Linear Inequalities

An inequality says one side is bigger or smaller than the other, instead of equal. Real limits are usually inequalities: you can spend at most $50, you need at least 80% to pass, an elevator holds no more than 1000 kg. Solving inequalities is almost the same as solving linear equations, with one important twist.

Key ideas

The symbols

Symbol	Means	Words that signal it
$x \lt 5$	less than	fewer than, below
$x \le 5$	less than or equal to	at most, no more than, maximum
$x \gt 5$	greater than	more than, above, exceeds
$x \ge 5$	greater than or equal to	at least, no less than, minimum

The solution of an inequality is usually a whole range of numbers, not just one. For example, $x \gt 5$ includes $5.1$, $6$, $100$, and every other number bigger than $5$.

Solving: like an equation, with one twist

You can add or subtract the same number on both sides, and multiply or divide both sides by the same positive number, without changing the inequality.

But if you multiply or divide by a negative number, flip the inequality sign. Here’s why. Start with a true statement and multiply both sides by $-1$:

$$2 \lt 5 \quad\longrightarrow\quad -2 \;?\; -5$$

On a number line, $-2$ is to the right of $-5$, so $-2 \gt -5$. Multiplying by a negative reverses the order of numbers, so the sign has to flip to stay true.

Graphing on a number line

• An open dot means the endpoint is not included ($\lt$ or $\gt$).
• A closed dot means the endpoint is included ($\le$ or $\ge$).
• Shade the side that contains the solutions, with an arrow if it goes on forever.

Compound inequalities

An “and” inequality needs both conditions to be true at once. It’s often written as one chain, like $-4 \lt x \le 3$, which means $x \gt -4$ and $x \le 3$: everything between $-4$ and $3$. To solve a chain like $-5 \lt 2x + 3 \le 9$, do the same step to all three parts.

An “or” inequality needs at least one condition to be true, like $x \lt -2$ or $x \ge 3$. Its graph is usually two pieces pointing away from each other. Solve each part separately.

Writing inequalities from word problems

1. Name the variable (and its units).
2. Write an expression for the quantity that has a limit.
3. Choose the symbol from the key words (“at most” is $\le$, “at least” is $\ge$).
4. Solve, then answer the question in context. If the variable counts things (tickets, boxes, classes), the answer must be a whole number, so round in the direction that keeps the inequality true.

Two-variable inequalities (a bridge)

An inequality like $3x + 2y \ge 120$ has two variables. Its solutions are ordered pairs $(x, y)$, and its graph is a half-plane: one side of the boundary line $3x + 2y = 120$, shaded. To test a point, substitute it: if the inequality is true, the point is a solution. The SAT often asks “which ordered pair satisfies this inequality (or system of inequalities)?”, and the fastest method is to test each choice. For graphing these regions, see graphing lines and regions.

Desmos tips

• Type a one-variable inequality like 4 - 3x >= 19. Desmos shades everything to the left of the vertical line $x = -5$ (the region $x \le -5$), so you can read the boundary from the graph. (Typing >= gives $\ge$.)
• Type a two-variable inequality like y > 2x - 3. Desmos shades the half-plane, with a dashed boundary for $\gt$ or $\lt$ and a solid one for $\ge$ or $\le$. Type two inequalities and the solution of the system is where the shadings overlap.
• To test a point, type it, like (1, 1), and see whether it’s in the shaded region. (A point exactly on the boundary is easier to check by substituting.)

Worked examples

Example 1: Dividing by a negative

Solve $4 - 3x \ge 19$ and graph the solution.

Solution.

$$\begin{aligned} 4 - 3x &\ge 19 \\ -3x &\ge 15 && \text{subtract } 4 \\ x &\le -5 && \text{divide by } -3 \text{ and flip the sign} \end{aligned}$$

The solution is $x \le -5$. On a number line, put a closed dot at $-5$ (because $-5$ is included) and shade to the left. See the first number line in the figure below.

Check with a number on each side. Try $x = -6$ (a solution): $4 - 3(-6) = 22$, and $22 \ge 19$ ✓. Try $x = 0$ (not a solution): $4 - 0 = 4$, and $4 \ge 19$ is false ✓.

Example 2: Compound inequalities

Solve each inequality and graph the solution.

• (a) $-5 \lt 2x + 3 \le 9$
• (b) $3x - 1 \lt -7$ or $2x + 5 \ge 11$

Solution.

(a) This is an “and” chain, so do each step to all three parts:

$$\begin{aligned} -5 &\lt 2x + 3 \le 9 \\ -8 &\lt 2x \le 6 && \text{subtract } 3 \\ -4 &\lt x \le 3 && \text{divide by } 2 \end{aligned}$$

Open dot at $-4$, closed dot at $3$, and shade between them.

(b) This is an “or” inequality, so solve each part on its own:

$$3x - 1 \lt -7 \;\Rightarrow\; 3x \lt -6 \;\Rightarrow\; x \lt -2 \qquad\qquad 2x + 5 \ge 11 \;\Rightarrow\; 2x \ge 6 \;\Rightarrow\; x \ge 3$$

The solution is $x \lt -2$ or $x \ge 3$: an open dot at $-2$ shaded left, and a closed dot at $3$ shaded right.

The solutions from Examples 1 and 2. Open dots are excluded endpoints; closed dots are included.

Example 3: A budget

A climbing gym charges a $35 sign-up fee plus $12 per class. Jordan has $200 to spend. What is the greatest number of classes Jordan can take?

Solution. Let $n$ be the number of classes. The total cost is $35 + 12n$ dollars, and it can be at most 200:

$$\begin{aligned} 35 + 12n &\le 200 \\ 12n &\le 165 \\ n &\le 13.75 \end{aligned}$$

Jordan can’t take $13.75$ classes. The number of classes must be a whole number that is at most $13.75$, so the greatest is $13$. (Rounding up to $14$ would break the budget.)

Check: $13$ classes cost $35 + 12(13) = 35 + 156 = 191$ dollars, which fits. $14$ classes would cost $203$ dollars, which doesn’t.

Example 4: A two-variable inequality

A bake-sale team sells muffins for $3 each and cookies for $2 each. They want to raise at least $120.

• (a) Write an inequality for $m$ muffins and $c$ cookies.
• (b) Does selling $20$ muffins and $25$ cookies meet the goal? What about $30$ muffins and $20$ cookies?

Solution.

(a) Muffins bring in $3m$ dollars and cookies bring in $2c$ dollars. The total must be at least 120:

$$3m + 2c \ge 120$$

(b) Substitute each pair:

• $(20, 25)$: $3(20) + 2(25) = 60 + 50 = 110$. Since $110 \ge 120$ is false, this does not meet the goal.
• $(30, 20)$: $3(30) + 2(20) = 90 + 40 = 130$. Since $130 \ge 120$ is true, this does meet the goal.

On a graph, $(30, 20)$ is in the shaded half-plane on or above the line $3m + 2c = 120$, and $(20, 25)$ is just outside it.

Common mistakes

Forgetting to flip the sign. Whenever you multiply or divide both sides by a negative number, flip $\lt$ to $\gt$ (or $\le$ to $\ge$), and the other way around. Check with a test value from your answer; if it doesn’t work in the original inequality, you probably missed a flip.

Flipping when you didn’t divide by a negative. Subtracting a number, or having a negative answer, doesn’t flip the sign. In $x + 7 \lt 3$, subtract $7$ to get $x \lt -4$, with no flip.

Mixing up open and closed dots. Closed dots go with $\le$ and $\ge$ (the endpoint is a solution). Open dots go with $\lt$ and $\gt$.

Rounding the wrong way in context. In Example 3, $n \le 13.75$ means the answer is $13$, not $14$. For a minimum, like “at least $8.2$ buses”, you round up to $9$. Always ask: does my whole number still satisfy the inequality?

Choosing the wrong symbol from the words. “At least” means $\ge$ and “at most” means $\le$, even though “least” sounds like “less”. “No more than $50$” means $\le 50$.

Only doing a step to two parts of a chain. In $-5 \lt 2x + 3 \le 9$, subtract $3$ from all three parts, not just the middle and one end.

Practice

1. (Warm-up) Solve $7x - 4 \gt 24$.

Solution

$$7x - 4 \gt 24 \quad\Rightarrow\quad 7x \gt 28 \quad\Rightarrow\quad x \gt 4$$

Dividing by positive $7$ doesn’t flip the sign.

2. (Warm-up) Solve $-\dfrac{x}{3} + 2 \le 6$.

Solution

$$\begin{aligned} -\frac{x}{3} + 2 &\le 6 \\ -\frac{x}{3} &\le 4 && \text{subtract } 2 \\ x &\ge -12 && \text{multiply by } -3 \text{ and flip} \end{aligned}$$

Check with $x = 0$: $0 + 2 = 2 \le 6$ ✓.

3. (Warm-up) Which value of $x$ is a solution of $5 - 2x \lt 1$?

• A) $-3$
• B) $0$
• C) $2$
• D) $5$

Solution

D. Solve: $-2x \lt -4$, so $x \gt 2$ (divide by $-2$ and flip). Only $5$ is greater than $2$. Choice C fails because $5 - 2(2) = 1$, and $1 \lt 1$ is false.

4. (Core) What is the greatest integer that satisfies $3(x - 4) \ge 5x + 6$?

Solution

$$\begin{aligned} 3x - 12 &\ge 5x + 6 \\ -2x &\ge 18 && \text{subtract } 5x \text{ and add } 12 \\ x &\le -9 && \text{divide by } -2 \text{ and flip} \end{aligned}$$

The greatest integer is $-9$.

Check: $3(-9 - 4) = -39$ and $5(-9) + 6 = -39$, and $-39 \ge -39$ ✓.

5. (Core) Solve $-7 \le 1 - 4x \lt 13$. How many integers satisfy the inequality?

Solution

Do each step to all three parts:

$$\begin{aligned} -7 &\le 1 - 4x \lt 13 \\ -8 &\le -4x \lt 12 && \text{subtract } 1 \\ 2 &\ge x \gt -3 && \text{divide by } -4 \text{ and flip both signs} \end{aligned}$$

Written in the usual order: $-3 \lt x \le 2$. The integers are $-2, -1, 0, 1, 2$, so there are $5$.

6. (Core) Priya’s first three test scores are $82$, $76$, and $91$. What is the lowest score she can get on the fourth test so that her average for the four tests is at least $85$? (Student-produced response.)

Solution

Let $x$ be the fourth score.

$$\begin{aligned} \frac{82 + 76 + 91 + x}{4} &\ge 85 \\ 249 + x &\ge 340 && \text{multiply by } 4 \\ x &\ge 91 \end{aligned}$$

The lowest score is $91$. Check: $\dfrac{249 + 91}{4} = \dfrac{340}{4} = 85$ ✓.

7. (Core) A freight elevator can carry at most $1000$ kg. The operator has a mass of $80$ kg, and each box has a mass of $23$ kg. What is the greatest number of boxes that can go up with the operator in one trip?

Solution

Let $b$ be the number of boxes.

$$80 + 23b \le 1000 \quad\Rightarrow\quad 23b \le 920 \quad\Rightarrow\quad b \le 40$$

The greatest number is $40$ boxes. Check: $80 + 23(40) = 80 + 920 = 1000$, which is exactly the limit, and “at most” allows it.

8. (Challenge) Which ordered pair $(x, y)$ is a solution of the system below?

$$y \gt 2x - 3 \qquad\text{and}\qquad x + y \le 4$$

• A) $(3, 2)$
• B) $(0, 5)$
• C) $(1, 1)$
• D) $(2, 0)$

Solution

C. Test each pair in both inequalities:

• A) $(3, 2)$: is $2 \gt 2(3) - 3 = 3$? No.
• B) $(0, 5)$: is $5 \gt -3$? Yes. Is $0 + 5 \le 4$? No.
• C) $(1, 1)$: is $1 \gt 2(1) - 3 = -1$? Yes. Is $1 + 1 \le 4$? Yes. ✓
• D) $(2, 0)$: is $0 \gt 2(2) - 3 = 1$? No.

In Desmos, type both inequalities (Desmos turns a typed >= into $\ge$, and the same works for less than or equal to). Only $(1, 1)$ lies in the region where the two shadings overlap.

9. (Challenge) In the inequality $ax - 4 \le 11$, $a$ is a constant. The solution of the inequality is $x \ge -5$. What is the value of $a$?

Solution

Add $4$: $ax \le 15$. The solution has $\ge$ while the inequality has $\le$, so the sign flipped. That only happens when we divide by a negative number, so $a \lt 0$ and

$$x \ge \frac{15}{a}$$

Match this with $x \ge -5$: $\dfrac{15}{a} = -5$, so $a = -3$.

Check: $-3x - 4 \le 11 \Rightarrow -3x \le 15 \Rightarrow x \ge -5$ ✓.