# Lesson 5; Solving any Linear Equations

- Author:
- Arpit Kesharwani

- Topic:
- Equations, Linear Equations

## Launch Into the Warm Up

The purpose of this warm-up is to elicit students’ strategies for solving an equation for the value of x. The negative integers and location of x in each equation were purposeful to spark a discussion about operations of integers and a negative coefficient of a variable.
Tell students to close their books or devices and that they are going to see some equations they are to solve mentally. Display each problem one at a time for all to see. Give students 30 seconds of quiet think time for each equation. Ask students to share their strategies for finding the value of x. Record and display their responses for all to see.

## "Think about this......."

Some students may reason about the value of x using logic. For example, in -3x=9, the x must be -3 since −3⋅-3=9. Other students may reason about the value of x by changing the value of each side of the equation equally by, for example, dividing each side of -3x=9 by -3 to get the result x=-3. Both of these strategies should be highlighted during the discussion where possible.To involve more students in the conversation, consider asking as the students share their ideas:

- “Can you explain why you chose your strategy?”
- “Can anyone restate ___’s reasoning in a different way?”
- “Did anyone reason about the problem the same way but would explain it differently?”
- “Did anyone reason about the problem in a different way?”
- “Does anyone want to add on to _____’s strategy?”
- “Do you agree or disagree? Why?”

## Warm Up!

Solve the equations;

- 5−x=8
- -1=x−2
- -3x=9
- -10=-5x

## Activity 1 and 2

MaterialsThe goal of this activity is for students to build fluency solving equations with variables on each side. Students describe each step in their solution process to a partner and justify how each of their changes maintains the equality of the two expressions (MP3).Look for groups solving problems in different, but efficient, ways. For example, one group may distribute the 12 on the left side in problem 2 while another may multiply each side of the equation by 2 in order to re-write the equation with less factors on each side.Arrange students in groups of 2. Instruct the class that they will receive 4 cards with problems on them and that their goal is to create a solution to the problems.For the first two cards they draw, students will alternate solving by stating to their partner the step they plan to do to each side of the equation and why before writing down the step and passing the card. For the final two problem cards, each partner picks one and writes out its solution individually before trading to check each others’ work.To help students understand how they are expected to solve the first two problems, demonstrate the trading process with a student volunteer and a sample equation. Emphasize that the “why” justification should include how their step maintains the equality of the equation. Use MLR 3 (Critique, Correct, and Clarify) by reminding students to push each other to explain how their step guarantees that the equation is still balanced as they are working. For example, a student might say they are combining two terms on one side of the equation, which maintains the equality as the value of that side does not change, only the appearance.Distribute 4 slips from the blackline master to each group. Give time for groups to complete the problems, leaving at least 5 minutes for a whole-class discussion. If any groups finish early, make sure they have checked their solutions and then challenge them to try and find a new solution to one of the problems that uses less steps than their first solution. Conclude with a whole-class discussion.If time is a concern, give each group 2 cards rather than all 4 and have them only doing the trading steps portion of the activity, but make sure that all 4 cards are distributed throughout the class. Give 6–7 minutes for groups to complete their problems. Make sure each problem is discussed in a final whole-group discussion. Alternatively, extend the activity by selecting more problems for students to solve with their partners.

## Student Work

Your teacher will give you 4 cards, each with an equation.

- With your partner, select a card and choose who will take the first turn.
- During your turn, decide what the next move to solve the equation should be, explain your choice to your partner, and then write it down once you both agree. Switch roles for the next move. This continues until the equation is solved.
- Choose a second equation to solve in the same way, trading the card back and forth after each move.
- For the last two equations, choose one each to solve and then trade with your partner when you finish to check one another’s work.

Select previously identified students to share their representations with the class. Record and display the different ways Tyler’s number puzzle can be written as an expression. To highlight the connection between the different expressions, ask:

- “What do all these expressions have in common that make the number puzzle work?” (All of them are equivalent to x−6.)
- “How would you justify that, for example, 16((3x−7)⋅2−22)=x−6?” (I would simplify the left side of the equation until it was x−6.)
- “What does it mean if we have an equation that says x−6=x−6?” (The two sides of this equation are always the same.)

*Lighter Support: MLR 7 (Compare and Connect).*Have students meet with a partner to discuss the method that they use to solve the problem. Prompt students to share and explain their solution strategy and then discuss what is the same or different about their approaches. Lesson SynthesisGive students 2–3 minutes to think about all the equations they solved in today’s lesson and to write down any errors they made or observed. Discuss and consider creating a permanent display showing:- different approaches for different structures of equations
- types of errors to look out for

## Exit Ticket (Must be Taken)

Noah wanted to check his solution of x=145 for the equation 12(7x−6)=6x−10. Substituting 145 for x, he writes the following:12(7(145)−6)(7(145)−6)5(7(145)−6)7⋅14−698−692=6(145)−10=12(145)−20=5(12(145)−20)=12⋅14−20=168−20=148Find the incorrect step in Noah's work and explain why it is incorrect.