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| 10 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Tiffaniqua's class is taking part in a simulated stock market project to learn about investing money. Every student starts with the same amount of money and their task is to make as much money as possible in three weeks!
Tiffaniqua invested all her money into one stock. After the first week, her stock was worth $65 more! After the second week, her stock was worth twice its value after the first week. During the last week of the project, the value of her stock fell by $50, making Tiffaniqua's stock worth $280.
When solving equations, sometimes more than one step is needed. Both the number of steps and the operations required depend on the complexity of the given equation. For example, consider the following pair of equations.
The general idea is to simplify both sides of the equation and then isolate the variable on one side of the equation. This is usually done by collecting all the variable terms on one side of the equations and combining them. Then, the operations applied to the variable are undone in reverse order.
Distribute 23
2a⋅2=a
Add terms
LHS⋅32=RHS⋅32
Commutative Property of Multiplication
ba⋅ab=1
1⋅a=a
a⋅cb=ca⋅b
Multiply
Calculate quotient
LHS+2y=RHS+2y
Commutative Property of Addition
Add terms
LHS+4=RHS+4
Add terms
LHS/5=RHS/5
Calculate quotient
Tiffaniqua and her friend Kevin are talking about the stock market project. Kevin is telling her how his investment is doing after two weeks.
Distribute 3
Multiply
Add terms
LHS+10=RHS+10
Add terms
LHS/3=RHS/3
Cross out common factors
Simplify quotient
Calculate quotient
Tearrik and Ramsha both invested the same amount in a single stock. After two weeks, they compared how their investments were doing.
It turns out that after two weeks, both investments are worth exactly the same amount of money!
LHS−34m=RHS−34m
Commutative Property of Addition
Rewrite 2m as 36m
Subtract terms
LHS+20=RHS+20
Add terms
LHS⋅23=RHS⋅23
Commutative Property of Multiplication
ba⋅ab=1
a⋅cb=ca⋅b
Multiply
Calculate quotient
Solve the given equation for the indicated variable. If necessary, round the answer to two decimal places.
Equations with linear terms may have zero, one, or infinitely many solutions. This lesson has already shown how to solve equations with one solution. Now it is time to focus on the other two possibilities.
There are three possible results when solving an equation.
Distribute 32
ca⋅b=ca⋅b
Multiply
Calculate quotient
LHS−32t=RHS−32t
Subtract terms
To solve the challenge presented at the beginning of the lesson, write and solve an equation that models the situation. The challenge stated that during her class simulated stock market project, Tiffaniqua invested all her money into a single stock.
After the first week, Tiffaniqua's investment was worth $65 more than it was at the beginning of the project. After the second week, her stock was worth twice its value after the first week. During the last week of the project, the value of her stock fell by $50, making Tiffaniqua's stock worth $280.
LHS/2=RHS/2
Cross out common factors
Simplify quotient
Calculate quotient
Distribute 2
Multiply
Subtract term
During a simulated stock market project, Emily invested some money into a stock.
We want to write an equation that represents the given situation. We will represent the unknown quantity with a variable. Here, the unknown quantity is the starting value of Emily's investment, which we will call v. We know that after the first week, the value of the investment fell by $25. This means that the value after the first week is v-25. Value After First Week: v-25 During the second week, the value of the investment doubled, so the value of the investment after the second week is 2(v-25). Value After First Week:& v-25 Value After Second Week:& 2(v-25) Finally, the value rose by $30 during the last week, meaning that the final value of the investment is 2(v-25)+30. Value After First Week:& v-25 Value After Second Week:& 2(v-25) Value After Third Week:& 2(v-25)+30 We know that the final value of the investment is $130. Let's use this information to write an equation. 2(v-25)+30=130 This equation can help us find the original value of the investment. Note that we can write different equations, but they will all be equivalent to the equation we wrote here. These equivalent equations can also be obtained by applying Properties of Equality to the above equation.
Ignacio and his friend are doing a four-day running challenge. Each person tries to run the farthest distance in four days. Ignacio ran a certain distance on the first day. On the second day, he ran 4 miles. On the third day, Ignacio doubled the distance he ran so far during the challenge, and then on the fourth day, he ran 3 miles.
We want to write an equation that represents the given situation. We first represent the unknown quantity with a variable. Here, the unknown quantity is the distance that Ignacio ran on the first day, which we will call d. Total Distance After One Day:& d On the second day, Ignacio ran 4 miles. This means that the total distance after two days was d+4 miles. Total Distance After One Day:& d Total Distance After Two Days:& d+4 We know that Ignacio doubled the distance he ran on the two previous days on Day 3. The total distance after three days was 2(d+4) miles. Total Distance After One Day:& d Total Distance After Two Days:& d+4 Total Distance After Three Days:& 2(d+4) On the final day, Ignacio ran 3 miles. Total Distance After One Day:& d Total Distance After Two Days:& d+4 Total Distance After Three Days:& 2(d+4) Total Distance After Four Days:& 2(d+4) + 3 We also know that Ignacio ran 21 miles over the whole challenge, so let's set our expression equal to 21. 2(d+4)+3=21 Note that we can write different equations, but they will all be equivalent to the equation we wrote. These equivalent equations can also be found by applying the Properties of Equality to the one we found.
Tearrik and Ramsha invested the same amount into their favorite stocks. After two weeks, they compared how their investments were doing.
We want to write an equation that represents the given situation. We will represent the unknown quantity with a variable. Here, the unknown quantity is the starting value of each investment, which we will call v. Initial Value:& v Tearrik's investment fell in value by $20 in the first week, so its value was v-20 at that point. The value then tripled over the next week. The final value of Tearrik's investment is 3(v-20). Initial Value:& v Value of Tearrik's Investment:& 3(v-20) Ramsha's investment grew by a quarter during the first week, so the value of the investment was 54v. In the second week, the value grew by $10, so the final value is 54v+10. Initial Value:& v [0.5em] Value of Tearrik's Investment:& 3(v-20) [0.5em] Value of Ramsha's Investment:& 5/4v+10 We know that the final values of both investments were the same, so let's set these expressions equal to each other to write an equation. 3(v-20)=5/4v+10 Note that we can write different equations, but they will all be equivalent to the equation we wrote. These equivalent equations can also be obtained by applying the Properties of Equality to our equation.
Maya enjoys running. On Thursday, she decided to run twice the distance she usually runs. However, she had to cut her run short by 3 miles. On Saturday, she decided to make up for it by running 3 miles more than she usually does.
We want to write an equation that represents the given situation. We first represent the unknown quantity with a variable. Here, the unknown quantity is the usual distance that Maya runs, which we will call d. Usual Distance:& d On Thursday, Maya wanted to run twice the distance she usually runs, or 2d miles. However, she ended up running 3 miles less than planned, so she ran 2d-3 miles. Usual Distance:& d Distance on Thursday:& 2d-3 On Saturday, Maya ran 3 miles more than she usually does. This means that she ran d+3 miles. Usual Distance:& d Distance on Thursday:& 2d-3 Distance on Saturday:& d+3 We know that the two distances are equal, so let's set them equal to each other to write an equation. 2d-3=d+3 Note that we can write different equations, but they will all be equivalent to the equation we wrote. These equivalent equations can also be written by applying the Properties of Equality to our equation here.