decide whether $y > x$ is true. What does that tell you about which side of the line the region $y > x$ occupies?

Draw the boundary line, decide whether it is solid or dashed, then test one point not on the line and shade the side that works.

Graphing $x + y < 3$ takes three decisions, and no more.

1. Where is the boundary? Replace the inequality sign with $=$ and draw that line. Here, $x + y = 3$, which passes through $(3,0)$ and $(0,3)$.

2. Is the boundary included? The sign is strict, so points on the line do not satisfy the inequality. Draw the line dashed. A sign of $\leq$ or $\geq$ would include them, and the line would be solid.

3. Which side? Pick any point not on the line and substitute it. Taking $(0,0)$: $0 + 0 = 0$, and $0 < 3$ is true, so the origin's side is the one to shade.

That is the whole method, and it works for every linear inequality in two variables.

Any point off the boundary works, so choose the one that makes the arithmetic easiest.

The origin, whenever possible. Substituting $(0,0)$ usually reduces the inequality to a comparison of two numbers, which needs no working at all. For $2x - 5y \geq 8$: $0 - 0 = 0$, and $0 \geq 8$ is false, so shade the side away from the origin.

Something else, when the line passes through the origin. For $y > 2x$ the boundary $y = 2x$ contains $(0,0)$, so the origin gives $0 > 0$, which is false but tells you nothing about which side to shade, since the origin is on the line rather than in either half. Use a point clearly off the line instead, such as $(0,1)$: $1 > 0$ is true, so shade the side containing $(0,1)$, which is above the line.

Testing a point on the line is the one real trap here. It always produces equality, which is consistent with neither strict half, so it decides nothing. Look at your chosen point and confirm it is genuinely off the boundary before using it.

Inequalities are often given in the form $ax + by + c < 0$ rather than solved for $y$. Rearranging is allowed, with the same rule as Lesson 8: dividing by a negative reverses the sign.

Take $2x - y > 4$. Solving for $y$:

$$-y > 4 - 2x \quad \Longrightarrow \quad y < 2x - 4$$

Multiplying by $-1$ reversed the direction, and the region is therefore below the line $y = 2x - 4$, not above it.

Once an inequality is in the form $y > \ldots$ or $y < \ldots$, there is a shortcut worth knowing:

$$y > \text{(line)} \ \Rightarrow \ \text{shade above} \qquad y < \text{(line)} \ \Rightarrow \ \text{shade below}$$

because a larger $y$ at the same $x$ means a point higher up. The shortcut is only safe after rearranging correctly, and the rearrangement is where the sign can turn. Testing a point costs a few seconds and needs no rearranging at all, which is why it remains the recommended method.

Two special cases: $x > 3$ is a vertical boundary with the region to the right, and $y \leq -2$ is a horizontal boundary with the region below, including the line.

Several inequalities may be imposed at once. The region satisfying all of them is their overlap, and is called the feasible region.

To find it, graph each inequality in turn and keep only what survives every one. Consider:

$$x \geq 0, \qquad y \geq 0, \qquad x + y \leq 5$$

The first two restrict to the quarter of the plane where both coordinates are non-negative. The third cuts that quarter with the line $x + y = 5$, keeping the side containing the origin, since $0 \leq 5$ is true.

The result is a triangle with corners at $(0,0)$, $(5,0)$ and $(0,5)$, and all three boundaries are solid because every sign is inclusive.

Two practical points. Shade lightly, or shade only the final region, since three overlapping shadings quickly become unreadable; many people mark each unwanted half-plane instead and keep what is left unmarked. And the corners are worth finding, by solving the two boundary equations that meet there as simultaneous equations, because in applications the best solution is usually at a corner.

A system can also have an empty feasible region, if the conditions conflict: $y \geq 4$ together with $y \leq 1$ leaves nothing at all.

Five items on two-variable inequalities. Work each one, then reveal the answer.

1. 1 Is the boundary of $2x + y \leq 6$ solid or dashed?

   Show answer
   The sign includes equality.Solid

2. 2 Does $(0,0)$ satisfy $x - y > 1$?

   Show answer
   Substitute: $0 - 0 = 0$, and $0 > 1$.No

3. 3 Which side of $y = 5$ is the region $y < 5$?

   Show answer
   Smaller $y$ means lower on the page.Below

4. 4 Rearrange $-2y > 6 - 4x$ into the form $y < \ldots$ or $y > \ldots$.

   Show answer
   Divide by $-2$ and reverse the sign.$y < 2x - 3$

5. 5 Why can $(2,2)$ not be a test point for $y = x$?

   Show answer
   Check whether it satisfies the boundary equation.It lies on the line