Graphical Linear Inequalities

From an equation to a region

A linear equation in two variables, such as $y=2x-1$, is represented by a straight line. A linear inequality such as

$y>2x-1$

has many more solutions. Every coordinate pair whose $y$-value is greater than $2x-1$ satisfies it, so its graph is a whole region on one side of the line $y=2x-1$.

The line separates points that satisfy the inequality from points that do not. It is called the boundary line and is found by replacing the inequality sign with an equals sign. OCR requires you to identify these graphical solution sets using the conventions for dashed and solid boundaries.

Decide whether the boundary is included

The symbol tells you how to draw the boundary:

• $<$ or $>$ is a strict inequality. Points on the boundary give equality, so they are not solutions. Draw a .

For example, the boundary of $y<2x+3$ is $y=2x+3$ and must be dashed. The boundary of $y\leq2x+3$ is the same straight line, but it must be solid. The line style is therefore part of the mathematical answer, not decoration.

Find the correct side of the line

When an inequality is written as $y>mx+c$, its solutions lie above the boundary; for , they lie below it. The same idea applies to and , with the boundary included.

A reliable method for any orientation is to use a test point:

1. Draw the boundary line using the corresponding equation.
2. Choose a point not on that line. $(0,0)$ is often convenient.
3. Substitute its coordinates into the original inequality.
4. If the statement is true, the side containing the test point is the solution region. If it is false, choose the opposite side.

For $y>2x-1$, testing $(0,0)$ gives $0>2(0)-1$, or , which is true. Therefore, the side containing the origin is the solution region. The boundary is dashed because the inequality is strict.

Vertical and horizontal boundaries

Not every inequality needs to be rearranged into the form $y>mx+c$.

For $x<3$, the boundary is the vertical line $x=3$. Points to its left have $x$-coordinates less than $3$, so the solution is the region to the left. The line is dashed. For , use the same vertical boundary, shade to the right and make the line solid.

Likewise, $y\geq4$ has the horizontal boundary $y=4$. Its solution lies above the line, including the line itself.

Satisfying several inequalities simultaneously

Several inequalities can be drawn on the same axes. Each one selects a half-plane, and the solution set is their overlap: the coordinates satisfying every inequality at the same time.

For instance, consider

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

The first boundary, $y=2x-1$, is solid and its solution lies above it. The second boundary, $x=3$, is dashed and its solution lies to the left. Only the region that is both above the first line and left of the second line belongs to the combined solution set.

A point can be checked algebraically. The point $(1,2)$ satisfies both inequalities because $2\geq2(1)-1$ and $1<3$. The point $$ satisfies the first inequality but not the second, so it is not in the combined solution set. Satisfying only one condition is not enough.

Occasionally, the required regions do not overlap at all. In that case, the system has no graphical solution. A point where solid boundaries meet may be included if it satisfies every inequality; a point on any dashed boundary is excluded.

Boundary first

Replace the inequality sign with $$ and draw the resulting straight line.