Lesson 5

Steps in Solving Equations

These materials, when encountered before Algebra 1, Unit 7, Lesson 5 support success in that lesson.

5.1: Explaining Equivalent Expressions (5 minutes)

Warm-up

In this warm-up, students determine how to rearrange an original equation into 3 different equivalent equations. The purpose is to remind students of moves allowed when rearranging equations. This will be helpful in this lesson when students rearrange other linear equations and in the associated Algebra 1 lesson when students rearrange quadratic equations.

Student Facing

Explain or show why each of these equations is equivalent to \(7(x-15) + 3 = 8\).

  1. \(7x - 105 + 3 = 8\)
  2. \(7(x-15) - 5 = 0\)
  3. \(7x - 102 - 8 = 0\)

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

The purpose of the discussion is to recall some of the moves for rearranging equations to create new equations. Select students to share their reasoning for why the equations are equivalent. To continue the discussion, ask students:

  • “What are some other equivalent equations we could get by rearranging this one? Explain how you know they are equivalent to the original equation.” (\(7x - 102 = 8\) since it is the first solution with the constant terms combined and that solution is equivalent to the original equation. There are many other reasonable answers.)
  • “Two of the equations given have 0 on one side of the equation. Why might it be useful to have an equation in this form?” (In this form, we can set the other side of the equation equal to \(y\), then graph it to find the solutions as the \(x\)-coordinates of the \(x\)-intercepts.)
  • “Andre rearranges the equation to \(7x = 110\). Is this equation equivalent to the original? If not, find an equivalent equation that has a similar form. Then, explain why this form might be useful.” (It is equivalent to the original equation. This could be useful for solving the equation. After this, dividing both sides of the equation by 7 would solve for \(x\).)

5.2: Checking Work (15 minutes)

Activity

In this activity, students examine work for common mistakes when solving equations. In the associated Algebra 1 lesson, students solve quadratic equations and must be mindful of possible mistakes in their work. Students construct viable arguments and critique the reasoning of others (MP3) when they identify the errors and explain their reasons to agree or disagree.

Student Facing

Here is Clare’s work to solve some equations. For each problem, do you agree or disagree with Clare’s work. Explain your reasoning.

  1. \(2(x-1)+4 = 3x - 2\)
    \(2x - 2 + 4 = 3x - 2\)
    \(2x + 2 = 3x - 2\)
    \(2x = 3x\)
    \(\text{-}x = 0\)
    \(x = 0\)
  2. \(3(x-1) = 5x + 6\)
    \(3x - 1 = 5x + 6\)
    \(\text{-}1 = 2x + 6\)
    \(\text{-}7 = 2x\)
    \(-3.5 = x\)
  3. \((x-2)(x+3) = x+10\)
    \(x^2 + x - 6 =x + 10 \)
    \(x^2 - 6 = 10\)
    \(x^2 = 16\)
    \(x = 4\)

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

The purpose of the discussion is to highlight some common errors students may have while solving equations. Select students to share their solutions and reasoning. Ask students,

  • “What other common errors do you have to be careful to avoid when solving equations?” (I sometimes forget to expand something like \((x+1)^2\) and accidentally write \(x^2 + 1\).)

5.3: Row Game: Rewriting Equations (20 minutes)

Activity

In this activity, students rewrite equations into equivalent versions with a given property. To transform the equations, students will need to be able to distribute binomials and factor quadratics. Students work in pairs and each partner is responsible for answering the questions in either column A or column B. Although each row has two different problems, they share the same answer. Ensure that students work their problems out independently and collaborate with one another when they do not arrive at the same answers. Students construct viable arguments and critique the reasoning of others (MP3) when they resolve errors by critiquing their partner’s work or explaining their reasoning.

Launch

Arrange students in groups of 2. In each group, ask students to decide who will work on column A and who will work on column B.

Student Facing

Work independently on your column. Partner A completes the questions in column A only and partner B completes the questions in column B only. Your answers in each row should match. Work on one row at a time and check if your answer matches your partner’s before moving on. If you don’t get the same answer, work together to find any mistakes.

Partner A: Write an equivalent equation so that the given condition is true.

  1. \(5x+10 = -35\)

    • The expression on the right side is 0

  2. \(x^2 - 9x = 42\)

    • The left side is a product

  3. \(x(x+3) + 9 = 1\)

    • The right side is 0

  4. \(8(x+1) = 5x\)

    • The left side is 0 and there are no parentheses

  5. \(11+x = \frac{12}{x}\)

    • The equation is quadratic and the right side is zero.

  6. \((3x-5)(x-2) = 0\)

    • One side of the equation has a term with \(3x^2\)

  7. \(4x^2 - 4 = 8\)

    • The right side is 0 and the left side is a product

Partner B: Write an equivalent equation so that the given condition is true.

  1. \(5(x+9) = 0\)

    • The left side is expressed as the sum of two terms

  2. \(x(x-9) - 42 = 0\)

    • The left side is a product and the right side is not 0

  3. \(x(x+3) + 6 = -2\)

    • The right side is 0

  4. \(3x = -8\)

    • The left side is 0

  5. \((x+12)(x-1) = 0\)

    • The left side involves \(x^2\)

  6. \(3x - 11 = \frac{10}{x}\)

    • One side of the equation has a term with \(3x^2\)

  7. \(4(x^2 - 1) = 8\)

    • The right side of is 0 and the left side is a product

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

The purpose of the discussion is to recognize different ways to rewrite equations. Ask students what they notice about the starting equations and final, rewritten equations. Students may notice:

  • a lot of the final forms end with one side equal to zero.
  • distributing can create a form that can be easier to compare to other forms.
  • multiplying an equation by a denominator (as long as it is not zero) helps simplify the equation by removing the fractions.