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Track Meet: Results Board Crossings worksheet preview

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Track Meet: Results Board Crossings

Interpret graph intersections on a track meet results board and verify that shared points satisfy both linear equations.

Grade 8MathInteractive · Printable

Aligned to8.EE.C.8.a

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What’s included

  • Single worksheet
  • Auto-graded online
  • 1 Common Core standard
  • Printable answer key
  • 13 questions
  • Classroom-ready

Details

Format
Interactive · Printable
Grade
Grade 8
Subject
Math
Question types
Fill in the blank, Multiple choice, Circle the answer, Short answer, Drawing

About this worksheet

What students practice

  • Students identify a system's solution as an ordered pair that lies on both graphs.
  • Students verify shared points using equations and tables.
  • Students interpret graph intersections in track meet situations.

How to use

Provide a straightedge for connecting the plotted points. Have students use both graphs or both equations as evidence before discussing the final error analysis.

Curriculum

Common Core standard covered

8.EE.C.8.a

Understand that a solution to a system of two linear equations is a point where their graphs cross.

Preview

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Name Date

Track Meet: Results Board Crossings

Use shared graph points to check when two linear models agree.

Results Waiting to Be Posted

Two linear models predict how many results are waiting to be posted. Use a straightedge to draw the graphs, then answer the questions.

On this board, x is minutes relative to noon: negative values mean before noon. The y-coordinate is the number of results waiting. The sprint model is y = −x + 5. The field-event model is y = x + 1.
1.
Connect J to K for the sprint model. Connect L to M for the field-event model. Circle the point that represents a solution of both equations.−6−5−4−3−2−11234560−6−5−4−3−2−1123456JKLMMinutes from noonResults waiting
2.
At x = 1, the sprint graph has y = 4 and the field-event graph has y = 2. What should the board reviewer conclude?
  • A. The system has no solution with x = 1 because the two outputs differ.
  • B. The system has a solution at (1, 3) because 3 is halfway between 4 and 2.
  • C. The system has a solution at (1, 6) because 4 + 2 = 6.
  • D. The system has two solutions with x = 1 because each graph has an output.

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