Systems of Inequalities

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A system of inequalities is two or more inequalities with the same variables. Its solutions are all the points that make every inequality true. To solve it, graph each inequality on the same grid. The solution is the region where the shaded areas overlap.

Before this lesson: You should know how to do graphing linear inequalities, solve systems of equations by graphing, and use slope-intercept form.

What Is a System of Inequalities?

Say you want a phone plan that costs under $50 and has at least 10 GB of data. Each rule is an inequality. A plan has to pass both rules to work for you.

A system of inequalities works the same way. One inequality, like y \ge x - 2, has a whole half of the grid as solutions. Add a second inequality, and only the points that pass both tests are left.

On a graph, each inequality shades one side of its line. The solution region is where the two shadings overlap. Every point there is a solution. Points outside it fail at least one inequality.

The system y ≥ x − 2 and y < −x + 4 on a grid from −5 to 5. The line y = x − 2 is solid blue. The line y = −x + 4 is dashed orange. The lines cross at (3, 1), shown with an open circle. The region to the left, above the blue line and below the orange line, is shaded green and labeled solutions. The test point (0, 0) is in the shaded region.
The green region is above the solid blue line and below the dashed orange line. Every point in it solves both inequalities.

How to Solve a System of Inequalities

  1. Solve each inequality for y. If you multiply or divide by a negative, flip the sign.
  2. Graph each boundary line. Use a solid line for \le or \ge. Use a dashed line for \lt or \gt.
  3. Shade each inequality. For y \gt or y \ge, shade above the line. For y \lt or y \le, shade below.
  4. Find the overlap. That region is the solution.
  5. Check a point from the overlap in every inequality.

Not sure which side to shade? Test a point not on the line, like (0, 0). If it makes the inequality true, shade its side.

Worked Examples

Example 1: Is the Point a Solution?

Decide if each point solves the system y \gt 2x - 1 and y \le -x + 5: (a) (1, 3), (b) (2, 2), (c) (-2, 7).

(a) 3 \gt 2(1) - 1 = 1 is true. 3 \le -1 + 5 = 4 is true. Both are true, so (1, 3) is a solution.

(b) 2 \gt 2(2) - 1 = 3 is false. One false is enough, so (2, 2) is not a solution.

(c) 7 \gt 2(-2) - 1 = -5 is true. 7 \le 2 + 5 = 7 is true, because \le allows equal. So (-2, 7) is a solution. It sits on the solid line.

Example 2: Graph a System

Graph the system y \ge x - 2 and y \lt -x + 4. This is the system in the picture above.

Step 1: Both are already solved for y.

Step 2: Graph y = x - 2 as a solid line (slope 1, y-intercept −2). Graph y = -x + 4 as a dashed line (slope −1, y-intercept 4).

Step 3: Shade above the solid line for \ge. Shade below the dashed line for \lt.

Step 4: The overlap is the region to the left of where the lines cross. Check (0, 0): 0 \ge -2 is true and 0 \lt 4 is true.

Answer: The solution is the green region in the graph. The lines cross at (3, 1), but that point is on the dashed line, so it is not a solution.

Example 3: Solve for y First

Graph the system 2x + y \le 6 and x - 2y \lt 4.

Step 1: Solve the first for y: subtract 2x to get y \le -2x + 6.

Step 2: Solve the second for y. Subtract x: -2y \lt -x + 4. Divide by −2 and flip the sign.

y \gt \tfrac{1}{2}x - 2

Step 3: Graph y = -2x + 6 solid and shade below. Graph y = \tfrac{1}{2}x - 2 dashed and shade above.

Answer: The solution is the overlap: below the solid line and above the dashed line. The point (0, 0) is in it, since 0 \le 6 and 0 \lt 4 are both true.

Example 4: Three Inequalities

Describe the solution of x \ge 0, y \ge 0 and x + y \le 4. Is (1, 2) a solution? Is (3, 2)?

Step 1: x \ge 0 keeps the right side of the y-axis. y \ge 0 keeps the part above the x-axis.

Step 2: x + y \le 4 becomes y \le -x + 4. Shade below that solid line.

Step 3: All three overlap in a triangle with corners (0, 0), (4, 0) and (0, 4). The edges are included, because all three signs allow equal.

Answer: (1, 2) is a solution, since 1 + 2 = 3 \le 4. (3, 2) is not, since 3 + 2 = 5 is more than 4.

Example 5: Word Problem

Jada tutors for $15 an hour and works as a cashier for $10 an hour. She can work at most 25 hours a week. She wants to earn at least $300. Write a system. Can she tutor 15 hours and work 8 hours as a cashier?

Step 1: Let t = hours tutoring and c = hours as a cashier.

\begin{aligned} t + c &\le 25 \\ 15t + 10c &\ge 300 \\ t \ge 0, \; c &\ge 0 \end{aligned}

Step 2: The last two inequalities say hours can’t be negative.

Graph with tutoring hours t from 0 to 30 across and cashier hours c from 0 to 30 up. The solid blue line t + c = 25 runs from (0, 25) to (25, 0). The solid orange line 15t + 10c = 300 runs from (0, 30) to (20, 0). The lines cross at (10, 15). The green triangle with corners (10, 15), (20, 0) and (25, 0) is the solution region. The point (15, 8) is marked inside it.
Jada’s choices are in the green triangle: at most 25 hours and at least $300. The point (15, 8) is inside.

Step 3: Test (15, 8). Hours: 15 + 8 = 23 \le 25. Pay: 15(15) + 10(8) = 225 + 80 = 305 \ge 300.

Answer: Yes. She works 23 hours and earns $305, so both goals are met. On the graph, the point is inside the green region.

Common Mistakes

MistakeExample of the errorFix
Not flipping the sign when dividing by a negative-2y \lt -x + 4 becomes y \lt \tfrac{1}{2}x - 2Flip it: y \gt \tfrac{1}{2}x - 2
Shading everything that is shaded at allUsing the area shaded by either inequalityKeep only the overlap, where both are shaded
Mixing up solid and dashed linesA solid line for y \lt -x + 4Dashed for \lt and \gt; solid for \le and \ge
Counting a corner on a dashed lineCalling (3, 1) a solution in Example 2A point on a dashed line is not a solution
Testing a point in only one inequality(2, 2) works in y \le -x + 5, so it is a solutionA solution must make every inequality true

Special Cases

Sometimes the shaded regions never overlap. Then the system has no solution. This can happen when the boundary lines are parallel and each one is shaded away from the other.

SystemWhat the graph showsSolutions
y \gt 2x + 3 and y \lt 2x - 1Parallel lines, shaded away from each otherNone
y \ge 2x - 1 and y \le 2x + 3Parallel lines, shaded toward each otherA strip between the lines
y \ge x - 2 and y \lt -x + 4Lines that crossA wedge-shaped region

Systems of inequalities are used to plan with limits on time, money or space, as in Example 5. A system of equations has a point as its answer. A system of inequalities has a whole region.

Practice Problems

  1. Is (2, 1) a solution of y \lt x + 1 and y \ge -2x + 3?
  2. Is (0, 4) a solution of y \le 3x + 4 and y \gt x + 4?
  3. Describe the graph of the system y \gt 1 and x \le 2.
  4. Which of these points solve y \ge x and y \lt 4: (1, 3), (5, 6), (-2, -3)?
  5. Solve each inequality for y: 3x + y \le 2 and x - y \gt 3. Is (0, -5) a solution?
  6. Where do the boundary lines of y \ge 2x - 3 and y \le -x + 6 cross? Is that point a solution?
  7. Does the system y \gt -x + 2 and y \lt -x - 1 have any solutions?
  8. Write a system for the region above the dashed line y = x and on or below the solid line y = 5.
  9. Adult tickets cost $10 and child tickets cost $5. A group can spend at most $60 and needs at least 8 tickets. Write a system. Do 3 adults and 6 children work? Do 4 adults and 5 children?
  10. Use Jada’s system from Example 5. Can she tutor 18 hours and work 4 hours as a cashier?
Show answers
  1. Yes. 1 \lt 3 is true and 1 \ge -1 is true.
  2. No. 4 \le 4 is true, but 4 \gt 4 is false.
  3. A dashed horizontal line at y = 1 shaded above, and a solid vertical line at x = 2 shaded left. The solution is the upper-left region where they overlap.
  4. Only (1, 3). For (5, 6), 6 \lt 4 is false. For (-2, -3), -3 \ge -2 is false.
  5. y \le -3x + 2 and y \lt x - 3 (dividing by −1 flips the sign). Yes, (0, -5) works: -5 \le 2 and -5 \lt -3.
  6. 2x - 3 = -x + 6 gives x = 3, y = 3. Both lines are solid, so (3, 3) is a solution.
  7. No. The lines are parallel. One region is above the higher line and the other is below the lower line, so they never overlap.
  8. y \gt x and y \le 5
  9. a + c \ge 8 and 10a + 5c \le 60. 3 and 6: 9 \ge 8 and 60 \le 60, so yes. 4 and 5: 40 + 25 = 65, which is more than 60, so no.
  10. Yes. 18 + 4 = 22 \le 25 and 15(18) + 10(4) = 310 \ge 300.

FAQ

How do you solve a system of inequalities?

Graph each inequality on the same grid and shade its side. The solution is the region where all the shaded areas overlap.

How do you know which side to shade?

After solving for y, shade above the line for \gt or \ge and below it for \lt or \le. You can also test a point like (0, 0).

Is the point where the lines cross a solution?

Only if both lines are solid. If either line is dashed, the crossing point is not included.

Can a system of inequalities have no solution?

Yes. If the shaded regions never overlap, there is no solution. This happens with parallel lines shaded away from each other.

Summary

  • A system of inequalities has two or more inequalities. A solution makes all of them true.
  • Graph each inequality: solid line for \le or \ge, dashed for \lt or \gt.
  • Shade each one. The solution region is the overlap.
  • Points on a dashed boundary are not solutions.
  • If the regions never overlap, the system has no solution.

Previous lesson: Graphing Linear Inequalities

Next lesson: Exponent Rules

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