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Systems Checkpoint

A free Algebra I lesson from the “Systems of Linear Equations and Inequalities” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples all five lessons in Systems of Linear Equations and Inequalities: systems and their graphs, substitution, elimination, word problems, and systems of inequalities. Questions are mixed, so start by looking at the structure: an isolated variable points to substitution, equal or opposite terms point to elimination, and a graph shows where two conditions hold together. If the variables disappear, a false statement means no solution and a true statement means infinitely many. In context, define your variables, check the answer in both conditions, and ask whether it makes sense: counts must be whole numbers, and times and lengths cannot be negative. For inequalities, decide which boundaries are included and which points lie in the overlap. The 25 review questions and the 25 checkpoint questions were written separately from each other and from the lesson practice and quizzes. Two review questions show a graph; checkpoint questions describe any graph in words, and many are multiple choice so that systems, classifications, and graph descriptions are graded reliably. A score reports performance on this sample of proposed outcome A1.6; it is not an external certification. The written tasks below, including drawing graphs, are reviewed with a rubric and do not change the score.

What you'll learn

Why it matters: Systems appear whenever two conditions must hold at once: a budget and a head count for a trip, two price plans that cross at a break-even point, or a mixture that must reach a target amount and strength.

Worked example

Problem. Solve the system 2x + 3y = 12 and 5x − 3y = 9, then check the solution in both equations.

  1. The y-terms 3y and −3y are opposites, so adding the equations eliminates y: 7x = 21, so x = 3.
  2. Substituting x = 3 into 2x + 3y = 12 gives 6 + 3y = 12, so 3y = 6 and y = 2.
  3. Check the first equation: 2(3) + 3(2) = 12. Check the second equation: 5(3) − 3(2) = 9.
  4. Both equations are true, so the lines cross at exactly one point, (3, 2).

Answer: (3, 2)

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. The graph shows line A, y = 2x − 1, and line B, y = −x + 5, crossing at one point. Which ordered pair is the solution of the system?

Choices: (2, 3) · (3, 2) · (0, −1) · (5, 0)

Show solution
  1. The solution of a system is the point that lies on both lines, which is where the graphs cross.
  2. Reading the graph, the lines cross at x = 2 and y = 3.
  3. Check in both equations: 2(2) − 1 = 3 and −2 + 5 = 3. The point (0, −1) is only line A's y-intercept, and (5, 0) is only line B's x-intercept.

Answer: (2, 3)

2. Without finding a solution, decide how many solutions the system y = 3x − 2 and 6x − 2y = 4 has.

Choices: Infinitely many, because both equations describe the same line · None, because the lines are parallel · Exactly one, because the equations are written in different forms · Exactly one, the point (0, −2)

Show solution
  1. Rewrite the second equation in slope-intercept form: −2y = −6x + 4, so y = 3x − 2.
  2. Both equations are the same line, so every point on y = 3x − 2 satisfies both.
  3. The point (0, −2) is one solution, but it is not the only one; parallel lines would need equal slopes with different y-intercepts.

Answer: Infinitely many, because both equations describe the same line

3. Which system has no solution?

Choices: y = −2x + 3 and 4x + 2y = 9 · y = −2x + 3 and 4x + 2y = 6 · y = −2x + 3 and y = 2x + 3 · y = −2x + 3 and x = 3

Show solution
  1. Rewrite 4x + 2y = 9 as y = −2x + 4.5: it has the same slope as y = −2x + 3 but a different y-intercept, so the lines are parallel and never meet.
  2. 4x + 2y = 6 is y = −2x + 3 itself, so that system has infinitely many solutions.
  3. y = 2x + 3 has a different slope, and x = 3 is a vertical line; each crosses y = −2x + 3 exactly once.

Answer: y = −2x + 3 and 4x + 2y = 9

4. A student says (4, 1) is a solution of the system x − 2y = 2 and 3x + y = 14 because it works in the first equation. Which statement is correct?

Choices: (4, 1) is not a solution: it satisfies x − 2y = 2, but 3(4) + 1 = 13, not 14 · (4, 1) is a solution, because one true equation is enough · (4, 1) is a solution, because it satisfies both equations · (4, 1) is not a solution, because it satisfies neither equation

Show solution
  1. A solution of a system must make every equation true at the same time.
  2. First equation: 4 − 2(1) = 2, which is true. Second equation: 3(4) + 1 = 13, which is not 14.
  3. So (4, 1) lies on the first line only. The actual solution is (30/7, 8/7), where both equations hold.

Answer: (4, 1) is not a solution: it satisfies x − 2y = 2, but 3(4) + 1 = 13, not 14

5. The equations d = 3 + 2t and d = 6t − 5 give the distances in kilometers of two cyclists from a trail marker t hours after noon. The models apply from 1:00 p.m. onward (t ≥ 1). The graphs intersect at (2, 7). What does this point mean?

Choices: At 2:00 p.m., both cyclists are 7 km from the marker · At 7:00 p.m., both cyclists are 2 km from the marker · The cyclists ride at 2 km per hour and 7 km per hour · Both cyclists stay 7 km from the marker from 1:00 p.m. to 2:00 p.m.

Show solution
  1. The ordered pair is (t, d): t = 2 hours after noon and d = 7 kilometers. t = 2 lies in the models' interval, t ≥ 1.
  2. Check both models at t = 2: 3 + 2(2) = 7 and 6(2) − 5 = 7. At 1:00 p.m. the distances are 5 km and 1 km, so they are not 7 km before 2:00 p.m.
  3. So at 2:00 p.m. the cyclists are the same distance, 7 km, from the marker. Their speeds are the rates 2 and 6 km per hour, and (7, 2) would reverse the coordinates.

Answer: At 2:00 p.m., both cyclists are 7 km from the marker

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Graph a system, then confirm it algebraically

Graph y = −(1/2)x + 4 and y = x − 2 on one coordinate grid. Label both axes and both lines, and mark each line's y-intercept. Estimate the intersection from your graph. Then find the exact solution by substitution, check it in both equations, and explain why a graph alone might give only an estimate.

Review rubric and solution
  • Graphs both lines accurately from their y-intercepts (4 and −2) and slopes (−1/2 and 1), with labeled axes and lines.
  • Estimates the intersection near (4, 2) from the graph.
  • Solves by substitution: −(1/2)x + 4 = x − 2 gives x = 4 and y = 2.
  • Checks (4, 2) in both equations.
  • Explains that graph readings depend on scale and drawing accuracy, so a non-grid intersection may only be estimated; algebra gives the exact point.
  1. Line 1 starts at (0, 4) and falls 1 unit for every 2 units right; line 2 starts at (0, −2) and rises 1 unit for every 1 unit right.
  2. The lines appear to cross at (4, 2).
  3. Substitution: −(1/2)x + 4 = x − 2, so 6 = (3/2)x and x = 4. Then y = 4 − 2 = 2.
  4. Check: −(1/2)(4) + 4 = 2 and 4 − 2 = 2. A sketch can misplace a crossing such as (4.3, 1.9); the algebra confirms the exact point.

Graph a feasible region and judge plans

A theater sells x adult tickets at $15 and y student tickets at $8 for a show with at most 120 seats, and it needs at least $1,200 in ticket sales. Write the system of inequalities, including x ≥ 0 and y ≥ 0. Graph it with labeled axes and intercepts, explain whether each boundary is solid or dashed, and shade the feasible region. Then decide whether the plans (40, 60), (60, 50), and (80, 0) are feasible, and explain why only whole-number points make sense.

Review rubric and solution
  • Writes x + y ≤ 120, 15x + 8y ≥ 1,200, x ≥ 0, and y ≥ 0.
  • Graphs x + y = 120 through (120, 0) and (0, 120) and 15x + 8y = 1,200 through (80, 0) and (0, 150), both solid because the symbols include equality.
  • Shades the region below x + y = 120 and above 15x + 8y = 1,200 in the first quadrant, justified with a test point.
  • Finds (40, 60) infeasible ($1,080), (60, 50) feasible (110 seats, $1,300), and (80, 0) feasible on the boundary ($1,200 exactly).
  • Explains that tickets are counted in whole numbers, so only lattice points in the region are actual plans.
  1. Seats: x + y ≤ 120. Sales: 15x + 8y ≥ 1,200. Counts: x ≥ 0 and y ≥ 0.
  2. Both boundaries are solid, because ≤ and ≥ include equality. Testing (0, 0): 0 ≤ 120 is true and 0 ≥ 1,200 is false, so shade toward the origin from the seat line and away from the origin from the sales line.
  3. (40, 60): 100 seats but 15(40) + 8(60) = 1,080 dollars, which is too little. (60, 50): 110 seats and 900 + 400 = 1,300 dollars, which is feasible. (80, 0): 80 seats and exactly 1,200 dollars, which is feasible on the included boundary.
  4. Tickets are whole numbers, so the actual plans are the whole-number points inside or on the boundary of the shaded region.

Choose a method, justify it, and classify

For each system, choose graphing, substitution, or elimination, explain why that method suits its structure, and solve it. State whether the system has one solution, no solution, or infinitely many, and explain how your work shows this. (a) y = 3x − 5 and 2x + y = 10. (b) 4x − 6y = 8 and −6x + 9y = −12. (c) 3x + 2y = 7 and 6x + 4y = 9.

Review rubric and solution
  • (a) Chooses substitution because y is isolated, finds (3, 4), and checks it in both equations.
  • (b) Chooses elimination, scales the equations (for example by 3 and 2) to reach 0 = 0, and concludes infinitely many solutions because the equations describe the same line.
  • (c) Chooses elimination or compares slopes, reaches a false statement such as 0 = −5, and concludes no solution because the lines are parallel and distinct.
  • Justifies each method choice from the structure of the equations rather than by habit.
  • Connects each classification to the final statement or to the slopes and intercepts.
  1. (a) Substitution: 2x + (3x − 5) = 10, so 5x = 15, x = 3, and y = 4. Check: 3(3) − 5 = 4 and 2(3) + 4 = 10. One solution, (3, 4).
  2. (b) Elimination: 3(4x − 6y) = 24 gives 12x − 18y = 24, and 2(−6x + 9y) = −24 gives −12x + 18y = −24. Adding gives 0 = 0, so both equations are the line 2x − 3y = 4: infinitely many solutions.
  3. (c) Doubling the first equation gives 6x + 4y = 14. Subtracting 6x + 4y = 9 gives 0 = 5, which is false. Both lines have slope −3/2 with different intercepts, so there is no solution.
  4. Each classification follows from the final statement: a single point, an always-true statement, or a never-true statement.

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