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Math lesson plan · 9th grade (Year 10 in England) · 45 minutes

Solve a linear system by substitution

Algebra

CCSS HSA.REI.C.6England: GCSE Foundation AlgebraEngland: GCSE Higher Algebra

Objective and success criteria

Students will: Solve a linear system by substitution.

  • I can answer questions on today's skill on my own, without help.
  • I can explain each step and say why it works.
  • I can check my answer and spot a mistake.

Before and after

Needs first: Systems of linear equations (simultaneous equations)

Leads to: the next skill in this area.

Materials: Balance scales or a balance drawing, Algebra tiles, Function machine cards, Mini whiteboards and pens, Printed student sheet (below).

Warm-up · 5 min

  • Quick retrieval from earlier learning (Systems of linear equations (simultaneous equations)). Students answer on mini whiteboards.
  • Show answers together and note who needs support in the main task.
  1. Solve the system: y = 3x + 3 and x + 2y = 48. Find x. → x = 6Prompt: Substitute the first equation into the second.
  2. Solve: 4x + 5y = 54 and 3x − 5y = −12. Find y. → 6Prompt: The y terms have opposite coefficients: add the equations to eliminate y.
  3. Solve the system: y = 3x − 5 and 4x + 2y = 10. Find x. → x = 2Prompt: Substitute the first equation into the second.

Hook · 5 min

  • Show a "think of a number" puzzle and ask the class to work out the starting number.
  • Share the objective: Solve a linear system by substitution. Read the success criteria together.

I do: model it · 10 min

  • Work through the example out loud, saying why you do each step, not just what you do.
  • Example: Solve the system: y = −3x − 16 4x − 3y = −17 Type the solution like (2, −1).
  • Step 1: 4x − 3(−3x − 16) = −17
  • Step 2: 13x = −65 → x = −5
  • Step 3: y = −1
  • Step 4: (−5, −1)
  • Answer: (−5, −1)
  • Pause after each step and ask a student to say what comes next.

We do: guided practice · 10 min

  • Do these together. Students try each step on whiteboards before you reveal it.
  • Use the hints as prompts rather than giving the answer.
  1. Solve the system: y = 3x − 6 −5x + 4y = −10 Type the solution like (2, −1). → (2, 0)Prompt: Replace y in the second equation with the first expression.
  2. Solve the system: x = 3y − 24 x + 3y = 12 Type the solution like (2, −1). → (−6, 6)Prompt: x is already on its own: substitute it into the other equation.

You do: independent practice · 10 min

  • Students work on the student sheet. Circulate and check the first two answers of every student early.
  • Students who finish: write a question of their own for a partner, with the answer.

The 8 questions are on the student sheet below, with answers in the key.

Exit ticket · 5 min

  • Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.
  1. To solve y = −4x − 7 and −2x − 5y = −3 by substitution, which is the correct next line? (−2x − 5·−4x − 7 = −3 / −2x + 5(−4x − 7) = −3 / −4x − 7 − 5y = −3 / −2x − 5(−4x − 7) = −3) → −2x − 5(−4x − 7) = −3Prompt: Substitution replaces y with the whole expression it equals.
  2. Solve the system: x = −2y − 3 x − 5y = 11 Type the solution like (2, −1). → (1, −2)Prompt: x is already on its own: substitute it into the other equation.

Watch out for

  • Doing an operation to one side of an equation but not the other.
  • Reading 3x as 3 + x, or thinking x always stands for one particular number.

Support and stretch

Support

  • Review first: Systems of linear equations (simultaneous equations).
  • Use concrete materials: balance scales or a balance drawing.
  • Give the first step of each question, or a partly worked example to finish.
  • Pair with a confident partner for the guided practice.

Stretch

  • Explain a common mistake in writing and how to fix it.
  • Write a word problem that uses today's skill, then solve a partner's.

Homework and more practice

Free online practice with hints and printable worksheets for this skill: eurekami.com/math/grade-9.

Free lesson plan from eurekami.com/lesson-plan-builder

eurekami

Solve a linear system by substitution

Name

Date

Score/ 8

  1. Question 1

    To solve y = −x + 5 and 3x + 3y = −1 by substitution, which is the correct next line?

    • A. 3x + 3(−x + 5) = −1
    • B. −x + 5 + 3y = −1
    • C. 3x − 3(−x + 5) = −1
    • D. 3x + 3·−x + 5 = −1

  2. Question 2

    Solve the system: x = −2y + 4 2x − 2y = 8 Type the solution like (2, −1).

    Answer

  3. Question 3

    Solve the system: y = 3x − 1 5x + 2y = 20 Type the solution like (2, −1).

    Answer

  4. Question 4

    To solve y = −2x + 2 and 4x + 4y = −14 by substitution, which is the correct next line?

    • A. 4x + 4·−2x + 2 = −14
    • B. −2x + 2 + 4y = −14
    • C. 4x − 4(−2x + 2) = −14
    • D. 4x + 4(−2x + 2) = −14

  5. Question 5

    Solve the system: x = 3y − 11 3x + 3y = 15 Type the solution like (2, −1).

    Answer

  6. Question 6

    Solve the system: y = 3x + 13 −3x − 3y = 21 Type the solution like (2, −1).

    Answer

  7. Question 7

    To solve y = 2x + 1 and −2x + 5y = 8 by substitution, which is the correct next line?

    • A. 2x + 1 + 5y = 8
    • B. −2x − 5(2x + 1) = 8
    • C. −2x + 5(2x + 1) = 8
    • D. −2x + 5·2x + 1 = 8

  8. Question 8

    Solve the system: x = −3y + 18 −x + 4y = 24 Type the solution like (2, −1).

    Answer

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Answer key: Solve a linear system by substitution

  1. 1. A. 3x + 3(−x + 5) = −1

  2. 2. (4, 0)

  3. 3. (2, 5)

  4. 4. D. 4x + 4(−2x + 2) = −14

  5. 5. (1, 4)

  6. 6. (−5, −2)

  7. 7. C. −2x + 5(2x + 1) = 8

  8. 8. (0, 6)