HSA.REI.A.1: Explaining and Justifying the Steps in Solving an Equation
In plain English: HSA.REI.A.1 is the Common Core algebra standard that asks students to explain each step in solving a simple equation as following from the previous step, starting from the assumption that the equation has a solution, and to construct a viable argument that justifies a solution method. Each step states an equality that must hold for the same number. It is usually taught in Algebra I.
Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Understand solving equations as a process of reasoning and explain the reasoning Also written as HSA-REI.A.1 or A-REI.1 · Official standard
This lesson treats solving an equation as a chain of logical statements. Students start by assuming that some number x makes the original equation true. Each step then states a new equality that must also be true for that same number, because it follows from the previous one by a property of equality or by rewriting an expression in an equivalent form. When the chain ends at x = 16, the conclusion is: if a solution exists, it is 16. Checking in the original equation confirms that 16 really is a solution.
Students name the reason for every step, find and explain errors in worked solutions, and build arguments for why a method works. They also see what happens when the chain ends in a false statement such as 6 = 1 (no solution) or a statement that is always true such as 0 = 0 (every real number is a solution), and why dividing both sides by an expression that could be zero is not a safe move.
Learning Objectives
By the end of this lesson, students will be able to:
Justify each step in solving a linear equation with a named property of equality or an equivalent rewriting (distributive property, combining like terms)
Explain that each new equation is true for the same number that made the previous equation true, starting from the assumption that a solution exists
Identify and correct an invalid step in a worked solution and explain why it is invalid
Interpret a chain that ends in a false statement (no solution) or a statement that is always true (every real number is a solution)
Construct a viable argument that a chosen solution method is valid, including why it is safe to multiply or divide by a nonzero number
Prior Knowledge Required
Students should already be comfortable with:
Solving linear equations with variables on both sides, including distribution and like terms 8.EE.C.7
Using substitution to decide whether a number makes an equation true 6.EE.B.5
Applying properties of operations to write equivalent expressions 7.EE.A.1
Operations with integers and fractions, including negative numbers
Write the following on the board and give students 3 minutes to think alone before discussing with a partner:
Warm-Up Prompt
"Jordan says: If 4x + 3 = 23, then 4x = 20, so x = 5. Sam says: That only tells us that IF there is a solution, it must be 5. We still have to check. Who is right, and why does Sam say IF?"
Collect responses. Draw out the idea that each step says "if the previous equation is true for some number x, then this one is true for that same x." Substitute 5 into 4x + 3 to show 23, which confirms the solution. Tell students that today they will explain every step this way and give a reason for it.
Direct Instruction20 minutes
Post the reasons students will use all lesson. Each one either keeps the two sides equal or rewrites one side without changing its value:
Addition or subtraction property of equality: if a = b, then a + c = b + c and a - c = b - c.
Multiplication or division property of equality: if a = b, then ac = bc, and a/c = b/c when c is not 0.
Equivalent rewriting of one side: distributive property, combining like terms, commutative and associative properties. The value of the side does not change.
Conclusion: the last equation tells what the solution must be, if one exists. Substitute into the original equation to confirm.
Work through these examples on the board in a two-column format (statement | reason). Say the reasoning aloud: "If x makes this equation true, then x also makes the next one true, because..."
Two-column justification
Solve 3(x - 4) + 5 = 2x + 9 and give a reason for each step: distributive property, combine like terms, subtract 2x from both sides, add 7 to both sides.
A student divides both sides of 5x = 3x by x, gets 5 = 3, and writes "no solution." Explain the error.
Equation: Dividing by x assumes x ≠ 0. Subtracting 3x instead gives 2x = 0, so x = 0. Check: 5(0) = 3(0).
Reasoning from the assumption
Assume some number x makes 2(x + 3) = 2x + 1 true. Follow the steps and decide what the result says about that assumption.
Equation: 2x + 6 = 2x + 1 → 6 = 1, which is false, so no number makes the equation true: no solution.
Contrast Example 4 with 2(x + 3) = 2x + 6, which leads to 6 = 6. That statement is true for every x, so every real number is a solution. Stress that the steps did not "lose" x: they showed that the value of x does not matter.
Guided Practice15-20 minutes
Pairs solve 5 - 2(x - 3) = x + 2 and x/4 + 1 = x/3 - 1 in two columns, one partner writing statements and the other writing reasons, then switching. For the first, 5 - 2x + 6 = x + 2 gives 11 - 2x = x + 2, so 9 = 3x and x = 3. For the second, multiplying by 12 gives 3x + 12 = 4x - 12, so x = 24. Circulate and ask "Why is that step allowed?" Common issues: writing "moved the 6 over" instead of naming the property, distributing the -2 only to x, and forgetting to multiply every term by 12.
Independent Practice10-15 minutes
Students complete three items alone: (1) solve 6(x + 1) = 4x - 8 with a reason for each step (x = -7); (2) find and fix the error in a worked solution where a student wrote 9 - 4x = 1 → 5x = 1; (3) decide whether 3(x - 2) = 3x - 6 has one solution, no solution, or every real number as a solution, and explain using the final statement of the chain.
Closure5-10 minutes
Exit ticket: "Solve 2x + 9 = 5x - 3 and give a reason for each step. Then explain in one sentence why the answer only becomes certain after you check it." (x = 4; the steps show what x must be if a solution exists, and the check shows that 4 works.)
Differentiation Strategies
For Struggling Students
Give a reason bank on a card (add, subtract, multiply by a nonzero number, divide by a nonzero number, distribute, combine like terms) so students choose rather than recall names
Start with two-step equations such as 3x - 4 = 11 before adding distribution and variables on both sides
Accept plain-language reasons ("subtract 2x from both sides") before requiring the formal property names
For Advanced Students
Ask why multiplying both sides by 0 is never used as a solving step, and what it would do to the solutions
Have students write a general argument that ax + b = c has exactly one solution when a ≠ 0, and describe what happens when a = 0
Extension (beyond this standard): explain why squaring both sides of an equation can add solutions, which previews HSA.REI.A.2
Assessment Guidance
What to Look For
The standard assesses the explanation, not only the answer. A correct x = 16 with no reasons does not show mastery. Look for reasons that name what was done to both sides (or which side was rewritten), for students who can say why a step keeps the same solution, and for the check in the original equation at the end. In error analysis, students should point to the exact step where an equality stops following from the previous one.
02
Classroom Activities
3 Activities
1
Reason Card Sort
15 minPairs
Each pair gets a solved equation cut into strips, one strip per step, plus a separate set of reason cards. Pairs put the steps in order and attach the reason that justifies each one.
Equation set 2: (x - 4)/5 = 2 → x - 4 = 10 → x = 14
Reason cards: distributive property, combine like terms, subtraction property of equality, addition property of equality, multiplication property of equality, division property of equality, plus 2 extra cards that do not fit
Procedure
Pairs order the strips and attach reasons (8 minutes)
Each pair checks its final value in the original equation and writes the check on the last strip
Debrief: which reason cards were left over, and why do they not justify any step?
Modification for Distance Learning
Put the strips and reason cards on a shared slide so pairs can drag them into order in breakout rooms.
2
Error Detectives
20 minGroups of 3
Groups receive four worked solutions that each contain one invalid step. They find the step, explain why the new equation does not follow from the previous one, and write a corrected solution.
Worked Solutions to Analyze
7 - 3(x + 1) = 10 → 4(x + 1) = 10 (subtracted 3 from 7 before distributing; the correct solution gives x = -2)
-5x = 20 → x = 4 (the sign was dropped when dividing by -5; x = -4)
2x = 7x → 2 = 7, "no solution" (divided by x, which could be 0; x = 0)
x/3 + 2 = 5 → x + 2 = 15 (only one term was multiplied by 3; x = 9)
Discussion Questions
Which errors produced a wrong number, and which one lost a solution entirely?
How could substituting the claimed answer into the original equation have caught each error?
Why is "subtract 3x from both sides" safe, while "divide both sides by x" is not?
3
Two Methods, One Argument
20 minIndividual then share
Students solve the same equation two different ways and write a short argument that both methods are valid and must give the same answer.
Equations
4(x - 5) = 12: divide both sides by 4 first, or distribute first (both give x = 8)
x/2 + x/3 = 5: multiply every term by 6, or combine the fractions to 5x/6 first (both give x = 6)
9 - 2x = 3x - 6: collect x terms on the left, or on the right (both give x = 3)
Requirements for the Argument
Name the reason for every step in both methods
Explain why the two chains must end at the same value if the original equation has a solution
Confirm the value by substituting it into the original equation
Peer Review Variation
Students trade arguments and use a checklist: every step has a reason, no division by a quantity that could be 0, and the check is shown. Reviewers write one question for the author.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Two-Column Justification
Each statement is true for the same number x that made the statement above it true. The chain shows that if a solution exists, it is 16; the check at the bottom confirms that 16 works.
Diagram 2: Moves That Keep or Change the Solutions
Use this as an anchor chart during error analysis. The moves on the left can be undone, so the new equation has exactly the same solutions. The moves on the right can add solutions, lose solutions, or break the equality.
04
Homework Assignment
~30 min
HSA.REI.A.1 Homework: Explain Every Step
Directions: Write each solution in two columns: the statement on the left and the reason on the right. Finish every problem by checking your answer in the original equation. For error-analysis problems, name the incorrect step, explain why it does not follow from the step before it, and give a corrected solution.
Part 1: Solve and Justify (Problems 1-3)
Solve 4(x - 3) = 2x + 10. Give a reason for each step.
Solve x/4 + 3 = x/2 - 1. Explain why multiplying both sides by 4 keeps the same solution.
Solve 7 - 2(x + 1) = 3x - 15. Give a reason for each step.
Part 2: Find and Fix the Error (Problems 4-5)
A student solved 5 - 3(x - 2) = 20 like this: 2(x - 2) = 20, then x - 2 = 10, so x = 12. Identify the invalid step, explain the error, and solve the equation correctly.
A student solved (x - 5)/2 = 3x + 1 like this: x - 5 = 3x + 1, then -6 = 2x, so x = -3. Show that x = -3 does not check, find the invalid step, and solve correctly.
Part 3: Construct an Argument (Problem 6)
Assume some number x makes 3(2x - 1) = 6x + 5 true. Use steps with reasons to show that this assumption leads to a false statement, and state what that means about the solutions. Then explain what changes if the right side is 6x - 3 instead.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Reasons
Every step has a correct, specific reason
Most steps have reasons, or some are vague
Few or no reasons given
Valid Steps
Each equation follows from the one before it
One invalid step
Several invalid steps
Error Analysis
Names the exact step and explains why it fails
Finds the step but explanation is unclear
Error not found
Check and Conclusion
Checks in the original equation and states the solution set
Check or conclusion 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
Which reason justifies going from 2x + 5 = 17 to 2x = 12?
Answer: C
Subtracting 5 from both sides keeps the sides equal, which is the subtraction property of equality (some texts call it adding -5 with the addition property). Choice B is wrong because nothing was multiplied, and choice A does not apply because there are no parentheses.
Question 2 of 20 · Multiple Choice
Which reason justifies going from 3(x + 4) = 21 to 3x + 12 = 21?
Answer: A
The left side was rewritten as an equivalent expression by distributing 3 over x + 4; the right side did not change. Choice B would describe dividing both sides by 3, which gives x + 4 = 7 instead.
Question 3 of 20 · Multiple Choice
Which reason justifies going from x/5 = 6 to x = 30?
Answer: D
Both sides were multiplied by 5, a nonzero number, so the multiplication property of equality applies. Choice A is a common mix-up: the equation contains division, but the step undoes it by multiplying.
Question 4 of 20 · Multiple Choice
Assume 5(x - 2) = 3x + 4 has a solution. Working step by step, with each equation following from the one before, what must that solution be?
Answer: B
5x - 10 = 3x + 4, so 2x = 14 and x = 7. Check: 5(5) = 25 and 3(7) + 4 = 25. Choice A comes from distributing the 5 only to x (5x - 2), and choice C comes from subtracting 10 instead of adding it (2x = -6).
Question 5 of 20 · Multiple Choice
A student solved 4x - 7 = 2x + 9 like this. Step 1: 2x - 7 = 9. Step 2: 2x = 2. Step 3: x = 1. Which step is invalid?
Answer: C
Step 1 subtracts 2x from both sides correctly. Step 2 should add 7 to both sides, giving 2x = 16, but the student subtracted 7 from 9 on one side only. The correct solution is x = 8. Choice B is tempting because x = 1 is wrong, but Step 3 correctly follows from Step 2.
Question 6 of 20 · Multiple Choice
When you solve 3x + 1 = 10 by writing "if x is a solution, then 3x = 9," what are you assuming?
Answer: A
The standard says to start from the assumption that the original equation has a solution. The steps then show what that number must be. Choice B assumes the answer before finding it, and choice C describes an identity, not this equation.
Question 7 of 20 · Multiple Choice
Solving 4(x + 1) = 4x + 4 leads to 4 = 4. What does this mean?
Answer: C
4 = 4 is true no matter what x is, so every real number makes the original equation true (both sides are the same expression). Choice B confuses this with a false final statement such as 4 = 5, and choice A is a common guess when the variable disappears.
Question 8 of 20 · Multiple Choice
Solving 2x + 7 = 2(x + 3) leads to 7 = 6. What does this mean?
Answer: D
2x + 7 = 2x + 6 gives 7 = 6. If some x made the original true, then 7 = 6 would be true, which is impossible. So the assumption fails and there is no solution. Choice B mixes this up with a true statement such as 6 = 6.
Question 9 of 20 · Multiple Choice
Why is dividing both sides of 3x = 7x by x an invalid first step?
Answer: B
Dividing by x gives 3 = 7, which suggests no solution, but x = 0 makes 3x = 7x true. Subtracting 3x instead gives 4x = 0, so x = 0. Choice A is wrong because dividing both sides by a nonzero number is valid.
Question 10 of 20 · Multiple Choice
Which equation has exactly the same solutions as 6x - 9 = 15?
Answer: A
Dividing every term on both sides by 3 (a nonzero number) gives 2x - 3 = 5; both equations have the solution x = 4. Choice B subtracts 9 from 15 instead of adding it. Choice C is what multiplying both sides by 0 would give, and it is true for every x, so it does not have the same solutions.
Question 11 of 20 · Multiple Choice
Assume (2x - 1)/3 = x - 2 has a solution. Working step by step, with each equation following from the one before, what must that solution be?
Answer: D
Multiply both sides by 3: 2x - 1 = 3x - 6, so x = 5. Check: (10 - 1)/3 = 3 and 5 - 2 = 3. Choice A comes from multiplying only the left side by 3 (2x - 1 = x - 2), and choice C comes from multiplying x by 3 but not the -2 (2x - 1 = 3x - 2).
Question 12 of 20 · Multiple Choice
Which reason justifies going from -3x = 12 to x = -4?
Answer: B
Both sides were divided by -3, which is not zero, so the equality still holds: 12 ÷ (-3) = -4. Choice A is a common error that treats -3x as "x minus 3."
Question 13 of 20 · Multiple Choice
Assume 0.2(x + 15) = 0.5x has a solution. Working step by step, with each equation following from the one before, what must that solution be?
Answer: D
Distribute: 0.2x + 3 = 0.5x. Subtract 0.2x: 3 = 0.3x. Divide by 0.3: x = 10. Check: 0.2(25) = 5 and 0.5(10) = 5. Choice A comes from distributing 0.2 only to x (0.2x + 15 = 0.5x), and choice B drops the 0.2x term (3 = 0.5x).
Question 14 of 20 · Multiple Choice
Maya says, "In 2(x - 3) = 2x - 6, the x cancels out, so there is no solution." Which response is correct?
Answer: B
Subtracting 2x from both sides of 2x - 6 = 2x - 6 gives -6 = -6, which is true for every x. The variable disappearing tells you to look at the final statement: a false statement means no solution, a true one means every real number. Choice C tests only one value; x = 3 works, but so does every other number.
Question 15 of 20 · Short Answer
Solve 5x - 3(x - 4) = 20 and give a reason for each step.
x = 4. 5x - 3x + 12 = 20 (distributive property; note -3 times -4 is +12) 2x + 12 = 20 (combine like terms) 2x = 8 (subtraction property of equality: subtract 12) x = 4 (division property of equality: divide by 2) Check: 5(4) - 3(0) = 20.
Question 16 of 20 · Short Answer
From 2x + 3 = 15 a student concluded x = 6. Explain how reversing the steps shows that 6 really is a solution.
The steps were: subtract 3 (2x = 12), then divide by 2 (x = 6). Each step can be undone: start from x = 6, multiply both sides by 2 to get 2x = 12, then add 3 to get 2x + 3 = 15. So x = 6 leads back to the original equation, which means 6 makes it true. Check: 2(6) + 3 = 15. Steps that can be undone (adding, subtracting, multiplying or dividing by a nonzero number) never add or lose solutions.
Question 17 of 20 · Short Answer
A student solved 8 - 2x = 14 like this: -2x = 6, so x = 3. Find the error and give the correct solution.
The first step is correct (subtract 8 from both sides). The error is in the second step: dividing 6 by -2 gives -3, not 3. x = -3. Check: 8 - 2(-3) = 8 + 6 = 14. Substituting x = 3 gives 8 - 6 = 2, not 14, which would have caught the error.
Question 18 of 20 · Short Answer
Assume some number x makes 4x - 1 = 4(x + 2) true. Show what this assumption leads to and state the conclusion.
If 4x - 1 = 4(x + 2), then 4x - 1 = 4x + 8 (distributive property). Subtracting 4x from both sides gives -1 = 8, which is false. Since the assumption that a solution exists leads to a false statement, the assumption must be wrong: the equation has no solution.
Question 19 of 20 · Short Answer
Two students solve 3(x + 2) = 18. Ana distributes first; Ben divides both sides by 3 first. Show both methods and explain why both are valid.
Ana: 3x + 6 = 18 (distributive property), 3x = 12 (subtract 6), x = 4 (divide by 3). Ben: x + 2 = 6 (divide both sides by 3, which is not 0), x = 4 (subtract 2). Both methods only use equivalent rewriting or the properties of equality, so each new equation is true for exactly the same x as the original. They must reach the same solution, x = 4. Check: 3(6) = 18.
Question 20 of 20 · Short Answer
Solve x/2 + x/5 = 14. Justify the step that clears the fractions.
Multiply both sides by 10 (multiplication property of equality; 10 is not 0): 5x + 2x = 140. Combine like terms: 7x = 140. Divide by 7: x = 20. Check: 20/2 + 20/5 = 10 + 4 = 14. Every term must be multiplied by 10, because the whole left side is multiplied, not just one fraction.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does "following from the equality of numbers asserted at the previous step" mean?
Each equation in a solution is a claim that two numbers are equal for some value of x. The next equation must be true whenever the previous one is. For example, if 3x - 7 and 2x + 9 are the same number, then subtracting 2x from both gives two numbers that are still equal: x - 7 = 9. The reason for a step is the rule that guarantees this.
Why do we start by assuming the equation has a solution?
The steps of a solution are "if, then" statements: if x makes the original equation true, then x makes each later equation true. That logic tells you what the solution must be, but only if one exists. That is why the chain can end in a false statement (no solution) and why the final check in the original equation matters.
Do students have to use formal property names?
The standard asks students to explain each step, not to memorize vocabulary. A reason such as "subtract 2x from both sides" or "distribute 3 to both terms in the parentheses" is a valid explanation. Many teachers introduce the formal names (subtraction property of equality, distributive property) so the class shares a language, but a precise plain-language reason shows the same understanding.
Is a two-column format required?
No. Two columns are a convenient way to make each reason visible, especially at first. Students can also write reasons in the margin, annotate arrows between lines, or explain in sentences. What matters is that every step has a reason and each new equation follows from the one before it.
Which moves are safe when solving an equation?
Adding or subtracting the same quantity on both sides, multiplying or dividing both sides by the same nonzero number, and rewriting one side as an equivalent expression. These moves can all be undone, so the new equation has exactly the same solutions. Multiplying both sides by 0 or dividing by an expression that might be 0 can change the solution set.
What should students conclude when the variable disappears?
Look at the statement that is left. If it is false, like 6 = 1, then assuming a solution exists led to a contradiction, so there is no solution. If it is true, like 6 = 6, then every real number is a solution because the two sides were equivalent expressions. Students often write "no solution" in both cases, so ask them to explain which kind of statement they reached.
What mistakes do students often make on this standard?
Giving "move it to the other side" as a reason without saying what was done to both sides
Distributing a negative number to only the first term in parentheses
Multiplying only one term by the denominator when clearing fractions
Dividing both sides by x, which loses the solution x = 0
Skipping the check in the original equation
How is HSA.REI.A.1 usually assessed?
Expect items that show a worked solution and ask which property justifies a given step, which step contains an error, or whether two methods are both valid. Constructed-response items may ask students to write a justified solution or to explain why a method works. Getting the right value is not enough on these items; the explanation is what is being scored.
Why does checking matter if every step was justified?
For linear equations solved with the safe moves, the check mainly catches arithmetic errors. It becomes essential later: squaring both sides of a radical equation or multiplying by an expression containing x can introduce values that do not satisfy the original equation. Building the checking habit now prepares students for extraneous solutions in HSA.REI.A.2.
What comes after HSA.REI.A.1?
The same reasoning is used to rearrange formulas (HSA.CED.A.4), to solve linear equations and inequalities with letter coefficients (HSA.REI.B.3), to explain why some solution methods create extraneous solutions (HSA.REI.A.2), and to justify elimination for systems of equations (HSA.REI.C.5). It is also the first step toward writing proofs in geometry.
07
Related Standards
6 standards
These standards connect to HSA.REI.A.1: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
6.EE.B.5Prerequisite
Understand solving an equation as finding which values make it true
