An inequality describes a range of possible values rather than one exact answer. On a number line this is a set of points; on a graph with two variables, it becomes a whole shaded region of the plane.

Treat it exactly like an equation — with one crucial exception.

Apply the same operation to all three parts at once, keeping the unknown isolated in the middle.

Add $2$ to every part: $-1 \le 3x < 9$. Then divide every part by $3$: $-\dfrac{1}{3} \le x < 3$.

Use an open circle for a strict inequality ($<,>$) to show the boundary value is excluded, and a closed (filled) circle for an inclusive inequality ($\le,\ge$) to show it's included. For $x\le3$, draw a filled circle at $3$ with shading extending left. For a double inequality like $-\dfrac{1}{3}\le x<3$, draw a filled circle at $-\dfrac{1}{3}$, an open circle at $3$, and shade the segment joining them.

The same open/closed logic applies to lines instead of points:

• Broken (dashed) line — the boundary is not included (strict inequality, $<$ or $>$).
• Solid line — the boundary is included ($\le$ or $\ge$).
• Shade the unwanted side of the line, unless the question tells you to shade the required region instead.

Pick any point that is clearly not on the line (the origin $(0,0)$ is usually easiest, provided the line doesn't pass through it), substitute its coordinates into the inequality, and check whether the inequality is true or false. If true, that point's side is the required side; if false, shade the other side as the unwanted region.

Given a shaded (or unshaded) region bounded by straight lines, identify each boundary line's equation and decide which side is required by testing a point that is clearly inside the region.

A region $R$ is bounded by three solid lines: $x\ge1$, $y\ge2$, and $x+y\le8$. Testing the point $(5,2)$:

• $x\ge1$: $5\ge1$ ✓
• $y\ge2$: $2\ge2$ ✓ (on the boundary — included, since the line is solid)
• $x+y\le8$: $5+2=7\le8$ ✓

All three hold, so $(5,2)$ lies inside $R$.

Worked Example

A region $S$ satisfies $y < 2x+1$, $x+y\ge4$, and $x<5$.

1. Draw the boundary lines: $y=2x+1$ (dashed, since $<$), $x+y=4$ (solid, since $\ge$), and $x=5$ (dashed, since $<$).
2. For $y<2x+1$, points below the line $y=2x+1$ satisfy the inequality (test $(0,0)$: $0<1$ ✓), so shade above the line as the unwanted region.
3. Test whether the point $(4,1)$ lies in $S$:

• $y<2x+1$: $1<9$ ✓
• $x+y\ge4$: $4+1=5\ge4$ ✓
• $x<5$: $4<5$ ✓

1. All three conditions hold, so $(4,1)$ lies in region $S$.