4. Writing and Solving Multi-Step Equations
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Chapter 6
4.
Writing and Solving Multi-Step Equations
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.
Show less Show more expand_more| 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 |
Writing and Solving Multi-Step Equations
Slide of 10
Some equations require more than one or two steps to solve. We may need to use properties such as the Distributive Property to move terms and variables to either side of the equation. This lesson focuses on writing and solving multi-step equations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Simulated Stock Market
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.
a
Write an equation in terms of w that represents the situation.
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b
Solve the equation to find the amount of money Tiffaniqua invested at the start.
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Discussion
Solving Multi-Step Equations
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.
Solving Equation (I)
Equation (I) contains a fraction and parentheses. 3/2(x-2)+5=14
1
Clear Parentheses and Combine Like Terms on Each Side
expand_more Apply the Distributive Property to clear the parentheses on the left-hand side of the equation. Distribute 32 to each term inside the parentheses, then combine like terms on the left-hand side.
3/2(x-2) + 5 = 14
Distr
Distribute 3/2
3/2x- 3/2* 2 + 5 = 14
FracMultDenomToNumber
a/2* 2 = a
3/2x-3+5=14
AddTerms
Add terms
3/2x+2=14
2
Isolate the Variable
expand_more Apply the Properties of Equality to isolate the variable on one side of the equation.
3/2x+2=14
SubEqn
LHS-2=RHS-2
3/2x+2-2=14-2
SubTerms
Subtract terms
3/2x=12
MultEqn
LHS * 2/3=RHS* 2/3
3/2x*2/3=12*2/3
CommutativePropMult
Commutative Property of Multiplication
3/2*2/3* x=12*2/3
MultFracByInverse
a/b* b/a=1
x = 12*2/3
MoveLeftFacToNum
a*b/c= a* b/c
x=12*2/3
Multiply
Multiply
x = 24/3
CalcQuot
Calculate quotient
x=8
As such, x=8 is the solution of Equation (I).
Solving Equation (II)
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.
1
Clear Parentheses and Combine Like Terms on Each Side
expand_more Apply the Distributive Property to clear the parentheses on each side of the equation. There are no parentheses to clear in this equation, so start by combining like terms instead.
2
Collect the Variable on One Side of the Equation
expand_more Apply the Properties of Equality to collect all the variables on one side of the equation. In this case, add 2y to both sides of the equation to group the y-terms on the left-hand side.
3y - 4 = 11-2y
AddEqn
LHS+2y=RHS+2y
3y - 4 + 2y = 11
CommutativePropAdd
Commutative Property of Addition
3y + 2y - 4 = 11-2y+2y
AddTerms
Add terms
5y - 4 = 11
3
Isolate the Variable
expand_more Apply the Properties of Equality again to isolate the variable on one side of the equation. In this case, add 4 to both sides, then divide both sides of the equation by 5.
5y - 4 = 11
AddEqn
LHS+4=RHS+4
5y = 11+4
AddTerms
Add terms
5y=15
DivEqn
.LHS /5.=.RHS /5.
5y/5 = 15/5
SimpQuot
Simplify quotient
y = 15/5
CalcQuot
Calculate quotient
y = 3
In conclusion, y=3 is the solution of Equation (II).
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Example
Books for the Classroom
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.
a
Write an equation using p to represent the situation.
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b
Solve the equation to find how many books the class had at the start.
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Hint
a
The community tripled the number of non-damaged books.
b
Use the Distributive Property and inverse operations to solve the equation.
Solution
a
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.
Non-damaged Books: p - 15 The community donated enough books to triple this amount. Non-damaged Books:& p - 15 After Community Donation:& 3(p - 15) Then the class receives another 35 books. Non-damaged Books:& p - 15 After Community Donation:& 3(p - 15) After Final Donation:& 3(p - 15)+35 The classroom library ends up with 140 books, which will complete the equation. 3 (p-15) + 35= 140
b
Solve the equation from Part A to find the number of books that the classroom library started with.
3 (p-15) +35= 140 First, apply the Distributive Property to clear the parentheses. Then, apply the Properties of Equality to isolate the variable p.
3 (p-15) + 35 = 140
Distr
Distribute 3
3p - 3 * 15 + 35 = 140
Multiply
Multiply
3p - 45 +35= 140
AddTerms
Add terms
3p - 10= 140
AddEqn
LHS+10=RHS+10
3p - 10+10= 140+10
AddTerms
Add terms
3p = 150
DivEqn
.LHS /3.=.RHS /3.
3p/3 = 150/3
SimpQuot
Simplify quotient
p=150/3
CalcQuot
Calculate quotient
p = 50
The solution to the equation from Part A is p = 50. This means that the classroom library started with 50 books.
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Example
Comparing Two Gardens
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.
a
Write an equation that models this situation. Let s represent the number of seedlings each garden had at the start.
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b
Solve the equation to find the starting number of seedlings.
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Hint
a
The starting number of seedlings is s.
b
Collect all the variable terms on one side of the equation.
Solution
a
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. This means that the number of seedlings in the garden can be expressed as 2(s-10).
East Garden 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, bringing their total number of plants to 43s+10. West Garden Seedlings: 4/3s+10 Since both gardens end up with the same number of seedlings, we can write an equation from these equivalent expressions. 2(s-10) = 4/3s+10
b
Solve the equation from Part A to find the starting number of seedlings for each garden.
2(s-10) = 4/3s+10 First, use the Distributive Property.
The resulting equation has variable terms on both sides. To solve the equation, gather all the variable terms on one side of the equation by using the Subtraction Property of Equality to subtract 43s from both sides.
2s-20=4/3s+10
SubEqn
LHS-4/3s=RHS-4/3s
2s-20-4/3s=4/3s+10-4/3s
CommutativePropAdd
Commutative Property of Addition
2s-4/3s-20=4/3s-4/3s+10
Rewrite
Rewrite 2s as 6/3s
6/3s-4/3s-20=4/3s-4/3s+10
SubTerms
Subtract terms
2/3s-20=10
Finally, isolate the variable using inverse operations and the Properties of Equality.
2/3s-20=10
AddEqn
LHS+20=RHS+20
2/3s-20+20=10+20
AddTerms
Add terms
2/3s = 30
MultEqn
LHS * 3/2=RHS* 3/2
2/3s *3/2= 30*3/2
CommutativePropMult
Commutative Property of Multiplication
2/3*3/2* s = 30 * 3/2
MultFracByInverse
a/b* b/a=1
s = 30* 3/2
MoveLeftFacToNum
a*b/c= a* b/c
s = 30* 3/2
Multiply
Multiply
s = 90/2
CalcQuot
Calculate quotient
s = 45
The solution to the equation from Part A is s = 45. This means that both the East and West Gardens started with 45 seedlings.
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Pop Quiz
Solving Different Equations
Solve the given equation for the indicated variable. If necessary, round the answer to two decimal places.
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Discussion
Equations With No Solutions or More Than One Solution
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.
Concept
Identity
There are three possible results when solving an equation.
- If an identify is obtained, then every number is a solution to the equation. In other words, the equation has infinitely many solutions.
- If a false statement is obtained, the equation has no solutions. In this case, the solution set is the null set.
- If the result is an equation in which the variable is isolated, then the equation has one solution.
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Example
Solving Equations With No Solutions or Infinitely Many Solutions
a
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.
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b
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.
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c
Select all the true sentences.
{"type":"multichoice","form":{"alts":["The equation from Part A has no solutions.","The equation from Part A has one solution.","The equation from Part A has infinitely many solutions.","The equation from Part B has no solutions.","The equation from Part B has one solution.","The equation from Part B has infinitely many solutions."],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":[2,3]}
Solution
a
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 amount of water in the tank now is 23t+20 liters.
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 means that the water remaining in the tank is 23(t+30). Water in Tank B: 2/3(t+30) Since the tanks have the same amount of water after these changes, we can set these expressions equal to each other to create an equation that models the given situation. 2/3t+20 = 2/3(t+30)
b
The first worker did not pack 10 of their items due to damage, so their items packed is g-10. Thy 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 write an equation that models the situation by setting these two expressions equal to each other. 2(g-10)=2g+15
c
To find the number of solutions of the equations from Parts A and B, solve the equations. There are three possible outcomes for these equations.
- If solving the equation results in a false statement, the equation has no solutions.
- If solving the equation results in a solution, the equation has one solution.
- If solving the equation results in an identity, the equation has infinitely many solutions.
First, solve the equation from Part A. The first step is to use the Distributive Property.
2/3t+20 = 2/3(t+30)
Distr
Distribute 2/3
2/3t+20 = 2/3t+2/3* 30
MoveRightFacToNum
a/c* b = a* b/c
2/3t+20 = 2/3t+2* 30/3
Multiply
Multiply
2/3t+20 = 2/3t+60/3
CalcQuot
Calculate quotient
2/3t+20 = 2/3t+20
SubEqn
LHS-2/3t=RHS-2/3t
2/3t+20-2/3t = 2/3t+20-2/3t
SubTerms
Subtract terms
20=20
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. 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.
- The equation from Part A has infinitely many solutions.
- The equation from Part B has no solutions.
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Pop Quiz
Determining the Number of Solutions to an Equation
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Closure
Simulated Stock Market
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.
a
Write an equation in terms of w that represents the situation.
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b
Solve the equation to find the initial amount of money Tiffaniqua had during the project.
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Hint
a
Assign the variable to the starting amount of money.
b
Undo the operations applied to the variable in reverse order.
Solution
a
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.
- After the first week, Tiffaniqua's investment was worth $65 more, or w+65.
- During the second week, the stock value doubled, so the investment was worth 2(w+65) at this point.
- During the last week of the project, the value of the stock fell by $50, making the final value 2(w+65)-50.
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
b
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.
2(w+65) = 330
DivEqn
.LHS /2.=.RHS /2.
2(w+65)/2 = 330/2
CrossCommonFac
Cross out common factors
2(w+65)/2 = 330/2
SimpQuot
Simplify quotient
w+65 = 330/2
CalcQuot
Calculate quotient
w+65 = 165
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.
Alternative Solution
The equation can be simplified using a different set of operations. This time the process will start by using the Distributive Property to distribute 2 to the parentheses, then we will combine like terms.
2(w+65)-50 = 280
Distr
Distribute 2
2w+2*65-50=280
Multiply
Multiply
2w+130-50=280
SubTerm
Subtract term
2w+80=280
In the resulting equation, the variable w is multiplied by 2 and 80 is added to the result. Use inverse operations to isolate w.
2w+80=280
SubEqn
LHS-80=RHS-80
2w+80-80=280-80
SubTerms
Subtract terms
2w=200
DivEqn
.LHS /2.=.RHS /2.
2w/2=200/2
CalcQuot
Calculate quotient
w=100
The solution to the equation is w = 100, the same result as before.
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Writing and Solving Multi-Step Equations
Exercise 3.1
On the first day, she covers a certain distance. On the second day, she cycles 30 miles. On the third day, she manages to cycle half of the total distance she cycled during the first two days of the challenge, and on the final day she cycles 25 miles.
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Let's write an equation that models this situation. The variable represents some unknown quantity in an equation. In our case, the unknown quantity is the number of miles Paulina cycled on the first day. We will use c as the variable. Distance After First Day: c We know that Paulina cycled 30 miles on the second day of the challenge. The total distance she cycled at the end of the second day is c+30. Distance After First Day:& c Distance After Second Day:& c+30 On the third day, she cycled half of the total distance that she cycled during the first two days. This brings the total distance after the third day to 32(c+30). Distance After First Day:& c Distance After Second Day:& c+30 [0.3em] Distance After Third Day:& 3/2(c+30) Finally, Paulina cycled 25 miles on the fourth day. Distance After First Day:& c Distance After Second Day:& c+30 [0.3em] Distance After Third Day:& 3/2(c+30) [0.8em] Distance After Fourth Day:& 3/2(c+30)+25 We know that the total distance that Paulina cycled during the challenge is 100 miles, so let's set our expression equal to 100 to write our equation. 3/2(c+30)+25 = 100
Now we need to solve the equation from Part A. 3/2(c+30)+25 = 100 First, let's use the Distributive Property and combine like terms to simplify the left-hand side of the equation.
When solving equations, we can use inverse operations and the Properties of Equality to undo
the operations applied to the variable. In this case, the variable c is multiplied by 32, then 70 is added to the product.
3/2c + 70 = 100
We use inverse operations to undo these operations in reverse order. In this case, this means first using the Subtraction Property of Equality to subtract 70 from both sides of the equation.
This is now a regular one-step equation. The coefficient next to the variable is the fraction 32. To solve the equation, let's multiply both sides by the reciprocal of the coefficient using the Multiplication Property of Equality.
The solution to the equation is c = 20.
Finally, we want to determine how many miles Paulina cycled on the first day of the challenge. We know from Part A that this situation is modeled by the following equation. 3/2(c+30)+25 = 100 Here, c represents the number of miles cycled on the first day. We know from Part B that the solution to our equation is c= 20. Therefore, Paulina cycled 20 miles on the first day.
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Solution
In the end, Maya cycled the same distance on Wednesday and on Sunday!
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Let's write an equation that models this situation. The variable represents some unknown quantity in an equation. In our case, the unknown quantity is the number of miles Maya usually cycles. We will use d for our variable. Usual Distance: d On Wednesday, Maya decided to cycle twice the usual distance, or 2d. However, she ended up riding 5 miles fewer than planned, or 2d-5. Usual Distance:& d Distance on Wednesday:& 2d-5 On Sunday, Maya cycled 5 miles more than she usually does, or d+5 miles. Usual Distance:& d Distance on Wednesday:& 2d-5 Distance on Sunday:& d+5 We know that Maya cycled the same distance on Wednesday and Sunday, so let's set the expressions equal to each other to write an equation. 2d-5=d+5
Now we want to solve the equation from Part A. 2d-5=d+5 There are variable terms on both sides of the equation, so let's collect them all on one side. We can use the Subtraction Property of Equality to subtract d from both sides of the equation.
Next, we add 5 to both sides of the equation using the Addition Property of Equality.
The solution to the equation is d = 10.
Finally, we want to determine how many miles Maya usually cycles. We know from Part A that the situation is modeled by the following equation. 2d-5=d+5 Here, d represents the number of miles Maya usually cycles. We know from Part B that the solution to our equation is d= 10. Therefore, Maya usually cycles 10 miles.
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Writing and Solving Multi-Step Equations
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∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
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