HSA.REI.C.5: Proving Why the Elimination Method Works
In plain English: HSA.REI.C.5 is the Common Core algebra standard that asks students to prove that replacing one equation in a system of two equations in two variables by the sum of that equation and a multiple of the other produces a system with the same solutions. This is why the elimination method works. It is usually taught in Algebra I.
Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Solve systems of equations Also written as HSA-REI.C.5 or A-REI.5 · Official standard
Students already use elimination to solve systems. In this lesson they prove why it works: if one equation in a system of two equations in two variables is replaced by the sum of that equation and a multiple of the other, the new system has exactly the same solutions as the original. The proof has two directions. Every solution of the original system is a solution of the new system, so no solutions are lost, and every solution of the new system is a solution of the original, so no solutions are gained.
Students test the claim numerically and graphically, write the general proof with the properties of equality, and then use it to justify each step when they solve a system. They also look at replacements that do not keep the same solutions, which shows why the theorem is stated so precisely.
Learning Objectives
By the end of this lesson, students will be able to:
Verify with a specific system that replacing one equation by the sum of that equation and a multiple of the other keeps the same solution
Prove, using properties of equality, that any solution of the original system satisfies the new system
Prove the reverse direction: any solution of the new system satisfies the original system
Justify each step of the elimination method by naming the replacement used
Identify replacements that do not preserve the solution set, such as multiplying an equation by zero
Prior Knowledge Required
Students should already be comfortable with:
Solving systems of two linear equations by elimination and substitution 8.EE.C.8
Explaining steps in solving an equation using properties of equality HSA.REI.A.1
Checking whether an ordered pair satisfies an equation
The distributive property and combining like terms
Give students the system x + y = 10 and x - y = 2. Most will find (6, 4) quickly. Then ask the question below.
Warm-Up Prompt
"Add the two equations to get 2x = 12. Now make a new system: keep x + y = 10 and replace x - y = 2 with 2x = 12. Does the new system have the same solution? Now start over and replace x + y = 10 with the first equation plus 2 times the second. What do you notice? Do you think this always works?"
Students should find that both new systems still have the solution (6, 4): 2(6) = 12, and x + y + 2(x - y) = 10 + 4 gives 3x - y = 14, and 3(6) - 4 = 14. Record the class conjecture on the board. Tell students that checking examples is not a proof and that today they will prove the conjecture for every system and every multiple.
Direct Instruction20 minutes
State the theorem in the class's own words, then name the parts: the original system {E1, E2}, a real number k, and the new system {E1, E2 + kE1}, in which E2 has been replaced by the sum of E2 and k times E1. Show Diagram 1: every combination of the two lines passes through the same intersection point. Then work through the examples below.
Verify with a known solution
System x + y = 10, x - y = 2 with solution (6, 4). Replace E2 by E2 + 1·E1.
Equation: New E2: 2x = 12, and 2(6) = 12, so (6, 4) still works
Forward direction (no solutions lost)
Suppose (s, t) satisfies Ax + By = C and Dx + Ey = F. Then As + Bt = C and Ds + Et = F are true statements about numbers.
Equation: (Ds + Et) + k(As + Bt) = F + kC, so (D + kA)s + (E + kB)t = F + kC
Reverse direction (no solutions gained)
Suppose (s, t) satisfies E1 and the new equation E2 + kE1. Subtract k times E1 from the new equation.
Equation: (F + kC) - kC = F, so Ds + Et = F and (s, t) satisfies E2
Using the theorem to solve
System 3x + 2y = 16, 5x - 4y = -10. Replace E2 by E2 + 2·E1 to eliminate y.
Equation: 11x = 22, so x = 2 and y = 5
Non-example
System x + y = 5, x - y = 1 (solution (3, 2)). Replace E2 by 0·E2.
Equation: New system x + y = 5, 0 = 0 has infinitely many solutions
Write the full forward proof as a two-column argument. Each step follows from a property: equations E1 and E2 are true at (s, t) (given); multiplying both sides of a true equation by k keeps it true (multiplication property of equality); adding equal quantities to both sides of a true equation keeps it true (addition property of equality); regrouping with the distributive property gives the new equation. Then write the reverse proof. Stress that the reverse step uses the same theorem with -k, which is why both systems have the same solution set. The last example shows the difference between this replacement, which is safe for every k, and multiplying an equation by 0, which throws away information.
Guided Practice15 minutes
Pairs work with the system 2x + y = 7 and x - 3y = -7, whose solution is (2, 3). They (1) replace E1 by E1 + 2·E2, (2) verify that (2, 3) satisfies the new equation 4x - 5y = -7, (3) solve the new system from scratch to confirm it has only the solution (2, 3), and (4) write the forward proof for this specific system with numbers instead of letters. Listen for students who check only the new equation and forget that the proof must also cover the equation that was kept. Debrief by asking which property justifies each line.
Independent Practice10 minutes
Students solve 4x + 3y = 5 and 6x - 5y = 17 by elimination and write, next to each step, the replacement they used and the value of k. (One path: replace E2 by E2 - (3/2)E1 to get -9.5y = 9.5, so y = -1 and x = 2.) Then they write the reverse argument showing that the solution of their final system also solves the original system.
Closure5-10 minutes
Exit ticket: "Explain in two or three sentences why replacing an equation by the sum of that equation and a multiple of the other gives a system with the same solutions. Then give one replacement that does not keep the same solutions." Look for both directions of the argument and a valid non-example, such as multiplying an equation by 0 or replacing both equations with the same sum.
Differentiation Strategies
For Struggling Students
Start the proof with a specific system and a specific k, using numbers in every step, before moving to letters
Provide a two-column proof template with the reasons listed in a word bank: given, multiplication property of equality, addition property of equality, distributive property
Color-code E1 and E2 so students can see which equation is kept and which is replaced
For Advanced Students
Ask students to prove that the theorem also holds for nonlinear systems, such as x² + y² = 25 and y = x + 1, since the proof never uses linearity
Ask students to show that replacing E2 by mE2 + kE1 keeps the same solutions when m ≠ 0, and to explain what goes wrong when m = 0
Ask students to explain, with the theorem, why elimination on a system with parallel lines leads to a false statement such as 0 = 4
Assessment Guidance
What to Look For
The standard asks for a proof, not only for correct elimination. A complete answer shows both directions: every solution of the original system solves the new one, and every solution of the new system solves the original one. Watch for students who verify one example and call it a proof, or who prove only the forward direction. Students should justify each step with a property of equality rather than with "you can do the same thing to both sides."
02
Classroom Activities
3 Activities
1
Lines Through One Point
20 minPairs
Students graph a system and several of its combinations to see the theorem geometrically: every equation of the form E2 + kE1 describes a line through the same intersection point.
Procedure
Graph E1: x + 2y = 8 and E2: 3x - y = 3 on graph paper and mark the intersection (2, 3)
Each partner picks two values of k (for example k = 1, -1, 2, -3) and writes E2 + kE1 in the form ax + by = c
For k = 1: 4x + y = 11. For k = -3: -7y = -21, a horizontal line y = 3. Graph each new line
Check algebraically that (2, 3) satisfies every new equation
Discussion Questions
Why must every new line pass through (2, 3)?
Which value of k makes a horizontal line? Why is that the one we choose when we eliminate?
Does any value of k give a line that misses (2, 3)? Explain.
Modification for Distance Learning
Use a graphing app with a slider for k. Students type (3x - y - 3) + k(x + 2y - 8) = 0 and watch the line turn around the point (2, 3) as k changes.
2
Build the Proof
20 minGroups of 3-4
Groups receive the statements of the forward and reverse proofs on cut-up strips, mixed with a few distractor strips. They put the strips in order, match each statement to its reason, and remove the statements that do not belong.
Setup
Forward proof strips: "(s, t) satisfies Ax + By = C and Dx + Ey = F (given)"; "k(As + Bt) = kC (multiplication property of equality)"; "(Ds + Et) + k(As + Bt) = F + kC (addition property of equality)"; "(D + kA)s + (E + kB)t = F + kC (distributive property)"
Reverse proof strips: "(s, t) satisfies E1 and E2 + kE1 (given)"; "subtract k(As + Bt) = kC from both sides"; "Ds + Et = F"
Distractors: "The solution is (0, 0)"; "k must be positive"; "Since the lines are parallel, the solutions are the same"
Procedure
Groups have 10 minutes to build both proofs and glue them on a poster
Each group writes one sentence explaining why the theorem needs both directions
Gallery walk: groups compare posters and note any differences in order that are still valid
Extension Variation
Give groups a nonlinear system, x² + y² = 25 and x - y = 1 (solutions (4, 3) and (-3, -4)), and ask whether the same strips prove the theorem there. Students should see that the proof never used the fact that the equations were linear.
3
Legal Move or Not?
15 minIndividual then share
Students judge proposed moves on a system and decide whether each one is guaranteed to keep the same solution set. For every illegal move, they give a system where it changes the solutions.
Moves to Judge
Replace E2 by E2 + 4E1 (legal)
Replace E1 by E1 - (1/2)E2 (legal)
Replace E2 by 0·E2 (not legal: x + y = 5 and x - y = 1 becomes x + y = 5 and 0 = 0)
Replace both E1 and E2 by E1 + E2 at the same time (not legal: you keep only one equation)
Add 3 to the left side of E1 only (not legal: the new equation is not equivalent)
Replace E2 by -2E2 (legal: multiplying one equation by a nonzero number keeps its solutions)
Share
Pairs compare answers and counterexamples. The class agrees on a rule: a move is safe if you can undo it with another allowed move.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Combinations of Two Equations Share the Solution
The lines x + y = 10 and x - y = 2 meet at (6, 4), drawn to scale. Adding the equations gives 2x = 12, and adding E1 to 2 times E2 gives 3x - y = 14. Both new lines pass through (6, 4), so replacing either original equation by one of them keeps the same solution.
Diagram 2: The Two Directions of the Proof
To show the systems have the same solutions, prove both arrows. The top arrow shows every solution of the original system solves the new system. The bottom arrow shows the replacement can be undone, so every solution of the new system solves the original system.
04
Homework Assignment
~30 min
HSA.REI.C.5 Homework: Proving That Elimination Keeps the Solutions
Directions: Show all work. When you replace an equation, write the replacement in the form "replace E2 by E2 + k·E1" and state the value of k. For the proofs, give a reason for every step.
Part 1: Verifying the Theorem (Problems 1-2)
The system 3x - y = 5 and x + 2y = 11 has the solution (3, 4). (a) Check that (3, 4) satisfies both equations. (b) Replace the first equation by the sum of the first equation and 3 times the second. (c) Show that (3, 4) satisfies the new equation, and solve the new system to show that (3, 4) is its only solution.
For the system x + 3y = 9 and 2x - y = 4, replace the second equation by the second equation plus -2 times the first. Write the new system, solve it, and check your solution in the original system.
Part 2: Writing the Proof (Problems 3-4)
Let (s, t) be a solution of the system Ax + By = C and Dx + Ey = F, and let k be any real number. Prove that (s, t) is a solution of (A + kD)x + (B + kE)y = C + kF. Give a reason for each step.
Prove the reverse direction: if (s, t) is a solution of the system Dx + Ey = F and (A + kD)x + (B + kE)y = C + kF, then (s, t) is a solution of Ax + By = C. Then explain in one or two sentences why both directions are needed to say the two systems have the same solutions.
Part 3: Applying and Critiquing (Problems 5-6)
Solve the system 3x + 4y = 6 and 5x - 6y = -28 by elimination. For each step, write the replacement you used and the value of k. Check your solution in the original system.
Sam starts with x + y = 7 and x - y = 3, whose solution is (5, 2). He replaces both equations by their sum, getting 2x = 10 and 2x = 10, and concludes that the system has infinitely many solutions. Explain what went wrong, using the exact wording of the theorem, and describe the solution set of Sam's new system.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Verification
Substitution correct in every equation, new system solved
Checks done but new system not solved
No checks
Forward Proof
Every step correct with a valid reason
Steps correct, reasons missing or vague
Example only, or missing
Reverse Proof
Correct, with a clear explanation of why it is needed
Correct steps, explanation missing
Missing
Elimination and Critique
Correct solution, each replacement and k named, error in Problem 6 explained
Correct solution but replacements not named, or partial critique
Incorrect or missing
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
In the system x + y = 11 and x - y = 5, the second equation is replaced by the sum of the second equation and the first. What is the new second equation?
Answer: A
(x - y) + (x + y) = 5 + 11 gives 2x = 16. Choice B subtracts the second equation from the first instead of adding them: (x + y) - (x - y) = 11 - 5, so 2y = 6. Choice C adds the left sides but subtracts the right sides. Choice D forgets that x + x is 2x.
Question 2 of 20 · Multiple Choice
The ordered pair (4, -1) is a solution of a system of two equations. The first equation is replaced by the sum of the first equation and 5 times the second. Which statement is true?
Answer: B
By the theorem, a solution of the original system satisfies both equations of the new system: the second equation is unchanged, and the new first equation is a sum of true statements. Choice D is wrong because the theorem holds for every real multiple, positive, negative, or zero.
Question 3 of 20 · Multiple Choice
Which replacement is guaranteed to produce a system with the same solutions as the original?
Answer: C
Replacing one equation by the sum of that equation and a multiple of the other (here the multiple is -4) is exactly the move in the theorem. Choice A changes E1 into an equation with different solutions. Choice B can add solutions: squaring x - y = 1 gives (x - y)² = 1, which is also true when x - y = -1. Choice D leaves only one distinct equation, so the information in E2 is lost.
Question 4 of 20 · Multiple Choice
In the system 2x + 3y = 12 and x - y = 1, the first equation is replaced by the first equation plus -2 times the second. What is the new first equation?
Answer: D
(2x + 3y) - 2(x - y) = 12 - 2(1). The left side is 2x + 3y - 2x + 2y = 5y, and the right side is 10, so 5y = 10. Choice A multiplies the left side of E2 by -2 but adds 2 on the right side. Choice C subtracts E2 once instead of twice.
Question 5 of 20 · Multiple Choice
In the forward proof, you know As + Bt = C and Ds + Et = F. Which property lets you conclude (Ds + Et) + k(As + Bt) = F + kC?
Answer: A
First, k(As + Bt) = kC by the multiplication property of equality. Then adding equal quantities to both sides of the true equation Ds + Et = F keeps it true. Choice C applies only to products equal to zero, and choice D is about inequalities.
Question 6 of 20 · Multiple Choice
Why does the proof need a second direction, showing that every solution of the new system is a solution of the original?
Answer: B
The forward direction shows no solutions are lost. Without the reverse direction, the new system could have additional solutions, as in the non-example where E2 is replaced by 0 = 0. Choice D is a common gap in student proofs.
Question 7 of 20 · Multiple Choice
A new system is {E1, E2 + kE1}. Which step recovers E2 from the new system?
Answer: C
(E2 + kE1) - kE1 = E2. This is the same kind of replacement with multiple -k, which is why the replacement can always be undone. Choice A gives E2 + 2kE1, and choice D fails when k = 0 and does not remove the kE1 part.
Question 8 of 20 · Multiple Choice
Replacing E2 by E2 + E1 turns the system 3x + y = 11 and 2x - y = 4 into the system 3x + y = 11 and 5x = 15. By the theorem, both systems have the same solution. What is it?
Answer: C
Replace E2 by E2 + E1: 5x = 15, so x = 3. Then 3(3) + y = 11 gives y = 2. Check: 2(3) - 2 = 4. Choice A switches x and y. Choice B has the wrong sign on y: 3(3) - 2 = 7, not 11.
Question 9 of 20 · Multiple Choice
For the system x + 2y = 4 and 2x + 4y = 8, replacing E2 by E2 - 2E1 gives 0 = 0. What does this tell you?
Answer: D
The new system {x + 2y = 4, 0 = 0} has the same solutions as the original. Since 0 = 0 is true for every pair, the solutions are all points on x + 2y = 4, so there are infinitely many. Choice B confuses the equation 0 = 0 with the point (0, 0).
Question 10 of 20 · Multiple Choice
For the system x + y = 3 and x + y = 7, replacing E2 by E2 - E1 gives 0 = 4. What does this tell you?
Answer: A
The new system has the same solutions as the original, and no pair makes 0 = 4 true, so neither system has a solution. The lines x + y = 3 and x + y = 7 are parallel. Choice D is tempting, but the arithmetic is correct: 7 - 3 = 4.
Question 11 of 20 · Multiple Choice
Which replacement eliminates x from the system 3x - 2y = 4 and 6x + 5y = 35?
Answer: B
E2 - 2E1: (6x + 5y) - 2(3x - 2y) = 35 - 8, which gives 9y = 27, so y = 3 and x = 10/3. Choice A doubles the x-term to 12x instead of removing it. Choice C leaves -3x.
Question 12 of 20 · Multiple Choice
Replacing E2 by E2 - 2E1 turns the system 5x + 2y = 1 and 3x + 4y = 9 into the system 5x + 2y = 1 and -7x = 7. What is the solution of the original system?
Answer: B
Replace E2 by E2 - 2E1: 3x - 10x = 9 - 2, so -7x = 7 and x = -1. Then 5(-1) + 2y = 1 gives y = 3. Check: 3(-1) + 4(3) = 9. Choice A satisfies the first equation but not the second: 3 - 8 = -5.
Question 13 of 20 · Multiple Choice
A student says: "The theorem only works for linear equations." Which response is correct?
Answer: D
The standard says "a system of two equations in two variables," not only linear equations. The proof adds true statements about numbers and never uses linearity. Choice A is wrong because slopes do not appear in the proof.
Question 14 of 20 · Multiple Choice
Which statement about replacing E2 by E2 + kE1 is true?
Answer: A
Nothing in the proof depends on the value of k. When k = 0, the new system is the same as the original, which is trivially true. Choice D confuses this move with multiplying an equation by a number, which must be nonzero.
Question 15 of 20 · Short Answer
Verify that (2, -3) solves 4x + y = 5 and x - 2y = 8. Then write the equation E1 + 3E2 and show that (2, -3) satisfies it.
Prove that any solution (s, t) of the system x + y = 6 and 2x - y = 3 also satisfies 3x = 9.
Since (s, t) is a solution, s + t = 6 and 2s - t = 3 are true. Adding equal quantities to equal quantities: (s + t) + (2s - t) = 6 + 3, so 3s = 9. Therefore (s, t) satisfies 3x = 9. (The solution is (3, 3).)
Question 17 of 20 · Short Answer
Explain why replacing E2 by 0·E2 does not always produce a system with the same solutions. Give an example.
0·E2 is the equation 0 = 0, which every pair satisfies, so the information in E2 is lost. Example: 2x + y = 9 and x - y = 3 has the single solution (4, 1). Replacing E2 by 0·E2 gives 2x + y = 9 and 0 = 0, which has infinitely many solutions, such as (0, 9). This move cannot be undone, which is why the proof fails.
Question 18 of 20 · Short Answer
Solve 2x + 5y = -1 and 3x - 2y = 8 by elimination. Name each replacement you use.
One path: replace E2 by E2 - (3/2)E1: -2y - (15/2)y = 8 + 3/2, so -(19/2)y = 19/2 and y = -1. Substituting, 2x - 5 = -1, so x = 2. Solution: (2, -1). Check: 6 + 2 = 8. Scaling first (3E1 and 2E2) and then subtracting also works.
Question 19 of 20 · Short Answer
A student replaced E2 of the system x - y = 2 and x + y = 8 with E2 + E1 and wrote 2x = 6. Find the error and finish solving.
The right sides must also be added: 8 + 2 = 10, so the new equation is 2x = 10. The student subtracted 8 - 2. Then x = 5 and y = 3. Solution: (5, 3). Check: 5 - 3 = 2 and 5 + 3 = 8.
Question 20 of 20 · Short Answer
Suppose (s, t) satisfies E1: Ax + By = C and the new equation (D + kA)x + (E + kB)y = F + kC. Show that (s, t) satisfies E2: Dx + Ey = F.
From the new equation, Ds + Et + k(As + Bt) = F + kC (distributive property). From E1, As + Bt = C, so k(As + Bt) = kC. Subtract equal quantities from both sides: Ds + Et = F. So (s, t) satisfies E2. This is the reverse direction of the proof.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
Does HSA.REI.C.5 ask students to solve systems or to prove something?
It asks students to prove the idea behind the elimination method. If a system has two equations and you replace one of them by the sum of that equation and a multiple of the other, the new system has exactly the same solutions. Students are not only solving systems here; they are explaining why the method is valid.
Why isn't checking a few examples a proof?
An example shows the idea works for one system and one multiple. A proof must cover every system of two equations in two variables and every real number k. The general proof uses letters for the coefficients, a solution (s, t), and the properties of equality, so it applies to all cases at once. Examples are a good way to build the conjecture before the proof.
Why does the proof need two directions?
"Same solutions" means the two solution sets are equal. The forward direction shows every solution of the original system solves the new system, so no solutions are lost. The reverse direction shows every solution of the new system solves the original, so no solutions are gained. Without the second part, the new system could have extra solutions, as happens when an equation is replaced by 0 = 0.
Which properties of equality appear in the proof?
The main ones are:
Multiplication property of equality: if As + Bt = C, then k(As + Bt) = kC
Addition property of equality: adding equal quantities to both sides of a true equation keeps it true
Distributive property: to regroup (Ds + Et) + k(As + Bt) as (D + kA)s + (E + kB)t
Subtraction property of equality: in the reverse direction, to remove kC from both sides
Does the theorem work if k is negative or a fraction?
Yes. The proof works for every real number k. In practice, subtracting an equation is the same as adding -1 times it, and fractional multiples such as -3/2 are sometimes the quickest way to eliminate a variable. Students often multiply both equations by integers first to avoid fractions; that is also valid, because multiplying one equation by a nonzero number keeps its solutions.
What is a common mistake students make with this standard?
Proving only one direction, or giving a numerical example instead of a general argument. In the computations, common errors are multiplying only the left side of an equation by k, and adding the left sides while subtracting the right sides. Ask students to write the whole equation k·E1 before adding it.
Does this only apply to linear systems?
No. The standard says "a system of two equations in two variables," and the proof never uses the fact that the equations are linear. For example, adding the equations x² + y² = 25 and x² - y² = 7 gives 2x² = 32, and the system with that replacement has the same four solutions (±4, ±3). Most classroom work uses linear systems, but advanced students can test the idea on nonlinear ones.
How does this connect to graphing?
Each equation of the form E2 + kE1 is a new line that passes through the intersection point of the two original lines (see Diagram 1). Choosing k to eliminate a variable turns the new line into a vertical line (x = a number) or a horizontal line (y = a number), which is why elimination makes the solution easy to read.
How is HSA.REI.C.5 usually assessed?
Expect three kinds of tasks: explaining or completing a proof, justifying the steps of an elimination solution, and deciding whether a proposed step keeps the same solutions. Written explanations matter here, so students should practice short arguments in complete sentences, not only computations.
What comes after HSA.REI.C.5?
Students apply the method in HSA.REI.C.6, solving systems of linear equations exactly and approximately. The same row operations appear again when systems are written as matrix equations in HSA.REI.C.8, and in later courses they are the basis for solving larger systems with row reduction.
07
Related Standards
6 standards
These standards connect to HSA.REI.C.5: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.EE.C.8Prerequisite
Analyze and solve pairs of simultaneous linear equations
