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This lesson focuses on teaching how to write and solve multi-step equations, a crucial skill in algebra. It delves into the use of properties of equality and the distributive property to simplify and solve these equations. The lesson is designed to help not just with academic exercises but also in understanding real-world applications. For instance, it uses investment scenarios to illustrate how these equations can be used to determine the value of stocks over time. By mastering these techniques, one can solve problems that have no solutions, one solution, or infinitely many solutions. The teaching approach includes both theoretical explanations and practical exercises, making it a comprehensive guide for anyone looking to enhance their algebra skills.
| Student Learning Objectives: |
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| | 10 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
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 invests it for three weeks.
Tiffaniqua invested all her money in one stock. After the first week, the value of her investment increases by $65. After the second week, its value doubles. During the final week of the project, its value decreases by $50, leaving her with $280.
Write an equation in terms of w that represents the situation.
Solve the equation to find the amount of money Tiffaniqua invested at the start.
When solving equations, both the number of steps and the operations required to solve it depend on the equation itself. Consider the following two equations.
rl Equation (I):& 3/2(x-2)+5 = 14 [0.8em] Equation (II):& y+2y-4 = 11-2y
The general process is to simplify both sides of the equation and isolate the variable on one side of the equation. This is usually done by collecting all the variable terms on one side of the equation and combining them. Then, the operations applied to the variable are undone in reverse order.
Equation (I) contains a fraction and parentheses. 3/2(x-2)+5=14
Distribute 3/2
a/2* 2 = a
Add terms
LHS-2=RHS-2
Subtract terms
LHS * 2/3=RHS* 2/3
Commutative Property of Multiplication
a/b* b/a=1
a*b/c= a* b/c
Multiply
Calculate quotient
As such, x=8 is the solution of Equation (I).
In this equation, the variable is on both sides of the equation. y+2y-4=11-2y Solving the equation will require the additional step of collecting the variables on one side of the equation.
LHS+2y=RHS+2y
Commutative Property of Addition
Add terms
LHS+4=RHS+4
Add terms
.LHS /5.=.RHS /5.
Simplify quotient
Calculate quotient
In conclusion, y=3 is the solution of Equation (II).
A class is collecting books for a new classroom library. They begin with p books. During sorting, 15 damaged books are removed. A community donation then triples the number of books they have. Finally, another 35 books are donated, bringing the total to 140 books.
Write an equation using p to represent the situation.
Solve the equation to find how many books the class had at the start.
The community tripled the number of non-damaged books.
Use the Distributive Property and inverse operations to solve the equation.
Let p be the number of books the class began with. They sorted out 15 damaged books, so the number of remaining, non-damaged books is the difference between p and 15.
p - 15 The community donated enough books to triple this amount. p - 15 ⇒ 3(p - 15) Then the class receives another 35 books. 3(p - 15) ⇒ 3(p - 15) + 35 The classroom library ends up with 140 books, which will complete the equation. 3 (p-15) + 35= 140
Let's solve the equation from Part A to find the number of books that the classroom library started with.
3 (p-15) +35= 140 First, we will apply the Distributive Property to clear the parentheses. Then, apply the Properties of Equality to isolate the variable p.
Distribute 3
Multiply
Add terms
LHS+10=RHS+10
Add terms
.LHS /3.=.RHS /3.
Simplify quotient
Calculate quotient
The solution to the equation from Part A is p = 50. This means that the classroom library started with 50 books.
Two community gardens start with the same number of seedlings. The East Garden loses 10 seedlings to a frost. Afterward, enough new seedlings are planted to double the number that remained after the frost.
The West Garden added 13 of its starting number of seedlings to the garden. It then receive 10 more seedlings from the North Garden. After all these changes, the East and West Gardens still have the same number of seedlings.
Write an equation that models this situation. Let s represent the number of seedlings each garden had at the start.
Solve the equation to find the starting number of seedlings.
The starting number of seedlings is s.
Collect all the variable terms on one side of the equation.
Let s be the starting number of seedlings at each garden. The East Garden lost 10 seedlings to a frost, then doubled the remaining number of seedlings. Let's write an expression for the final number of East Garden's seedlings.
2(s - 10) The West Garden planted an additional 13 of their starting number of seedlings, so their garden contains 1s+ 13s= 43s seedlings. Then they receive another 10 seedlings from the North Garden. Let's write an expression for the final number of West Garden's seedlings. 4/3s + 10 Since both gardens end up with the same number of seedlings, we can equate the two expressions and write the following equation. 2(s-10) = 4/3s+10
Let's solve the equation from Part A to find the starting number of seedlings for each garden.
2(s-10) = 4/3s+10 First, we will use the Distributive Property.
The resulting equation has variable terms on both sides. To solve it, let's gather all the variable terms on one side of the equation by using the Subtraction Property of Equality to subtract 43s from both sides.
LHS-4/3s=RHS-4/3s
Commutative Property of Addition
Rewrite 2s as 6/3s
Subtract terms
Finally, we will isolate the variable using inverse operations and the Properties of Equality.
LHS+20=RHS+20
Add terms
LHS * 3/2=RHS* 3/2
Commutative Property of Multiplication
a/b* b/a=1
a*b/c= a* b/c
Multiply
Calculate quotient
The solution to the equation from Part A is s = 45. This means that both the East and West Gardens started with 45 seedlings.
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, so now let's focus on the other two possibilities.
There are three possible results when solving an equation.
Two water tanks start with the same amount of water. Tank A is drained of 13 of its water. Then, 20 liters of water are added. Tank B receives 30 liters of water, but then 13 of its water is drained.
After these changes, both tanks contain the same amount of water. Let t represent the amount of water, in liters, that each tank had at the start. Write an equation that models the situation. Do not simplify the equation.
Two workers start with the same number of items to pack in boxes. The first worker removes 10 items because they are damaged. The remaining number of items is then doubled before being packed. For the second worker, the starting number of items is doubled, and then 15 more items are added. Both boxes ended up with the same number of items packed.
Let g represent the number of items each worker started with. Write an equation that models this situation.
Select all the true sentences.
Let t be the starting amount of water in each tank. The first tank lost 13 of its volume, so it now has 23t water. After that, 20 liters were added to the tank. The following expression describes the water in Tank A after these changes.
Water in Tank A: 2/3t + 20 Tank B had 30 liters of water added to it, but then 13 of the volume was drained. This gives us the following expression for the water remaining in Tank B. Water in Tank B: 2/3(t + 30) The tanks have the same amount of water after these changes. Therefore, we can equate the two expressions and write the following equation. 2/3t+20 = 2/3(t+30)
The first worker did not pack 10 of their items due to damage, so they packed g - 10 of their items. They then pack double this value, so 2(g-10) items were packed.
First Worker: 2(g - 10) The second worker packed twice the initial number of items, then added 15 more to their box. They packed 2g+15 items. Second Worker: 2g + 15 Since both boxes ended up with the same number of items packed, we can equate the two expressions writing the following equation. 2(g-10)=2g+15
To find the number of solutions of the equations from Parts A and B, let's solve them.
We will start with the equation from Part A. The first step is to use the Distributive Property.
Distribute 2/3
a/c* b = a* b/c
Multiply
Calculate quotient
LHS-2/3t=RHS-2/3t
Subtract terms
Solving the equation resulted in an identity, which means that every number is a solution to the equation. No matter how much water was initially in the tanks, they will end up with the same amount in the end.
Next we will solve the equation written in Part B, starting again with the Distributive Property.
The next step is to collect all the variable terms on one side and combine like terms. Let's use the Subtraction Property of Equality to subtract 2g from both sides of the equation.
This is a false statement, so this equation has no solutions. In other words, no starting number of items will make the two final amounts equal after the given changes.
The following statements are true.
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 in a single stock.
After the first week, the value of her investment increases by $65. After the second week, its value doubles. During the final week of the project, its value decreases by $50, leaving her with $280.
Write an equation in terms of w that represents the situation.
Solve the equation to find the initial amount of money Tiffaniqua had during the project.
Assign the variable to the starting amount of money.
Undo the operations applied to the variable in reverse order.
A variable is used to represent an unknown quantity when writing an equation that models a real-life situation. In this situation, the unknown quantity is the amount of money that Tiffaniqua invested initially, which we will represent with w.
Starting Amount of Money: w We can express the final value of Tiffaniqua's stock in terms of w.
The following expression represents final value of Tiffaniqua's stock. Final Value: 2(w+65)-50 Finally, write the equation by setting this expression equal to the final value of the investment, $280. 2(w+65)-50 = 280
The w-variable must be isolated to solve the equation. To do this, the operations applied to the variable are undone in reverse order.
2(w+65)-50 = 280 On the left-hand side of the equation, 65 is added to w and the result is then multiplied by 2. 50 is subtracted from the product. These operations can be undone using inverse operations. Undo the subtraction by using the Addition Property of Inequality.
Undo the multiplication using the Division Property of Equality.
.LHS /2.=.RHS /2.
Cross out common factors
Simplify quotient
Calculate quotient
Finally, use the Subtraction Property of Equality to undo the addition.
The solution to the equation is w = 100. This means that Tiffaniqua started the simulated stock market project with $100.
Distribute 2
Multiply
Subtract term
In the resulting equation, the variable w is multiplied by 2 and 80 is added to the result. Use inverse operations to isolate w.
LHS-80=RHS-80
Subtract terms
.LHS /2.=.RHS /2.
Calculate quotient
The solution to the equation is w = 100, the same result as before.
During a simulated stock market project, Emily invested some money into a stock.
At the end, Emily's investment was worth $130. Write an equation to represent this situation, using v as the variable.
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.
In the end, Ignacio ran 21 miles during the challenge! Write an equation to represent this situation, using d as the variable.
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.
After two weeks, both investments were worth exactly the same amount of money! Write an equation that models this situation.
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.
In the end, Maya ran the same distance on Thursday as she did on Saturday! Write an equation that models this situation. Use d as the variable.
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.