Eureka Math » Grade 9 » Module 1 » Topic C » Lesson 13

### Some Potential Dangers when Solving Equations

### Student Outcomes & Lesson Notes

**Student Outcomes**

- Students learn “if-then” moves using the properties of equality to solve equations. Students also explore moves that may result in an equation having more solutions than the original equation.

### Classwork

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**Section 1** Exercise 1 (4 Minutes)

Give students a few minutes to answer the questions individually. Then, elicit responses from students.

- Do we know for certain that
$x=4$
is the solution to every equation shown? Explain why.
*Have students verify this by testing the solution in a couple of the equations.*

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**Section 2** Exercise 2 (4 Minutes)

##### Consider:

Students may point out that they solved the equation in a different way but got the same answer. Consider allowing them to show their approach and discuss whether or not it was algebraically sound.

Read MoreWork through the exercise as a class. Perhaps have one student writing the problem on the board and one student writing the operation used in each step as the class provides responses.

Emphasize that the solution obtained in the last step is the same as the solution to each of the preceding equations.The moves made in each step did not change the solution set.

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**Section 3** Exercise 3 (8 Minutes)

Note with the class that students may have different approaches that arrived at the same answer.

Ask students how they handled the fraction in problem (c).

- Was it easier to use the distributive property first or multiply both sides by $6$ first?

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**Section 4** Discussion (10 Minutes)

Use the following sample dialog to inspire a similar exchange between you and your students where you play the part of Mike, suggesting ideas of actions you could perform on both sides of an equation that would not predictably preserve the solution set of the original equation. Start by asking students to summarize what they have been studying over the last two lessons and then make Mike’s first suggestion. Be sure to provide more than one idea for things that could be done to both sides of an equation that might result in solutions that are not part of the solution set for the original equation, and conclude with an affirmation that you can try anything, but you will have to check to see if your solutions work with the original equation.

- Fergus says, “Basically, what I’ve heard over the last two lessons is that whatever you do to the left side of the equation, do the same thing to the right side. Then solutions will be good.”
- Lulu says, “Well, we’ve only said that for the properties of equality – adding quantities and multiplying by non-zero quantities. (And associative, commutative, and distributive properties too.) Who knows if it is true in general?”
- Mike says, “Okay … Here’s an equation:

$\frac{x}{12}=\frac{1}{3}$

- If I follow the idea, “Whatever you do to the left, do to the right as well,” then I am in trouble. What if I decide to remove the denominator on the left and also remove the denominator on the right. I get $x=1$ . Is that a solution?”
- Fergus replies, “Well, that is silly. We all know that is a wrong thing to do. You should multiply both sides of that equation by $12$ . That gives $x=4$ , and that does give the correct solution.”
- Lulu says, “Okay Fergus, you have just acknowledged that there are some things we can’t do! Even if you don’t like Mike’s example, he’s got a point.”
- Mike or another student says, “What if I take your equation and choose to square each side. This gives

$\frac{{x}^{2}}{144}=\frac{1}{9}$

- Multiplying through by $144$ gives ${x}^{2}=\frac{144}{9}=16$ , which has solutions $x=4$ AND $x=-4$ .”
- Fergus responds, “Hmmm. Okay I do see the solution $x=4$ , but the appearance of $x=-4$ as well is weird.”
- Mike says, “Lulu is right. Over the past two days we have learned that using the commutative, associative, and distributive properties, along with the properties of equality (adding and multiplying equations throughout) definitely DOES NOT change solution sets. BUT if we do anything different from this we might be in trouble.”
- Lulu continues, “Yeah! Basically when we start doing unusual operations on an equation, we are really saying that **IF** we have a solution to an equation, then it should be a solution to the next equation as well. BUT remember, it could be that there was no solution to the first equation anyway!”
- Mike says, “So feel free to start doing weird things to both sides of an equation if you want (though you might want to do sensible weird things!), but all you will be getting are possible CANDIDATES for solutions. You are going to have to check at the end if they really are solutions.”

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**Section 5** Exercises 4-7 (12 Minutes)

##### Scaffold:

Have early finishers explore the idea of cubing both sides of an equation. If $x=2$ , then ${x}^{3}=8$ . If ${x}^{3}=8$ , can $x$ equal any real number besides $2$ ?

Read MoreAllow students to work through Exercises 4–7 either individually or in pairs. Point out that they are trying to determine what impact certain moves have on the solution set of an equation.

Review answers and discuss the following points:

- Does squaring both sides of an equation change the solution set?
*Sometimes but not always!*

- For Exercise
$6$
, was it just luck that Bonzo got one out of the three correct answers?
*Yes, in part ( $c$ ), the answer obtained is not a solution to the original equation.*

Consider having students make up another problem to verify.

- What effect did multiplying both sides by a variable factor have on the solution set?
*In our case, it added another solution to the solution set.*

- Can we predict what the second solution will be?

Have students make up another problem to test the prediction.

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**Section 6** Closing (2 Minutes)

- What moves have we seen that do not change the solution set of an equation?
- What moves did change the solution set?
- What limitations are there to the principle “whatever you do to one side of the equation, you must do to the other side?”

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**Section 7** Exit Ticket (5 Minutes)

1. Solve the equation for $x$ . For each step, describe the operation and/or properties used to convert the equation.

$5(2x-4)-11=4+3x$

2. Consider the equation $x+4=3x+2.$

a. Show that adding $x+2$ to both sides of the equation does not change the solution set.

b. Show that multiplying both sides of the equation by $x+2$ adds a second solution of $x=-2$ to the solution set.

### Solutions

**Exit Ticket Sample Solutions**

**Problem Set Sample Solutions**