HSA.REI.C.6: Solving Systems of Linear Equations Exactly and Approximately
In plain English: HSA.REI.C.6 is the Common Core algebra standard that asks students to solve systems of linear equations exactly and approximately, focusing on pairs of linear equations in two variables. Exact solutions come from substitution or elimination, and approximate solutions come from graphs, where the solution is the intersection point of the two lines. It is usually taught in Algebra I.
Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Solve systems of equations Also written as HSA-REI.C.6 or A-REI.6 · Official standard
Students solve pairs of linear equations in two variables in two ways: approximately, by graphing both lines and estimating the intersection point, and exactly, by substitution or elimination. The lesson puts the two side by side on purpose. A graph shows how many solutions a system has and roughly where the solution is, but when the intersection is not on a grid point, only algebra gives the exact answer.
Students compare their graphical estimates with exact answers, recognize systems with no solution or infinitely many solutions, and solve real-world problems that lead to a pair of linear equations, checking that the answer makes sense in context.
Learning Objectives
By the end of this lesson, students will be able to:
Estimate the solution of a system of two linear equations by graphing, and explain the limits of that estimate
Solve a system exactly by substitution and by elimination, and check the solution in both equations
Choose an efficient exact method based on the form of the equations
Recognize and interpret systems with no solution or infinitely many solutions, both graphically and algebraically
Solve a real-world problem that leads to a pair of linear equations and interpret the solution in context
Prior Knowledge Required
Students should already be comfortable with:
Solving linear equations in one variable HSA.REI.B.3
Graphing linear equations in two variables on coordinate axes HSA.CED.A.2
Understanding that a solution of a system is a point on both graphs 8.EE.C.8
Substituting an ordered pair into an equation to check it
Project a graph of y = x + 1 and y = -2x + 5 on a 1-unit grid (Diagram 1, without the labels on the right).
Warm-Up Prompt
"Where do these two lines cross? Write your best estimate as an ordered pair. Then substitute your estimate into both equations. Is it exactly right? How could you find the exact point?"
Collect estimates on the board; most will be close to (1.3, 2.3) or (1.5, 2.5). Substituting (1.3, 2.3) gives 1.3 + 1 = 2.3 in the first equation but -2(1.3) + 5 = 2.4 in the second, so the estimate is close but not exact. This motivates the lesson's question: when is a graph good enough, and when do we need an exact method?
Direct Instruction25 minutes
Model three methods, and for each one say what it is good for: graphing shows the number of solutions and gives an estimate; substitution is quick when one equation is already solved for a variable; elimination is quick when the equations are in standard form with matching or opposite coefficients. Every solution is checked in both original equations.
Graphing, exact grid point
Graph y = 2x - 1 and y = -x + 5. The lines cross on a grid point.
Graph y = x + 1 and y = -2x + 5. The graph gives about (1.3, 2.3). Set the expressions equal to find the exact point.
Equation: x + 1 = -2x + 5, so x = 4/3 and y = 7/3
Substitution
y = 3x - 4 and 2x + y = 11. The first equation is already solved for y.
Equation: 2x + (3x - 4) = 11, so x = 3 and y = 5
Elimination
4x + 3y = 10 and 2x - 3y = 14. The y-coefficients are opposites.
Equation: Add: 6x = 24, so x = 4 and y = -2
Real-world system
A school play sells 150 tickets for $1,520. Adult tickets cost $12 and student tickets cost $8. How many of each were sold?
Equation: a + s = 150 and 12a + 8s = 1520, so a = 80 and s = 70
Close with the special cases in Diagram 2. When elimination or substitution produces a false statement such as 0 = 7, the lines are parallel and the system has no solution. When it produces a true statement such as 0 = 0, the two equations describe the same line and there are infinitely many solutions. Show how to spot both cases from the graph and from the slopes and intercepts.
Guided Practice15 minutes
Pairs solve three systems, first by graphing to estimate and then exactly: y = -x + 4 and y = 0.5x - 2 (exact grid point (4, 0)); y = 2x - 3 and y = -x + 2 (estimate about (1.7, 0.3), exact (5/3, 1/3)); and 3x - y = 2 and 6x - 2y = 4 (same line, infinitely many solutions). For each system, pairs write one sentence comparing the graph with the algebra. Circulate and watch for sign errors when substituting a negative expression, and for students who solve for x and stop without finding y.
Independent Practice10-15 minutes
Students solve four systems on their own: one by graphing only (the answer is a grid point), one by substitution, one by elimination that requires multiplying one equation first, and one word problem. For the word problem, students define both variables, write the system, solve it, and write a sentence that answers the question with units.
Closure5-10 minutes
Exit ticket: A graphing calculator shows the intersection of y = 2x + 3 and y = -x + 5 as (0.6667, 4.3333). (1) Find the exact solution. (Answer: (2/3, 13/3).) (2) Explain in one sentence why the calculator's answer is an approximation. (3) Without solving, how many solutions does y = 4x - 2 and y = 4x + 5 have? (None: the lines are parallel.)
Differentiation Strategies
For Struggling Students
Start with systems whose solutions are grid points so the graph and the algebra give the same answer, then move to fractional solutions
Provide a checklist: solve for one variable, substitute or eliminate, find the other variable, check in both equations
Let students use a graphing app to see the intersection before solving algebraically, so they know what answer to expect
For Advanced Students
Ask for a system whose solution is (-3/2, 5/4), and ask how a graph would make that solution hard to read exactly
Give a system with decimal coefficients, such as 0.4x + 1.5y = 3.1 and 1.2x - 0.5y = 1.3, and ask students to decide when rounding in a middle step changes the final answer
Ask students to find the value of k for which kx + 2y = 6 and 3x + y = 4 has no solution (k = 6), and explain it with slopes
Assessment Guidance
What to Look For
The standard asks for both exact and approximate solutions, so check both. For graphs, look for accurate lines and a reasonable estimate stated as an approximation. For algebra, look for exact answers (fractions, not rounded decimals) checked in both original equations. For special cases, students should state the conclusion ("no solution" or "infinitely many solutions") and not stop at 0 = 7 or 0 = 0. In context problems, the answer should be a sentence with units.
02
Classroom Activities
3 Activities
1
Estimate, Then Nail It
20 minPairs
Partners graph systems by hand, record an estimate of the solution, and then solve exactly. They score their estimates by how close they were, which builds a sense of how accurate graphing can be.
Systems
y = 0.5x + 2 and y = -x + 6 (exact (8/3, 10/3), about (2.67, 3.33))
y = 3x - 2 and y = -x + 3 (exact (5/4, 7/4), about (1.25, 1.75))
x + 2y = 6 and 3x - y = 4 (exact (2, 2), a grid point)
y = -2x + 1 and y = x - 3 (exact (4/3, -5/3), about (1.33, -1.67))
Procedure
Partner A graphs and estimates; Partner B solves exactly without looking at the graph; they swap roles for the next system
For each system, compute how far off the estimate was in x and in y
Discuss: what made some estimates better than others (scale, steepness of the lines, neat graphing)?
Technology Variation
Repeat one system in a graphing app. Students compare the app's decimal intersection with their exact fraction and explain why the app shows a rounded decimal.
2
Method Choice Sort
20 minGroups of 3-4
Groups sort 9 system cards into three columns: best solved by substitution, best solved by elimination, and best solved by graphing. Then each group solves one card from each column and defends its sorting.
2x - y = 3 and 4x - 2y = 6 (any method: infinitely many solutions)
Discussion Questions
Which features of the equations made you choose each method?
Did any card belong in more than one column?
When would you use graphing even if you need an exact answer?
3
Plan Comparison
25 minGroups of 3-4
Groups compare two pricing plans, write a system, find the break-even point exactly and on a graph, and write a recommendation. This connects the intersection point to a decision.
Scenario
Bike Shop A charges $15 plus $4 per hour. Bike Shop B charges $5 plus $6 per hour. Let h be the number of hours and C the cost in dollars.
Tasks
Write the system C = 15 + 4h and C = 5 + 6h
Graph both lines for 0 ≤ h ≤ 8 and estimate the intersection
Solve exactly: 15 + 4h = 5 + 6h gives h = 5 and C = 35
Write a recommendation: Shop B is cheaper for fewer than 5 hours, both cost $35 at 5 hours, and Shop A is cheaper for more than 5 hours
Modification for Distance Learning
Groups build the graph in a shared graphing app and post their recommendation with a screenshot in the class discussion board.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: An Approximate Solution from a Graph and the Exact Solution
The lines y = x + 1 and y = -2x + 5, drawn to scale on a 1-unit grid. The intersection is not on a grid point, so the graph gives only an estimate. Solving algebraically gives the exact point (4/3, 7/3).
Diagram 2: One Solution, No Solution, Infinitely Many Solutions
Drawn to scale. Two lines with different slopes cross once. Parallel lines never meet, and elimination gives a false statement such as 0 = 3. Two equations for the same line (the dashed line lies on top of the solid one) give a true statement such as 0 = 0.
04
Homework Assignment
~30 min
HSA.REI.C.6 Homework: Solving Systems of Linear Equations
Directions: Show all work. Use graph paper for Part 1. Give exact answers as fractions where needed, and check every solution in both original equations. For Problem 6, define your variables and answer in a complete sentence.
Part 1: Graphing and Approximating (Problems 1-2)
Graph y = -x + 5 and y = 0.5x - 1 on the same grid. Write the solution and check it in both equations.
Graph y = 2x + 1 and y = -x + 5. Estimate the solution from your graph to the nearest tenth. Then solve the system exactly and compare the exact answer with your estimate.
Part 2: Exact Algebraic Methods (Problems 3-4)
Solve by substitution: x = 2y - 3 and 4x - 3y = 8.
Solve by elimination: 3x + 4y = 1 and 5x - 2y = 19.
Part 3: Special Cases and Context (Problems 5-6)
Without graphing, decide whether each system has one solution, no solution, or infinitely many solutions. Show the algebra that supports your answer. (a) 3x + 2y = 4 and 9x + 6y = 12 (b) y = 3x + 1 and 6x - 2y = 5
Two truck rental companies charge for a one-day move. Company A charges $40 plus $0.50 per mile. Company B charges $25 plus $0.80 per mile. Write a system, find the number of miles at which the two companies cost the same, and say which company is cheaper for an 80-mile move.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Graphing
Accurate lines, reasonable estimate stated as approximate
Minor graphing error or estimate not compared
Graph missing or inaccurate
Exact Solutions
Correct exact values, including fractions
Correct method, arithmetic error
Incorrect or missing
Special Cases
Correct classification with supporting algebra
Correct classification, no support
Incorrect
Context and Checking
Variables defined, answer in a sentence with units, solutions checked
Answer correct but not interpreted or not checked
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
What is the solution of the system y = x + 3 and y = -x + 7?
Answer: A
Set the expressions equal: x + 3 = -x + 7, so 2x = 4 and x = 2. Then y = 2 + 3 = 5. Check: -2 + 7 = 5. Choice B switches the coordinates: (5, 2) gives 5 + 3 = 8, not 2.
Question 2 of 20 · Multiple Choice
Solve by substitution: y = 2x and x + y = 12.
Answer: B
Substitute 2x for y: x + 2x = 12, so x = 4 and y = 8. Choice C switches x and y, and (8, 4) does not satisfy y = 2x. Choice D satisfies y = 2x but gives x + y = 9.
Question 3 of 20 · Multiple Choice
Solve by elimination: 3x + 2y = 7 and 3x - 2y = 5.
Answer: C
Add the equations: 6x = 12, so x = 2. Then 6 + 2y = 7 gives y = 1/2. Check: 6 - 1 = 5. Choice A solves 2y = 7 - 6 as y = 1 by forgetting to divide by 2. Choice D has the wrong sign on y.
Question 4 of 20 · Multiple Choice
What is the exact solution of y = 1.5x - 1 and y = -x + 3?
Answer: D
1.5x - 1 = -x + 3, so 2.5x = 4 and x = 1.6. Then y = -1.6 + 3 = 1.4. Choice A is a reasonable graphical estimate, but it is not exact: 1.5(1.5) - 1 = 1.25, not 1.5. Choice B switches the coordinates.
Question 5 of 20 · Multiple Choice
How many solutions does the system y = -2x + 3 and y = -2x - 4 have?
Answer: B
The lines have the same slope, -2, and different y-intercepts, so they are parallel. Setting -2x + 3 = -2x - 4 gives 3 = -4, which is false. Choice D would require the same slope and the same intercept. Two distinct lines can never cross exactly twice, so choice C is impossible.
Question 6 of 20 · Multiple Choice
How many solutions does the system 2x + 6y = 8 and x + 3y = 4 have?
Answer: C
The first equation is 2 times the second, so both describe the same line. Subtracting 2 times the second from the first gives 0 = 0, which is always true, so every point on the line is a solution. Choice B is the error of reading 0 = 0 as "no solution."
Question 7 of 20 · Multiple Choice
Which system has the solution (-1, 3)?
Answer: A
Substitute (-1, 3): -1 + 3 = 2 and 2(-1) - 3 = -5, so both equations of A are true. In B, 2(-1) + 3 = 1, not -5. In C, -1 - 3 = -4, not 2. In D, -1 + 3 = 2, not 4. A point is a solution only if it satisfies both equations.
Question 8 of 20 · Multiple Choice
A student graphs a system and reads the intersection as (2.5, 3). Substituting shows the point satisfies one equation but not the other. What should the student do?
Answer: D
A graph gives an approximate solution. When the check fails, the estimate is close but not exact, so the student should use substitution or elimination. Choice C is wrong because the lines clearly cross; the reading is only imprecise.
Question 9 of 20 · Multiple Choice
A jar has 20 coins, all nickels and dimes, worth $1.40 in total. A student writes the system n + d = 20 and 5n + 10d = 1.40, where n is the number of nickels and d is the number of dimes, and solves it. What is the result, and what should the student conclude?
Answer: B
Substitute n = 20 - d: 5(20 - d) + 10d = 1.40, so 100 + 5d = 1.40 and d = -19.72, which gives n = 39.72. A negative number of dimes is impossible, so the model is wrong: the left side is in cents and the right side is in dollars. The correct second equation is 5n + 10d = 140. Choice A does not satisfy the student's second equation: 5(14) + 10(6) = 130, not 1.40. Choice C is wrong because the slopes, -1 and -1/2, are different, so the lines cross once. Choice D rounds an impossible answer instead of fixing the model.
Question 10 of 20 · Multiple Choice
Solve the corrected coin system from the previous question: n + d = 20 and 5n + 10d = 140.
Answer: C
Replace the second equation by the second minus 5 times the first: 5d = 40, so d = 8 and n = 12. Check: 12(5) + 8(10) = 60 + 80 = 140 cents. Choice A switches the two answers: 8(5) + 12(10) = 160 cents.
Question 11 of 20 · Multiple Choice
Solve 4x + 5y = 7 and 2x - 5y = 11.
Answer: D
Add the equations to eliminate y: 6x = 18, so x = 3. Then 12 + 5y = 7 gives y = -1. Check: 6 + 5 = 11. Choice A has the wrong sign on y: 12 + 5 = 17, not 7. Choice C forgets to divide by 6 and uses x = 18. Choice B switches the coordinates.
Question 12 of 20 · Multiple Choice
Solve x = 3y - 1 and 2x + y = 12.
Answer: A
Substitute: 2(3y - 1) + y = 12, so 7y - 2 = 12, y = 2, and x = 3(2) - 1 = 5. Check: 10 + 2 = 12. Choice B switches the coordinates. Choice C satisfies x = 3y - 1 but not the second equation.
Question 13 of 20 · Multiple Choice
A graphing calculator shows the intersection of y = 4x - 1 and y = -2x + 3 as (0.6667, 1.6667). What is the exact solution?
Answer: B
4x - 1 = -2x + 3 gives 6x = 4, so x = 2/3 and y = 4(2/3) - 1 = 5/3. Check: -2(2/3) + 3 = 5/3. The calculator's decimals are rounded, so choice A is an approximation. Choice C comes from solving 6x = 4 as x = 6/4 = 3/2. Choice D adds 1 instead of subtracting it: 4(2/3) + 1 = 11/3.
Question 14 of 20 · Multiple Choice
For what value of k does the system y = kx + 2 and y = 3x - 1 have no solution?
Answer: C
The system has no solution when the lines are parallel: same slope, different intercepts. The slopes match when k = 3, and the intercepts 2 and -1 are different. Choice A gives lines with opposite slopes, and choice D gives a reciprocal slope; both cross the line y = 3x - 1 once.
Question 15 of 20 · Short Answer
Solve by substitution: y = -2x + 9 and 3x - y = 1.
3x - (-2x + 9) = 1, so 5x - 9 = 1 and x = 2. Then y = -4 + 9 = 5. Solution: (2, 5). Check: 6 - 5 = 1. Watch the parentheses: subtracting -2x + 9 changes both signs.
Question 16 of 20 · Short Answer
Solve by elimination: 2x + 5y = 16 and 3x - 2y = 5.
Multiply the first equation by 2 and the second by 5: 4x + 10y = 32 and 15x - 10y = 25. Add: 19x = 57, so x = 3. Then 6 + 5y = 16, so y = 2. Solution: (3, 2). Check: 9 - 4 = 5.
Question 17 of 20 · Short Answer
Graph y = -0.5x + 4 and y = x - 1. Estimate the solution, then find it exactly.
The lines cross between x = 3 and x = 4, near (3.3, 2.3). Exactly: -0.5x + 4 = x - 1, so 1.5x = 5 and x = 10/3. Then y = 10/3 - 1 = 7/3. The exact solution is (10/3, 7/3), about (3.33, 2.33).
Question 18 of 20 · Short Answer
While solving a system, a student gets 5 = -2. What does this mean, and what does the graph look like?
5 = -2 is false for every x and y, so the system has no solution. The two lines are parallel: they have the same slope and different y-intercepts, so they never meet. The student should write "no solution," not "x = 5" or "x = -2."
Question 19 of 20 · Short Answer
Plan A for a gym costs $60 to join plus $20 per month. Plan B has no joining fee and costs $35 per month. After how many months do the plans cost the same, and what is that cost?
Let m be the number of months and C the cost. C = 60 + 20m and C = 35m. Then 60 + 20m = 35m, so 15m = 60 and m = 4. Both plans cost $140 after 4 months. After that, Plan A is cheaper.
Question 20 of 20 · Short Answer
A theater sold 200 tickets for $1,640. Adult tickets cost $10 and child tickets cost $6. How many of each were sold?
Let a and c be the numbers of adult and child tickets. a + c = 200 and 10a + 6c = 1640. Subtract 6 times the first equation: 4a = 440, so a = 110 and c = 90. 110 adult tickets and 90 child tickets. Check: 1100 + 540 = 1640.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does "exactly and approximately" mean in HSA.REI.C.6?
An exact solution comes from algebra, usually substitution or elimination, and is written with exact values such as (4/3, 7/3). An approximate solution comes from a graph, a table, or a calculator, and is written as an estimate such as (1.3, 2.3). The standard expects students to do both and to know which one they have.
When should students use substitution and when elimination?
Substitution is efficient when one equation is already solved for a variable, such as y = 3x - 4 or x = 2y + 1. Elimination is efficient when both equations are in standard form, especially when one variable has matching or opposite coefficients. Both always give the same answer, so the choice is about efficiency and fewer arithmetic errors.
Why graph a system if algebra gives the exact answer?
A graph shows at a glance whether a system has one solution, none, or infinitely many, and where the solution is roughly located. It is also a check: if algebra gives (4/3, 7/3), the graph should show an intersection near (1.3, 2.3). In real-world problems, the graph shows which option is better on each side of the intersection.
What does it mean when a system has no solution or infinitely many solutions?
If the algebra ends in a false statement such as 0 = 7, the lines are parallel and the system has no solution. If it ends in a true statement such as 0 = 0, the equations describe the same line and every point on that line is a solution, so there are infinitely many solutions. Students can also tell from slope-intercept form: same slope and different intercepts means no solution; same slope and same intercept means infinitely many.
What are the common mistakes when solving systems?
Finding x and forgetting to find y
Sign errors when substituting an expression with a negative term, such as 3x - (-2x + 9)
Multiplying only one side, or only one term, of an equation before eliminating
Checking the answer in only one equation
Reporting a graphical estimate as if it were exact
How accurate does a graphical estimate need to be?
On a hand-drawn graph with a 1-unit grid, an estimate to about the nearest half unit is realistic, and to the nearest tenth with careful graphing. Students should label graphical answers as approximate and, when the problem asks for an exact solution, confirm with algebra. A graphing calculator gives more decimal places, but its answer is still rounded unless the solution happens to be a terminating decimal.
Does the standard include systems with three variables?
The standard says "focusing on pairs of linear equations in two variables," so the main work is 2-by-2 linear systems. Systems with a linear and a quadratic equation are the next standard, HSA.REI.C.7, and systems of three or more equations are usually treated later with matrices.
Is HSA.REI.C.6 on the SAT?
Yes. Solving systems of two linear equations in two variables, including word problems and questions about how many solutions a system has, is part of the Algebra domain of the digital SAT. The digital SAT includes a built-in graphing calculator, so students benefit from knowing both the algebraic and the graphical approach.
How do students check their answer?
Substitute the ordered pair into both original equations. For example, for (4, -2) in 4x + 3y = 10 and 2x - 3y = 14: 16 - 6 = 10 and 8 + 6 = 14. A pair that works in only one equation is a point on one line, not the intersection. In word problems, also check that the numbers make sense: whole numbers of tickets, positive costs, and so on.
How does this standard connect to later topics?
The idea that the solution is where two graphs meet extends to HSA.REI.D.11, where students solve f(x) = g(x) for many kinds of functions by finding intersections. Systems of linear inequalities (HSA.REI.D.12) and linear-quadratic systems (HSA.REI.C.7) build directly on it. The reason elimination works is proved in HSA.REI.C.5.
07
Related Standards
6 standards
These standards connect to HSA.REI.C.6: 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
