# Partner Chelpers: Equations in the form p - x = q

## Objective

SWBAT work with partner chelpers to solve two-step equations including negative one as a coefficient.

#### Big Idea

Students work in pairs to solve and check equations in the form p – x = q.

## Do Now

10 minutes

Students enter silently according to the Daily Entrance Routine. There are Do Now assignments at their desks. Our work with one step equations is complex and requires regular check-ins with student progress toward mastery. The use of technology helps students identify problem areas they need to improve before the end of week quiz. Students are encouraged to ask questions about the steps involved in solving for x in an equation. Now that we are solving equations using all 4 operations, students often get confused about the operation that needs to be used to isolate the variable. For example, in question 9,

–5y = 35

Some students may not understand why adding 5 is an incorrect procedure. The concept of like terms can be used here are –5y and +5 cannot be combined. Furthermore, y is being multiplied by –5, thus the opposite operation is division. The conceptual understanding of a balance is helpful in understanding this opposite operation as well. If we had a balance with 5 canisters, each with an equal but unknown amount of chips inside, they would be balanced by the weight of 35 positive chips on the opposite side.  We can rationalize that y, the number of chips in each canister, would need to be negative if the product is to be positive because a negative times a negative equals a positive. We would divide 35 by 5 to determine the amount of chips inside each canister.

## Class Notes

15 minutes

The coefficient in the equation 9 – x = 18 is important for the solution. Students must recognize the coefficient as –1 in order to understand why the sign flips to –18 when x is positive. Many students are still struggling with the vocabulary, sometimes mixing the definition of constant and coefficient. Thus, we begin class notes by reviewing these definitions. I explain to students that they need to push themselves to use the mathematical language when talking about and writing about the solutions to these equations. I attempt to inject joy in the lesson by explaining that lately I have been using a lot of new language and often I will look at a student’s face and the look is a “say wha’?!” look. The impression is the key to a quick laugh in the class. I explain that I will try to provide friendlier definition that students can access to understand a concept, but they are still expected to push themselves to use the proper language.

I then introduce the steps for solving a two-step equation: isolate the variable term, and then isolate the variable. A conversation about the difference between isolating the variable term and isolating the variable is necessary. In the example

9 – x = 18,

the variable term is identified as

– x

• Why is it not alone, or isolated? Because there is a +9 in the way.
• What is the opposite of +9? Minus 9.
• Therefore we must subtract 9 from both sides first. This leaves the expression:

– x = 9

Now we must isolate the variable. The variable is x.

• Why is it not alone, or isolated? Because there is a –1 as a coefficient, it is being multiplied by the variable.
• What is the opposite of multiply by –1? Divide by negative 1.

This is a series of questioning that students are to follow throughout the task today. The study tip at the bottom of the page is noted so that students have an alternate explanation to seek out at home should they need additional practice and explanations.

20 minutes

Worksheets are distributed. Half of the room will receive task A sheets and the other half will receive task B sheets. Students are instructed to solve silently for 6 minutes. At the end of 7-8 minutes, I will ask all students to stop working. Then, I will show everyone on the board how I use substitution to check the answer to a solution. This guidance is especially necessary when x equals a negative number. For example, in task A, the solution for question one is the following:

5 – y = 1

__–5        –5

_–y_ = _5_

–1      –1

y = –5

Since this is a negative answer, when substituted in the check step, it looks like this:

5 – y = 10

5 – (–5) = 10

5 + 5 = 10

10 = 10

After showing students how to check, the following directions will be given:

• if you are not finished solving continue working silently.
• if you are finished and have a task A worksheet, hold up a #1
• if you are finished and have a task B worksheet, hold up a #2
• If you are finished, look for a student holding up the OPPOSITE letter worksheet as yours. If there are booths available, you may take one. Use the right hand column to check your partner's work. After checking, if you find they have made a mistake, quietly show them the mistake and ask them to fix it. Make sure your partner understands why this is a mistake (MP3).

I will continue to help students complete the solving steps.

## Closing

10 minutes

At the end of 15-20 minutes of check steps and solving equations all students will be asked to return to their seats. If there are 10 or more minutes left in class I will answer questions for 2-3 minutes about the solutions and check steps and then students will be asked to complete the exit ticket on the back of their classwork. Exit tickets will be collected at the end of class as homework is being distributed. They will be graded and returned to students the next day to provide feedback as we get closer to the Friday quiz.