Graphical Inequalities and Shaded Regions

From an inequality to a region

An inequality in two variables usually has many solutions. For example, $y>2x-1$ is satisfied by $(0,0)$ because

$0>2(0)-1.$

It is not satisfied by $(0,-2)$ because $-2>-1$ is false. Instead of listing infinitely many coordinate pairs, we represent all the solutions as a region on coordinate axes.

The edge of that region is its boundary line. Obtain it by replacing the inequality sign with an equals sign. Thus, the boundary of $y>2x-1$ is

$y=2x-1.$

Plot this line using the usual methods for straight-line graphs: here the gradient is $2$ and the $y$-intercept is $-1$.

Dashed and solid boundaries

The type of inequality determines how the boundary is drawn:

• $<$ or $>$ is strict, so points on the boundary do not satisfy the inequality. Draw a dashed line.
• $\leq$ or $\geq$ includes equality, so points on the boundary do satisfy the inequality. Draw a solid line.

For example, every point on $y=2x-1$ makes the two sides equal. It therefore belongs to $y\geq2x-1$, but not to $y>2x-1$.

This is the two-dimensional equivalent of using an open circle for a strict inequality and a closed circle for an included endpoint on a number line.

Deciding which side to shade

A boundary line divides the plane into two regions. Choose a point away from the line and substitute its coordinates into the original inequality. The origin, $(0,0)$, is often convenient.

For $y\geq2x-1$, testing $(0,0)$ gives

$0\geq2(0)-1,$

which is true. Shade the side of the solid line containing $(0,0)$. If the test produces a false statement, shade the opposite side.

Do not test a point on the boundary: it cannot tell you which side is required. If the line passes through $(0,0)$, choose another simple point such as $(1,0)$ or $(0,1)$.

If the inequality is rearranged into $y>\text{expression}$, its solutions lie above the boundary; if it is $y<\text{expression}$, they lie below. Testing a point remains the safer general method, especially when the inequality has not been rearranged.

Vertical and horizontal boundaries

An inequality involving only $x$ has a vertical boundary. For example, $x<3$ has the dashed boundary $x=3$, and its solution lies to the left because those points have $x$-coordinates below $3$.

An inequality involving only $y$ has a horizontal boundary. For example, $y\leq4$ has the solid boundary $y=4$, with the solution on or below it.

Satisfying several inequalities simultaneously

When several inequalities are drawn on the same axes, a solution must satisfy all of them. Draw each boundary, identify the correct side of each one, and keep only the common overlap.

Consider

$y\geq2x-1 \qquad\text{and}\qquad x<3.$

The first boundary is solid and its region is the side containing $(0,0)$. The second boundary is the dashed vertical line $x=3$, with its region to the left. The solution set is where these two regions overlap. Points on $y=2x-1$ may be included when they also have $x<3$, but no point on $x=3$ is included.

A quick check confirms the interpretation. The point $(0,0)$ lies in the overlap and satisfies both inequalities. The point $(4,8)$ satisfies $y\geq2x-1$ because $8\geq7$, but it fails $x<3$, so it is not in the combined solution region.

Reading a shaded graph backwards

To write an inequality from a graph, first find the equation of the boundary line. Then inspect its style: dashed means a strict sign, while solid means equality is included. Finally, test a point in the shaded region to choose between the two possible directions. For a boundary $y=2x-1$, for instance, the shading must represent either values above the line or values below it; substitution identifies which one.

Boundary convention

• Replace the inequality sign with $=$ to plot the boundary.
• $<$ or $>$: dashed line; boundary excluded.
• $\leq$ or $\geq$: solid line; boundary included.

Find the shaded side

1. Choose a point not on the boundary, often $(0,0)$.
2. Substitute it into the original inequality.
3. True: shade the side containing the point. False: shade the other side.

For $x<a$, shade left of $x=a$; for $x>a$, shade right. For $y<a$, shade below $y=a$; for $y>a$, shade above.

Several inequalities

Draw every boundary and take the common overlap. A point is in the final solution region only if it satisfies every inequality. Retain solid boundary sections where all the other conditions also hold; exclude every dashed boundary.