HSA.REI.B.3: Solving Linear Equations and Inequalities in One Variable
In plain English: HSA.REI.B.3 is the Common Core algebra standard that asks students to solve linear equations and inequalities in one variable, including equations whose coefficients are letters, such as ax + b = c. Key ideas are reversing the inequality symbol when multiplying or dividing by a negative number and stating restrictions such as a ≠ 0. It is usually taught in Algebra I.
Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Solve equations and inequalities in one variable Also written as HSA-REI.B.3 or A-REI.3 · Official standard
Students solve multi-step linear equations and inequalities in one variable: equations with the variable on both sides, with parentheses, and with fraction or decimal coefficients, and inequalities whose solution sets they write and graph on a number line. They learn why multiplying or dividing both sides of an inequality by a negative number reverses the inequality symbol, and they recognize equations with no solution or with every real number as a solution.
The standard also asks for equations whose coefficients are letters, such as ax + b = c. Students solve these with exactly the same steps they use for numbers and state the conditions needed, for example a ≠ 0, before dividing by an expression. This builds directly toward rearranging formulas and working with parameters in later courses.
Learning Objectives
By the end of this lesson, students will be able to:
Solve multi-step linear equations in one variable, including those with parentheses, fractions, decimals and the variable on both sides
Recognize and explain linear equations with one solution, no solution, or every real number as a solution
Solve linear inequalities in one variable, reversing the symbol when multiplying or dividing by a negative number, and graph the solution set
Solve linear equations whose coefficients are letters and state any restriction needed, such as a ≠ 0
Interpret a solution or solution set in the context of a word problem
Prior Knowledge Required
Students should already be comfortable with:
Solving linear equations with rational coefficients and like terms 8.EE.C.7
Solving and graphing inequalities of the form px + q > r 7.EE.B.4
Justifying the steps in solving an equation HSA.REI.A.1
Integer and fraction operations, including the distributive property with negative numbers
Write two statements on the board and ask students to decide, without solving anything, whether each is true:
Warm-Up Prompt
"We know 2 < 6. Is 2 · 3 < 6 · 3 true? Is 2 · (-3) < 6 · (-3) true? What has to change to make the second statement true?"
Students find that 6 < 18 is true, but -6 < -18 is false; the true statement is -6 > -18. Mark 2, 6, -6 and -18 on a number line: multiplying by a negative number reflects points across 0, so their order reverses. This is the reason the inequality symbol flips, and it is the one rule that differs between solving equations and solving inequalities.
Direct Instruction20 minutes
Review the solving moves and the one difference for inequalities:
Simplify each side: distribute and combine like terms. Clear fractions or decimals by multiplying every term by the same positive number.
Collect the variable terms on one side and the constant terms on the other by adding or subtracting on both sides.
Divide by the coefficient. For an inequality, reverse the symbol if you multiply or divide by a negative number. For a letter coefficient, state that it cannot be 0.
Check and interpret: substitute into the original, test one value from an inequality's solution set, and answer the question in context.
Equation: -3x ≥ 12 → x ≤ -4 (symbol reversed when dividing by -3). Closed dot at -4, shaded to the left.
Inequality in context
A ride costs $4 plus $1.50 per mile. How many miles can you ride with at most $25?
Equation: 4 + 1.50m ≤ 25 → 1.50m ≤ 21 → m ≤ 14. You can ride at most 14 miles.
Coefficients represented by letters
Solve ax + 3 = bx + 7 for x, where a and b are constants.
Equation: ax - bx = 4 → x(a - b) = 4 → x = 4/(a - b), provided a ≠ b.
After Example 5, ask what happens when a = b. The equation becomes 3 = 7 after subtracting ax, so it has no solution. The restriction a ≠ b is part of the answer, not a detail. Then substitute a = 5 and b = 3 to show that the general answer gives x = 2, the same result as solving 5x + 3 = 3x + 7 directly.
Guided Practice15-20 minutes
Pairs solve three problems, checking each other's work at every line: (1) 3(x + 4) = 3x + 10, which leads to 12 = 10, so there is no solution; (2) the compound inequality -3 < 2x + 1 ≤ 7, which gives -2 < x ≤ 3 (open dot at -2, closed dot at 3); (3) solve m(x - 2) = 10 for x (x = 10/m + 2, m ≠ 0). Circulate and listen for students who flip the symbol when dividing by a positive number, or who forget to apply an operation to all three parts of a compound inequality.
Independent Practice10-15 minutes
Students work alone on four items: 0.4(x - 5) = 0.1x + 1 (x = 10); 2 - 5x < 17 (x > -3, graphed); 4(x - 1) = 2(2x - 2) (every real number); and solve p - qx = r for x (x = (p - r)/q, q ≠ 0). For the inequality, students test one value inside the solution set and one outside.
Closure5-10 minutes
Exit ticket: "Solve -2(x + 1) > 8 and graph the solution. Then write one sentence explaining when you reverse an inequality symbol and why." (-2x - 2 > 8 gives -2x > 10, so x < -5.)
Differentiation Strategies
For Struggling Students
Provide a number line strip for testing values: after solving an inequality, students substitute one shaded and one unshaded value into the original
Before letter coefficients, have students solve 3x + 5 = 17 and ax + b = c side by side, one step per row (see Diagram 2)
Highlight negative coefficients in a different color before students divide, as a reminder to reverse the symbol
For Advanced Students
Solve ax + b > c for x and describe the solution for a > 0, a < 0 and a = 0
Find a value of k for which kx + 6 = 2(x + 3) has every real number as a solution (k = 2), and one for which 2(x + 3) = 2x + k has no solution (any k ≠ 6)
Write a word problem whose inequality has a negative coefficient, such as a tank that drains at a constant rate
Assessment Guidance
What to Look For
Watch for three things: reversing the inequality symbol exactly when multiplying or dividing by a negative number (and not otherwise), stating restrictions such as a ≠ 0 when dividing by a letter, and correct interpretation of special cases. A student who writes "x = 0" for 3(x + 4) = 3x + 10 has not yet understood what a false statement such as 12 = 10 means. For inequalities, ask students to test a value from their solution set in the original inequality.
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Classroom Activities
3 Activities
1
Flip or Keep?
15 minPairs
Pairs get eight inequality cards. For each one they predict whether the inequality symbol will be reversed during solving, then solve it, graph it on a number line and test one value.
Card Set
3x - 4 > 11 (keep: x > 5)
-2x + 1 ≤ 9 (flip: x ≥ -4)
5 - x < 2 (flip: x > 3)
x/(-4) ≥ 2 (flip: x ≤ -8)
6 > 2x - 8 (keep: x < 7)
-3(x - 2) ≥ 12 (flip: x ≤ -2)
4x + 7 < x - 5 (keep: x < -4)
2x - 9 > 5x (flip: x < -3)
Procedure
Partners predict "flip" or "keep" for each card before solving
They solve, graph and test one value in the original inequality
Accept "keep" on a "flip" card when a pair collected the x terms on the side where the coefficient is positive (for example, 2x - 9 > 5x becomes -9 > 3x, so x < -3). When x appears on both sides, the labels assume the x terms are collected on the left.
Debrief: which cards were hardest to predict? Why does 6 > 2x - 8 not need a flip, even though the variable is on the right?
Modification for Distance Learning
Put the cards on a shared slide with a "Flip" and a "Keep" column and a number line under each card that students can mark with shapes.
2
Numbers to Letters
20 minGroups of 3
Groups solve a numeric equation and then the same equation with letters in place of the numbers, matching every step. They finish by stating the restriction the letter version needs.
Equation Pairs
4x - 7 = 13 and ax - b = c (x = 5; x = (b + c)/a, a ≠ 0)
6x + 2 = 2x + 14 and mx + p = nx + q (x = 3; x = (q - p)/(m - n), m ≠ n)
3(x + 5) = 21 and k(x + h) = d (x = 2; x = d/k - h, k ≠ 0)
Discussion Questions
Why does the letter version need a restriction but the numeric version does not?
Substitute the numbers into your letter answer. Do you get the numeric answer?
What happens to mx + p = nx + q when m = n and p = q? When m = n and p ≠ q?
3
Budget Constraints
20 minPairs
Pairs write and solve inequalities for realistic budget and planning situations and interpret the solution sets, including when only whole-number answers make sense.
Scenarios
A club has $180 for a trip. The bus costs $60 and each ticket costs $8. How many students can go? (60 + 8s ≤ 180, so s ≤ 15)
A phone plan costs $35 plus $0.25 per minute over the plan limit. How many extra minutes keep the bill at or under $50? (35 + 0.25m ≤ 50, so m ≤ 60)
A water tank holds 500 gallons and drains 20 gallons per minute. When does it hold fewer than 140 gallons? (500 - 20t < 140, so t > 18 minutes)
Requirements
Define the variable with units
Show the step where the symbol is reversed, if there is one
State the solution set in context, such as "0 to 15 students"
Create Your Own Variation
Each pair writes a scenario whose inequality needs a reversed symbol, trades with another pair, and checks the other pair's solution by testing a value.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Graphing Solution Sets on a Number Line
Three solution sets from this lesson, drawn to scale. The arrow shows that the set continues without end. For a compound inequality, the solution is the segment between the two endpoints, with each endpoint open or closed according to its symbol.
Diagram 2: The Same Steps With Numbers and With Letters
Solving an equation with letter coefficients uses the same moves as solving one with numbers. The new part is the condition: you can divide by a only if a is not 0.
04
Homework Assignment
~30 min
HSA.REI.B.3 Homework: Linear Equations and Inequalities
Directions: Show every step. Check each equation solution in the original equation. For each inequality, graph the solution set on a number line and test one value. For problems with letter coefficients, state any restriction on the letters.
Part 1: Linear Equations (Problems 1-2)
Solve 4(2x - 3) - 5 = 3(x + 1).
Solve 0.6x + 2.4 = 0.2(x + 30).
Part 2: Linear Inequalities (Problems 3-4)
Solve 3 - 4(x - 1) > x + 12 and graph the solution set.
A school club has $240 for a museum trip. The bus rental costs $72, and each student ticket costs $12. Write and solve an inequality to find how many students can go. Describe the solution set in context.
Part 3: Coefficients Represented by Letters (Problems 5-6)
Solve ax + b = c - dx for x and state the restriction. Then use your answer to solve 3x + 4 = 24 - 2x.
Solve a(x + b) = cx for x and state the restriction. Check your answer with a = 2, b = 3 and c = 5.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Solving Steps
All steps shown and correct
Steps shown with one error
Little or no work shown
Inequality Symbol and Graph
Symbol reversed correctly and graph matches (open or closed dot, direction)
Correct solution but graph has an error
Symbol or graph incorrect
Letter Coefficients
Correct expression and restriction stated
Correct expression, restriction missing
Incorrect
Check and Context
Answers checked and interpreted in context
Check or interpretation missing
Neither shown
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Solve 3x - 7 = 2x + 5.
Answer: B
Subtract 2x from both sides: x - 7 = 5. Add 7: x = 12. Check: 36 - 7 = 29 and 24 + 5 = 29. Choice A comes from subtracting 7 from 5 instead of adding (x = 5 - 7).
Question 2 of 20 · Multiple Choice
Solve 2(x + 5) = 4x - 6.
Answer: A
2x + 10 = 4x - 6, so 16 = 2x and x = 8. Check: 2(13) = 26 and 32 - 6 = 26. Choice B comes from distributing the 2 only to x (2x + 5 = 4x - 6).
Question 3 of 20 · Multiple Choice
Solve -4x + 3 > 15.
Answer: D
-4x > 12. Dividing by -4 reverses the symbol: x < -3. Test x = -4: 16 + 3 = 19 > 15, true. Choice A forgets to reverse the symbol.
Question 4 of 20 · Multiple Choice
Which describes the graph of x ≥ 2 on a number line?
Answer: B
The symbol ≥ includes 2, so the dot is closed, and the solutions are numbers greater than 2, to the right. Choice A would show x > 2, which leaves out 2.
Question 5 of 20 · Multiple Choice
Solve 6 - x ≤ 2x - 9.
Answer: C
Add x to both sides: 6 ≤ 3x - 9. Add 9: 15 ≤ 3x. Divide by 3: 5 ≤ x, that is, x ≥ 5. No reversal is needed because the division is by a positive number. Choice A reverses the symbol when it should not; students who move the variable to the right side sometimes misread 5 ≤ x as x ≤ 5.
Question 6 of 20 · Multiple Choice
Solve x/4 - 1 = x/6 + 2.
Answer: A
Multiply every term by 12: 3x - 12 = 2x + 24, so x = 36. Check: 9 - 1 = 8 and 6 + 2 = 8. Choice B comes from multiplying only the fractions by 12 and not the constants (3x - 1 = 2x + 2).
Question 7 of 20 · Multiple Choice
How many solutions does 3(x - 2) = 3x + 4 have?
Answer: D
3x - 6 = 3x + 4 leads to -6 = 4, which is false for every x, so there is no solution. Choice C would be the answer if the final statement were true, such as 4 = 4. Choice A is a common guess when the variable disappears.
Question 8 of 20 · Multiple Choice
How many solutions does 2(3x - 1) = 6x - 2 have?
Answer: C
Distributing gives 6x - 2 = 6x - 2. The two sides are the same expression, so every real number is a solution. Choice A confuses a true final statement (-2 = -2) with a false one.
Question 9 of 20 · Multiple Choice
Solve x/a + b = c for x, where a ≠ 0.
Answer: B
Subtract b from both sides: x/a = c - b. Multiply both sides by a: x = a(c - b). Choice A divides by a instead of multiplying, and choice C multiplies only c by a, not the whole right side c - b.
Question 10 of 20 · Multiple Choice
Solve px + q = rx for x, where p ≠ r.
Answer: A
Subtract rx and q from both sides: px - rx = -q, so x(p - r) = -q and x = -q/(p - r) = q/(r - p). Choice D has the sign wrong: it comes from moving q to the other side without changing its sign.
Question 11 of 20 · Multiple Choice
A parking garage charges $6 plus $2.50 per hour. Which describes the number of hours h you can park if you spend at most $21?
Answer: B
6 + 2.50h ≤ 21, so 2.50h ≤ 15 and h ≤ 6. You can park for at most 6 hours. Choice A stops after subtracting 6 and forgets to divide by 2.50. Choice C uses the wrong direction: "at most" means ≤. Choice D adds 6 instead of subtracting it.
Question 12 of 20 · Multiple Choice
Solve -5 ≤ 2x + 3 < 9.
Answer: C
Subtract 3 from all three parts: -8 ≤ 2x < 6. Divide all three by 2: -4 ≤ x < 3. The symbols keep their positions because 2 is positive. Choice B switches which end is included, and choice A comes from adding 3 to all three parts instead of subtracting it.
Question 13 of 20 · Multiple Choice
Which inequality has the solution set x < -2?
Answer: D
Dividing -3x > 6 by -3 reverses the symbol: x < -2. Choice A gives x > -2, choice B gives x > -2 (after reversing), and choice C gives x > 2.
Question 14 of 20 · Multiple Choice
Solve 5x + k = 2x - 7 for x.
Answer: C
Subtract 2x and k from both sides: 3x = -7 - k, so x = -(k + 7)/3. Check with k = 2: 5x + 2 = 2x - 7 gives x = -3, and -(2 + 7)/3 = -3. Choice D adds 5x and 2x instead of subtracting.
Question 15 of 20 · Short Answer
Solve 7x - 2(x - 3) = 3(x + 4).
7x - 2x + 6 = 3x + 12 (distribute; note -2 times -3 is +6). 5x + 6 = 3x + 12. 2x = 6. x = 3. Check: 21 - 2(0) = 21 and 3(7) = 21.
Question 16 of 20 · Short Answer
Solve 4 - 3x ≥ 2x + 29 and graph the solution set.
Subtract 2x: 4 - 5x ≥ 29. Subtract 4: -5x ≥ 25. Divide by -5 and reverse the symbol: x ≤ -5. Graph: closed dot at -5, shaded to the left. Test x = -6: 4 + 18 = 22 and -12 + 29 = 17, and 22 ≥ 17 is true.
Question 17 of 20 · Short Answer
Gym A charges $40 per month for unlimited classes. Gym B charges $25 per month plus $3 per class. For how many classes per month is Gym B cheaper?
Let c = number of classes. Gym B is cheaper when 25 + 3c < 40, so 3c < 15 and c < 5. Since classes come in whole numbers, Gym B is cheaper for 0 to 4 classes per month. At exactly 5 classes both gyms cost $40.
Question 18 of 20 · Short Answer
Solve a(x + 4) = 9 for x, where a ≠ 0.
Divide both sides by a (allowed because a ≠ 0): x + 4 = 9/a. Subtract 4: x = 9/a - 4, which can also be written x = (9 - 4a)/a. Check with a = 3: 3(x + 4) = 9 gives x = -1, and 9/3 - 4 = -1.
Question 19 of 20 · Short Answer
Solve mx - 3 = nx + 9 for x. State the restriction and explain what happens when the restriction is not met.
Subtract nx from both sides and add 3: mx - nx = 12, so x(m - n) = 12 and x = 12/(m - n), with m ≠ n. If m = n, the x terms cancel and the equation becomes -3 = 9, which is false, so there is no solution. Check with m = 7, n = 3: 7x - 3 = 3x + 9 gives x = 3, and 12/4 = 3.
Question 20 of 20 · Short Answer
A student solved -2x < 8 and wrote x < -4. Explain the error and give the correct solution.
Dividing both sides by -2 reverses the order of the two sides, so the symbol must be reversed: x > -4. Test x = 0: -2(0) = 0 < 8 is true, and 0 is in x > -4 but not in x < -4. Test x = -5: 10 < 8 is false, so -5 cannot be in the solution set.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
Why does the inequality symbol flip when you multiply or divide by a negative number?
Multiplying by a negative number reflects every point on the number line across 0, which reverses the order of any two numbers. For example, 2 < 6, but -2 > -6. So the side that was smaller becomes larger, and the symbol must reverse to stay true. Adding or subtracting, even a negative number, only shifts points and never changes their order.
Do I flip the symbol when the variable ends up on the right side?
No. Flipping depends only on multiplying or dividing by a negative number. If you get 15 ≤ 3x, divide by 3 to get 5 ≤ x, which means the same as x ≥ 5. Rewriting 5 ≤ x as x ≥ 5 swaps the sides and the symbol together, so the meaning does not change.
What does "coefficients represented by letters" mean?
It means equations such as ax + b = c or mx + 4 = nx + 10, where letters stand for fixed numbers. You solve for x with the same steps you use for numbers, and the answer is an expression such as x = (c - b)/a. Because you may divide by an expression, you also state when that is allowed, such as a ≠ 0.
Why do I have to write restrictions like a ≠ 0?
Dividing by 0 is undefined. If a = 0, the equation ax + b = c becomes b = c, which is either always true or never true, and the formula x = (c - b)/a does not apply. The restriction tells the reader exactly when the answer is valid, so it is part of a complete solution.
How can a linear equation have no solution or infinitely many solutions?
If the variable terms cancel, look at the statement that is left. A false statement such as -6 = 4 means no value of x works, so there is no solution. A true statement such as -2 = -2 means the two sides are the same expression, so every real number is a solution.
What mistakes do students often make on this standard?
Distributing a negative number to only the first term in parentheses
Forgetting to reverse the symbol after dividing by a negative, or reversing it after dividing by a positive
Multiplying only some terms by the common denominator when clearing fractions
Using an open dot for ≤ or ≥, or a closed dot for < or >
Leaving out the restriction when dividing by a letter
How should students check an inequality solution?
Substitute the boundary value to confirm the two sides are equal there (if you replace the symbol with =), then test one value inside the solution set, which should make the original inequality true, and one value outside, which should make it false. This quick test catches most direction errors.
Is HSA.REI.B.3 on the SAT?
Yes. Linear equations and linear inequalities in one variable are part of the Algebra domain of the digital SAT, including word problems and equations with constants represented by letters. Questions may ask for the value of a constant that gives an equation no solution or infinitely many solutions, which draws directly on this lesson.
Are compound inequalities part of HSA.REI.B.3?
Yes. A compound inequality such as -3 < 2x + 1 ≤ 7 is a pair of linear inequalities in one variable, so solving it uses exactly the skills of this standard. Perform each step on all three parts and reverse both symbols if you multiply or divide by a negative number. Many Algebra I courses teach them with this standard.
How does this standard connect to later topics?
Solving with letter coefficients leads straight into rearranging formulas (HSA.CED.A.4), and the same equation-solving moves are used for systems of linear equations (HSA.REI.C.6). Graphing inequality solution sets on a number line prepares students for graphing linear inequalities in two variables as half-planes (HSA.REI.D.12).
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Related Standards
6 standards
These standards connect to HSA.REI.B.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.EE.C.7Prerequisite
Solve linear equations with one, no or infinitely many solutions
