HSA.REI.C.7: Solving Linear-Quadratic Systems Algebraically and Graphically
In plain English: HSA.REI.C.7 is the Common Core algebra standard that asks students to solve a simple system of one linear and one quadratic equation in two variables, both algebraically and graphically. Substituting the linear equation into the quadratic gives a quadratic in one variable, so the system has zero, one or two solutions, seen as intersection points. It is taught in Algebra I or Algebra II, depending on the course.
Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. For example, find the points of intersection between the line y = -3x and the circle x2 + y2 = 3.
Common Core State Standards for Mathematics · Domain: Reasoning with Equations and Inequalities (REI) · Cluster: Solve systems of equations Also written as HSA-REI.C.7 or A-REI.7 · Official standard
In this lesson, students solve systems made of one linear equation and one quadratic equation in two variables, such as a line and a parabola, or a line and a circle centered at the origin. They solve each system two ways: algebraically, by substituting the linear expression into the quadratic equation and solving the resulting quadratic in one variable, and graphically, by sketching both graphs on one grid and reading the intersection points.
The key idea is that a solution is an ordered pair that makes both equations true, and that substitution turns the system into a single quadratic, so there can be 0, 1 or 2 real solutions. Students connect the number of real roots of that quadratic to what they see on the graph: two crossing points, a tangent line, or no intersection.
Learning Objectives
By the end of this lesson, students will be able to:
Solve a linear-quadratic system algebraically by substitution and report each solution as an ordered pair
Solve the same system graphically by graphing a line and a parabola (or circle) and identifying the intersection points
Explain why a line and a parabola or circle meet in 0, 1 or 2 points, and connect this to the real roots of the substituted quadratic
Check each solution in both original equations and interpret solutions in a context
Prior Knowledge Required
Students should already be comfortable with:
Solving systems of two linear equations by substitution and by graphing 8.EE.C.8
Solving quadratic equations by factoring and with the quadratic formula HSA.REI.B.4
Graphing linear and quadratic functions, including intercepts and the vertex HSF.IF.C.7
Knowing that a point is on a graph exactly when its coordinates satisfy the equation HSA.REI.D.10
Project the prompt below. Students answer on mini whiteboards or scratch paper, then compare with a partner.
Warm-Up Prompt
"Sketch the parabola y = x² and then draw one straight line that crosses it twice, one line that touches it once, and one line that misses it completely. Could any line cross the parabola three times?"
Collect a few sketches. Students usually find all three pictures quickly. Push on the last question: nobody can draw a non-vertical line that crosses y = x² three times, and the lesson will explain why with algebra. (A vertical line such as x = 1 crosses the parabola exactly once; note it and set it aside.)
Direct Instruction20 minutes
Model the substitution method, then confirm each answer on a graph:
Isolate a variable in the linear equation: write it as y = mx + b (or x in terms of y).
Substitute that expression into the quadratic equation, so only one variable is left.
Rearrange to standard form ax² + bx + c = 0 and solve by factoring, the quadratic formula or square roots.
Back-substitute each x-value into the linear equation to get the matching y-value, and write ordered pairs.
Check graphically: graph both equations and confirm the intersection points match.
Work these examples on the board. For each one, graph both equations on the same grid after solving.
Line and parabola, two solutions
Solve y = x² - 4x + 3 and y = x - 1. Substituting gives x² - 4x + 3 = x - 1, so x² - 5x + 4 = 0 and (x - 1)(x - 4) = 0.
Equation: Solutions: (1, 0) and (4, 3)
Line and circle (the standard's own example)
Find the points of intersection of the line y = -3x and the circle x² + y² = 3. Substituting gives x² + (-3x)² = 3, so 10x² = 3, x² = 3/10 and x = ±√30/10.
Equation: Solutions: (√30/10, -3√30/10) and (-√30/10, 3√30/10), about (0.55, -1.64) and (-0.55, 1.64)
Tangent line, one solution
Solve y = x² and y = 2x - 1. Substituting gives x² - 2x + 1 = 0, so (x - 1)² = 0 and x = 1 is a double root.
Equation: One solution: (1, 1)
No real solution
Solve y = x² + 2 and y = x. Substituting gives x² - x + 2 = 0. The discriminant is (-1)² - 4(1)(2) = -7, which is negative.
Equation: No real solution: the graphs do not intersect
In context
A sprinkler at the foot of a hill sprays water along y = -0.25x² + 2x, and the hillside rises along y = 0.5x (x and y in feet). Substituting gives -0.25x² + 1.5x = 0, so x(-0.25x + 1.5) = 0.
Equation: x = 0 (the sprinkler head) or x = 6: the water lands on the hill at (6, 3)
Guided Practice15-20 minutes
Pairs solve three systems: one partner solves algebraically while the other graphs on graph paper or a graphing app, then they swap roles for the next system. Use y = x² - 2 with y = x (solutions (2, 2) and (-1, -1)), x² + y² = 10 with y = x + 2 (solutions (1, 3) and (-3, -1)), and y = x² + 4 with y = 4x (one solution, (2, 8)). Circulate and listen for two common errors: stopping after finding the x-values, and pairing an x-value with the wrong y-value. Debrief by asking each pair whether their algebra and their graph agreed, and what they did when they did not.
Independent Practice10-15 minutes
Students solve four systems on their own: two with a parabola, one with a circle centered at the origin, and one that has no real solution. For each, they write the substituted quadratic, the solutions as ordered pairs, and a one-line graphical check ("the line crosses the parabola twice"). Early finishers write a line that is tangent to y = x² at the point (2, 4) and prove it is tangent by showing the substituted quadratic has a double root.
Closure5-10 minutes
Exit ticket: "After substituting, Jordan gets x² - 6x + 9 = 0. (a) How many solutions does the system have? (b) Describe what the graphs look like. (c) Why can a line and a parabola never meet in exactly three points?" Look for: one solution, a tangent line, and the reasoning that substitution always produces a quadratic, which has at most two real roots.
Differentiation Strategies
For Struggling Students
Start with systems where the substituted quadratic factors easily, and give a template with boxes for "substitute," "standard form," "solve," and "find y"
Provide pre-drawn parabolas on graph paper so students only need to draw the line and read off intersections
Have students check each ordered pair in both original equations and circle the pair only after both checks work
For Advanced Students
Find all values of k for which y = x² and y = x + k have two, one, or no solutions, and explain the result with the discriminant
Solve systems where substitution for x is easier than for y, such as x = y² - 1 with x + y = 1
Write their own system whose solutions are two given points, such as (-1, 3) and (2, 6)
Assessment Guidance
What to Look For
Check that students report solutions as ordered pairs, not as x-values alone, and that each y-value comes from the correct x-value. Ask students to explain what their graph shows when the substituted quadratic has a negative discriminant. A student who can connect "no real roots" to "the graphs do not meet" has understood why both methods give the same answer.
02
Classroom Activities
3 Activities
1
Two Methods, One Answer
20 minPairs
Each pair gets four system cards. One partner solves by substitution while the other graphs by hand on the same grid window. They compare answers, then switch roles for the next card.
System Cards
y = x² - 1 and y = x + 1 (solutions (-1, 0) and (2, 3))
y = -x² + 6 and y = 2 (solutions (-2, 2) and (2, 2))
x² + y² = 5 and y = 2x (solutions (1, 2) and (-1, -2))
y = x² + 3 and y = 2x + 2 (one solution, (1, 4))
Procedure
Before solving, each partner predicts the number of solutions from a quick sketch
The algebra partner writes the substituted quadratic in standard form and solves it; the graphing partner marks intersection points
If the answers disagree, the pair finds the error together and writes one sentence about what went wrong
Modification for Distance Learning
The graphing partner uses a free graphing app and shares a screenshot in the breakout room; the algebra partner types the substitution steps in a shared document.
2
Tangent Hunt
20 minGroups of 3-4
Groups investigate the family of lines y = 2x + k with the parabola y = x² + 2 and find the value of k where the line switches from crossing twice to missing the parabola.
Procedure
Each group member tests one value of k (k = 3, 2, 1, 0) by solving the system algebraically
Record the substituted quadratic x² - 2x + (2 - k) = 0, its discriminant 4 - 4(2 - k) = 4k - 4, and the number of solutions
Use the pattern to predict the tangent line (k = 1, touching at (1, 3)), then confirm on a graph
Discussion Questions
How does the sign of the discriminant match what you see on the graph?
Why does a double root correspond to a line that touches the parabola without crossing it?
Is there a value of k that gives three solutions? Explain using algebra.
3
Road and Coverage Zone
20-25 minIndividual then share
A cell tower at the origin covers every point within 5 miles, so the edge of coverage is the circle x² + y² = 25. Students find where straight roads enter and leave the coverage zone.
Roads to Test
Road A: y = -x + 7 (enters and leaves at (3, 4) and (4, 3))
Road B: y = 5 (touches the edge only at (0, 5))
Road C: y = x + 8 (substituting gives 2x² + 16x + 39 = 0 with discriminant -56, so the road never enters the zone)
Requirements
Solve each system algebraically and state the points as ordered pairs
Sketch the circle and all three roads on one grid and label the intersection points
Write one sentence per road interpreting the result for a driver
Extension Variation
Students find the length of the stretch of Road A inside the coverage zone, using the distance formula between (3, 4) and (4, 3). The length is √2, about 1.4 miles.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Solving y = x² - 4x + 3 and y = x - 1 Graphically
Drawn to scale, 1 unit per grid square. The line crosses the parabola at (1, 0) and (4, 3), which are exactly the solutions found by substitution: x² - 5x + 4 = 0 gives x = 1 and x = 4.
Diagram 2: A Line and a Parabola Meet in 0, 1 or 2 Points
The same parabola y = x² with three different lines, drawn to scale. The number of real roots of the substituted quadratic matches the number of intersection points: two roots, one double root, or no real roots.
04
Homework Assignment
~30 min
HSA.REI.C.7 Homework: Lines Meeting Parabolas and Circles
Directions: Solve each system algebraically and write every solution as an ordered pair. Then check your answer with a graph (by hand or with technology) and state how many times the graphs intersect. Show all work.
Part 1: Solve Algebraically (Problems 1-3)
Solve the system y = x² + 2x - 3 and y = 2x + 1.
Solve the system x² + y² = 10 and y = 3x.
Solve the system y = x² - 6x + 11 and y = 2x - 5. How many solutions are there, and what does this tell you about the line?
Part 2: Graph and Apply (Problems 4-6)
Graph y = -x² + 4 and y = x + 2 on the same grid for -3 ≤ x ≤ 3. Estimate the intersection points from your graph, then verify them algebraically.
A bakery sells x dozen specialty cookies per week. Its weekly revenue is R = -x² + 80x dollars and its weekly cost is C = 20x + 800 dollars. Solve the system to find the break-even points, and explain what they mean for the bakery.
A radio transmitter's signal covers the region inside the circle x² + y² = 169 (units in miles). A straight highway follows y = x + 7. Find the points where the highway enters and leaves the coverage zone.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Substitution
Correct substituted quadratic in standard form
Correct setup with a sign or combining error
No substitution or incorrect setup
Solving
All real roots found correctly
One root missing or one arithmetic error
Roots not found
Ordered Pairs
Every solution given as a correct (x, y) pair
x-values only, or one y-value wrong
No ordered pairs
Graphical Check and Interpretation
Graph matches solutions; context explained where asked
Graph or interpretation incomplete
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 are the solutions of the system y = x² and y = x + 6?
Answer: A
Substituting gives x² = x + 6, so x² - x - 6 = 0 and (x - 3)(x + 2) = 0. Then x = 3 gives y = 9 and x = -2 gives y = 4. Choice B comes from factoring as (x + 3)(x - 2), which gives the wrong signs. Choice D uses the wrong y-value for x = -2: (-2)² = 4 and -2 + 6 = 4, not -4.
Question 2 of 20 · Multiple Choice
After substituting y = 2x + 3 into y = x² + 4x - 5, which equation in standard form do you get?
Answer: B
Set 2x + 3 = x² + 4x - 5 and move everything to one side: x² + 4x - 2x - 5 - 3 = 0, so x² + 2x - 8 = 0. Choice A adds 2x + 3 to the quadratic instead of subtracting it. Choice C subtracts 2x but adds 3 instead of subtracting it.
Question 3 of 20 · Multiple Choice
How many real solutions does the system y = x² + 1 and y = 2x have?
Answer: C
Substituting gives x² - 2x + 1 = 0, which is (x - 1)² = 0. The only root is x = 1, so the only solution is (1, 2) and the line is tangent to the parabola. Choice B assumes every quadratic has two roots; a double root gives only one point.
Question 4 of 20 · Multiple Choice
Solve the system x² + y² = 20 and y = 2x.
Answer: A
Substituting gives x² + (2x)² = 20, so 5x² = 20 and x = ±2. Using y = 2x, the solutions are (2, 4) and (-2, -4). Choice C forgets the negative square root. Choice D lies on the circle but not on the line, because y = 2x is not true for (4, 2).
Question 5 of 20 · Multiple Choice
The graphs of y = x² + 1 and y = 2x + 4 cross at (-1, 2) and (3, 10). Which statement is true?
Answer: D
A solution of a system is any ordered pair that makes both equations true. For (-1, 2): (-1)² + 1 = 2 and 2(-1) + 4 = 2. For (3, 10): 9 + 1 = 10 and 6 + 4 = 10. Choice A mixes coordinates from two different points. Choice B wrongly keeps only one intersection.
Question 6 of 20 · Multiple Choice
For which value of k do y = x² and y = x + k have exactly one solution?
Answer: B
Substituting gives x² - x - k = 0. There is exactly one solution when the discriminant is zero: (-1)² - 4(1)(-k) = 1 + 4k = 0, so k = -1/4. Then x = 1/2 and the tangent point is (1/2, 1/4). Choice A gives x² - x = 0, which has two roots (0 and 1). Choice C makes the discriminant 2, so there are two solutions.
Question 7 of 20 · Multiple Choice
Solve the system y = -x² + 5 and y = x - 1.
Answer: D
Substituting gives -x² + 5 = x - 1, so x² + x - 6 = 0 and (x + 3)(x - 2) = 0. Then x = 2 gives y = 1 and x = -3 gives y = -4. Choice A comes from factoring as (x - 3)(x + 2), which flips the signs of the roots. Choice C uses the wrong y-value for x = -3.
Question 8 of 20 · Multiple Choice
A ball's height is h = -16t² + 32t + 6 and a drone's height is h = 16t + 6, both in feet after t seconds. After t = 0, when are they at the same height?
Answer: C
Set -16t² + 32t + 6 = 16t + 6, so -16t² + 16t = 0 and -16t(t - 1) = 0. The roots are t = 0 (both start at 6 feet) and t = 1. At t = 1, h = 16(1) + 6 = 22 feet. Choice D forgets to add the starting height of 6 feet. Choice A: at t = 2 the ball is back at 6 feet, not at the drone's height.
Question 9 of 20 · Multiple Choice
After substituting, a student gets x² + 8x + 16 = 0. What does this tell you about the graphs of the two equations?
Answer: B
x² + 8x + 16 = (x + 4)², so x = -4 is a double root. One real root means one intersection point, so the line is tangent to the parabola there. Choice A would need two different real roots. Choice C would need a negative discriminant, but here the discriminant is 64 - 64 = 0.
Question 10 of 20 · Multiple Choice
Which system has no real solutions?
Answer: A
For A, substituting gives x² - x + 3 = 0 with discriminant 1 - 12 = -11, which is negative, so the line and parabola never meet. B has two solutions (x = ±2). C gives x² - 2x + 1 = 0, a double root, so the line is tangent (one solution). D gives x² = 8, which has two real roots.
Question 11 of 20 · Multiple Choice
A graph of y = x² - 3 and y = x shows intersections near x ≈ 2.3 and x ≈ -1.3. Which expression gives the exact x-coordinates?
Answer: C
Substituting gives x² - x - 3 = 0. The quadratic formula gives x = (1 ± √(1 + 12))/2 = (1 ± √13)/2, which is about 2.303 and -1.303, matching the graph. Choice A uses +b instead of -b. Choice B forgets to divide by 2a. Choice D divides by 4 instead of 2a = 2.
Question 12 of 20 · Multiple Choice
A company's revenue is R = -x² + 50x and its cost is C = 10x + 300, where x is the number of units sold (in hundreds) and R and C are in thousands of dollars. At what values of x does the company break even?
Answer: B
Break-even means R = C: -x² + 50x = 10x + 300, so x² - 40x + 300 = 0 and (x - 10)(x - 30) = 0. The company breaks even at 1,000 and 3,000 units, and makes a profit in between. Choice D has the wrong signs from factoring; negative sales also make no sense in context. Choice A misses the second break-even point.
Question 13 of 20 · Multiple Choice
Solve the system x² + y² = 16 and y = -4.
Answer: D
Substituting y = -4 gives x² + 16 = 16, so x² = 0 and x = 0. The only solution is (0, -4): the horizontal line touches the bottom of the circle. Choice B includes (0, 4), which is on the circle but not on the line y = -4. Choice A makes x² = 16 by forgetting to subtract 16.
Question 14 of 20 · Multiple Choice
A student solves y = x² - 4 and y = 3x, finds x = 4 and x = -1, and writes the solutions (4, 12) and (-1, 5). Which statement is true?
Answer: C
The x-values are right: x² - 3x - 4 = 0 factors as (x - 4)(x + 1) = 0. For x = -1, y = 3(-1) = -3, and the parabola agrees: (-1)² - 4 = -3. The student likely computed (-1)² + 4 = 5. Choice A misses the error; (-1, 5) is on neither graph.
Question 15 of 20 · Short Answer
Solve the system y = x² - 2x and y = x + 10 algebraically. Give each solution as an ordered pair.
Solutions: (5, 15) and (-2, 8) Substitute: x² - 2x = x + 10, so x² - 3x - 10 = 0 and (x - 5)(x + 2) = 0. For x = 5, y = 15; for x = -2, y = 8. Check in the parabola: 25 - 10 = 15 and 4 + 4 = 8.
Question 16 of 20 · Short Answer
Solve the system x² + y² = 13 and y = x + 1.
Solutions: (2, 3) and (-3, -2) Substitute: x² + (x + 1)² = 13, so 2x² + 2x + 1 = 13, 2x² + 2x - 12 = 0 and x² + x - 6 = 0. Then (x + 3)(x - 2) = 0 gives x = 2 or x = -3. Check: 4 + 9 = 13 and 9 + 4 = 13.
Question 17 of 20 · Short Answer
Explain why a non-vertical line and a parabola can meet in 0, 1 or 2 points, but never in 3 points.
Substituting y = mx + k into y = ax² + bx + c gives one quadratic equation in x. A quadratic has at most two real roots, and each real root gives exactly one intersection point. So there are 2 points (positive discriminant), 1 point (discriminant zero, a tangent line) or 0 points (negative discriminant). Three points would need a quadratic with three roots, which is impossible.
Question 18 of 20 · Short Answer
Show that (2, 1) is a solution of the system y = x² - 3 and y = 4x - 7. Is it the only solution? Explain.
Yes, it is the only solution. Check: 2² - 3 = 1 and 4(2) - 7 = 1, so (2, 1) is on both graphs. Substituting gives x² - 3 = 4x - 7, so x² - 4x + 4 = (x - 2)² = 0. The only root is x = 2, so the line is tangent to the parabola at (2, 1).
Question 19 of 20 · Short Answer
Water from a sprinkler follows y = -0.5x² + 3x, and the ground slopes up along y = 0.5x (both in meters, with the sprinkler at the origin). Where does the water land on the slope? What does the other solution mean?
The water lands at (5, 2.5). Substitute: -0.5x² + 3x = 0.5x, so -0.5x² + 2.5x = 0 and -0.5x(x - 5) = 0. The roots are x = 0 and x = 5, and y = 0.5(5) = 2.5. The solution (0, 0) is the sprinkler head, where the water starts, so it is not a landing point.
Question 20 of 20 · Short Answer
Find the positive value of b for which the line y = bx is tangent to the parabola y = x² + 9. Give the point of tangency.
b = 6, tangent at (3, 18) Substituting gives x² - bx + 9 = 0. For exactly one solution the discriminant must be zero: b² - 36 = 0, so b = 6 (positive). Then x² - 6x + 9 = (x - 3)² = 0 gives x = 3 and y = 6(3) = 18.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What counts as a "simple" linear-quadratic system for HSA.REI.C.7?
A system with one linear equation and one quadratic equation in x and y, where substitution leads to a quadratic you can solve with standard methods. Typical examples are a line with a parabola such as y = x² - 4x + 3, or a line with a circle centered at the origin such as x² + y² = 25. The official standard's own example is the line y = -3x with the circle x² + y² = 3.
Why is substitution the usual method instead of elimination?
The linear equation is easy to solve for one variable, and substituting that expression into the quadratic leaves one equation in one variable. Elimination by adding equations rarely removes the squared term cleanly. Substitution also shows why there are at most two solutions: you always end up with a quadratic.
Why do students need to solve both algebraically and graphically?
The standard asks for both. The algebra gives exact answers, including irrational ones such as x = 2 ± √3. The graph shows how many solutions to expect and catches mistakes: if the algebra gives two points but the line clearly misses the parabola, something went wrong. Each method checks the other.
What is a common mistake on these problems?
Stopping after finding the x-values. A solution of a system is an ordered pair, so each x-value needs its matching y-value. Another common mistake is pairing a y-value with the wrong x-value, or computing y with a sign error such as writing (-3)² as -9. Have students check every pair in both original equations.
How do I know how many solutions there will be before I solve?
Look at the discriminant of the substituted quadratic ax² + bx + c = 0. If b² - 4ac is positive there are two solutions, if it is zero there is one (the line is tangent), and if it is negative there is no real solution. A quick sketch gives the same information visually.
What happens with a vertical line, like x = 2?
A vertical line crosses a parabola y = ax² + bx + c exactly once, at the point where x = 2, so substitution gives the single point (2, f(2)) directly. With a circle, a vertical line can still meet it in 0, 1 or 2 points. Vertical lines are a good discussion case but not the focus of this standard.
Should the answers ever be decimals?
Yes. If the substituted quadratic does not factor, use the quadratic formula for exact answers and a calculator for decimal approximations. A graph read by hand usually gives estimates to the nearest tenth or so, which is why the algebraic method is still needed for exact solutions.
Where do these systems show up in real situations?
Break-even problems where revenue is quadratic and cost is linear, the path of a thrown object compared with a slope or a rising object, and a straight road or flight path crossing a circular coverage zone. In each case, students should check which solutions make sense: negative sales or times before the event starts are rejected or explained.
Is this skill tested on the SAT?
Yes. Systems made of a linear and a quadratic equation appear in the Advanced Math domain of the digital SAT, often asking for the number of solutions or for one coordinate of an intersection point. The graphing calculator built into the test can be used to check answers graphically.
What comes after HSA.REI.C.7?
Students extend the idea that intersection points solve systems to any pair of functions in HSA.REI.D.11, where they solve f(x) = g(x) for polynomial, rational, absolute value, exponential and logarithmic functions, often approximately with technology. Work with circles also leads into writing circle equations in geometry (HSG.GPE.A.1).
07
Related Standards
6 standards
These standards connect to HSA.REI.C.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.EE.C.8Prerequisite
Solve systems of two linear equations in two variables, algebraically and by graphing
