Unit 3 · Lesson 5 of 5
Graphing linear inequalities
You'll be able to: graph an inequality like \(y \lt 2x + 1\) with the right kind of boundary line and shading, and tell whether a point is a solution.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
Write the equation of the line with slope \(2\) through \((3, 1)\), in \(y = mx + b\) form.
Point-slope: \(y - 1 = 2(x - 3)\), so \(y - 1 = 2x - 6\).
Add 1: \(y = 2x - 5\). Check: \(2(3) - 5 = 1\).
Solve \(-2x + 3 \gt 9\). Give the answer in interval notation.
Subtract 3: \(-2x \gt 6\).
Divide by \(-2\) and flip the sign: \(x \lt -3\).
Answer: \((-\infty, -3)\).
Step 2 of 5The idea
An inequality with two letters, like \(y \gt 2x - 1\), has a whole region of solutions: every point on one side of a line. That line is the boundary line. You get it by pretending the sign is \(=\).
To graph the inequality, you draw the boundary line and shade the side where the inequality is true.
1. Boundary line. Graph the line with \(=\) in place of the sign. Use a solid line for \(\le\) or \(\ge\) (points on the line count). Use a dashed line for \(\lt\) or \(\gt\) (they don't).
2. Test point. Pick a point not on the line. \((0, 0)\) is easiest. Put it into the inequality. True: shade its side. False: shade the other side.
Shortcut when \(y\) is alone: \(y \gt\) or \(y \ge\) means shade above the line. \(y \lt\) or \(y \le\) means shade below.
Is a point a solution? Put it in. A true statement means yes. A point on a dashed line is never a solution.
A vertical boundary works the same way: \(x \gt 3\) is the region to the right of the dashed line \(x = 3\).
Why this shows up on the placement test: you may be asked which graph matches an inequality, or whether a point is a solution, and both need the solid/dashed and shading rules.
Step 3 of 5Worked examples
Example 1: graph \(y \le \tfrac{1}{2}x + 1\)
- Boundary: \(y = \tfrac{1}{2}x + 1\). Start at \((0, 1)\), then up 1 and right 2.
- The sign is \(\le\), so the line is solid. Points on the line are solutions too.
- Test \((0, 0)\): \(0 \le \tfrac{1}{2}(0) + 1\) means \(0 \le 1\). True, so shade the side with the origin.
- The origin is below the line, so the shading goes below. That matches the shortcut: \(y \le\) means below.
Example 2: graph \(3x - 2y \gt 6\), then check \((4, 1)\)
- Boundary \(3x - 2y = 6\). Intercepts: \(y = 0\) gives \(x = 2\), so \((2, 0)\). \(x = 0\) gives \(-2y = 6\), so \((0, -3)\).
- The sign is \(\gt\), so draw a dashed line through \((2, 0)\) and \((0, -3)\).
- Test \((0, 0)\): \(3(0) - 2(0) \gt 6\) means \(0 \gt 6\). False, so shade the side without the origin. Here that is below the line.
- Check \((4, 1)\): \(3(4) - 2(1) = 12 - 2 = 10\), and \(10 \gt 6\) is true. So \((4, 1)\) is a solution, and it sits in the shaded part. The graph and the arithmetic agree.
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went.
Is \((2, 3)\) a solution of \(y \gt 2x - 4\)?
Put in \(x = 2\), \(y = 3\): \(3 \gt 2(2) - 4\) means \(3 \gt 0\).
That is true, so the answer is yes.
For \(6x - 2y \gt 4\), do you shade above or below the boundary line?
Solve for \(y\): \(-2y \gt -6x + 4\). Divide by \(-2\) and flip the sign: \(y \lt 3x - 2\). So shade below.
Test-point check: \(6(0) - 2(0) \gt 4\) means \(0 \gt 4\), false. The origin is above the line, so shade the other side: below.
How do you graph \(2x + y \lt 4\)?
- solid line, shade below it
- dashed line, shade above it
- dashed line, shade below it
- solid line, shade above it
The sign is \(\lt\), so the line is dashed. That rules out (a) and (d).
Solve for \(y\): \(y \lt -2x + 4\), so shade below. Test point: \(0 \lt 4\) is true, and the origin is below the line. Both agree.
Answer: (c).
Which inequality does this graph show?
- \(y \gt 2x + 1\)
- \(y \ge 2x + 1\)
- \(y \lt 2x + 1\)
- \(y \le 2x + 1\)
The line goes through \((0, 1)\) with slope \(\dfrac{3 - 1}{1 - 0} = 2\), so the boundary is \(y = 2x + 1\).
Dashed means \(\lt\) or \(\gt\). Shaded above means \(y \gt\).
Answer: (a) \(y \gt 2x + 1\).
On paper, graph \(y \ge -2x + 3\). Then open the answer and compare.
Boundary \(y = -2x + 3\): start at \((0, 3)\), down 2 and right 1 to \((1, 1)\). The sign is \(\ge\), so the line is solid.
Test \((0, 0)\): \(0 \ge 3\) is false, so shade the side away from the origin. That is above the line.
On paper, graph \(x - 3y \lt 6\). Then open the answer and compare.
Boundary \(x - 3y = 6\). Intercepts: \(y = 0\) gives \((6, 0)\); \(x = 0\) gives \(-3y = 6\), so \((0, -2)\).
The sign is \(\lt\), so the line is dashed.
Test \((0, 0)\): \(0 - 0 \lt 6\) is true, so shade the side with the origin, which is above the line.
A club sells adult tickets for $5 and child tickets for $2. To meet its goal, \(5a + 2c \ge 100\), where \(a\) is adult tickets and \(c\) is child tickets. Does selling 12 adult and 22 child tickets meet the goal?
Put in \(a = 12\), \(c = 22\): \(5(12) + 2(22) = 60 + 44 = 104\).
\(104 \ge 100\) is true, so yes.
Is \((3, 0)\) a solution of \(2x - 3y \lt 6\)?
Put it in: \(2(3) - 3(0) = 6\), and \(6 \lt 6\) is false.
So no. The point lies exactly on the dashed boundary line, and points on a dashed line are not solutions.
Step 5 of 5Wrap-up
You can now graph a linear inequality: draw the boundary (solid for \(\le, \ge\); dashed for \(\lt, \gt\)), use a test point to pick the side, and check whether any point is a solution.
Read or watch next: OpenStax Intermediate Algebra 2e, 3.4 Graph Linear Inequalities in Two Variables (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Graphing two-variable inequalities (short videos). Quick rules: Lines and systems cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why can't I use (0, 0) as a test point when the line goes through the origin?" It knows this course and can explain things another way.