4. Writing and Solving Multi-Step Equations
+ Customize lesson
0
0
Sign In
An error occurred, try again later!
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.
| Student Learning Objectives: |
|---|
|
| | 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.
Share
Report error
Translate
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 $280.
a
Write an equation in terms of w that represents the situation.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["w"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2(w+65)-50=280","280=2(w+65)-50","2w+130-50=280","280=2w+130-50","2w+80=280","280=2w+80","2w=280-80","280-80=2w","2w=200","200=2w"]}}
b
Solve the equation to find the amount of money Tiffaniqua invested at the start.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"$","formTextAfter":null,"answer":{"text":["100"]}}
Share
Report error
Translate
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).
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["p"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["3(p-15)+35 = 140","140=3(p-15)+35","3p-45 +35= 140","140=3p-45+35","3p-10 = 140","140=3p-10","3p = 140+10","140+10=3p","3p = 150","150=3p"]}}
b
Solve the equation to find how many books the class had at the start.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"books","answer":{"text":["50"]}}
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.
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
b
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.
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["s"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2(s-10)=\\frac{4}{3}s+10","\\frac{4}{3}s+10=2(s-10)","2s-2*10=\\frac{4}{3}s+10","\\frac{4}{3}s+10=2s-20","\\frac{4}{3}s+10=2s-20"]}}
b
Solve the equation to find the starting number of seedlings.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"seedlings","answer":{"text":["45"]}}
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. 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
b
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.
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, we will 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.
Share
Report error
Translate
Pop Quiz
Solving Different Equations
Solve the given equation for the indicated variable. If necessary, round the answer to two decimal places.
Share
Report error
Translate
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.
Share
Report error
Translate
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["t"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["\\frac{2}{3}t+20=\\frac{2}{3}(t+30)","\\frac{2}{3}(t+30)=\\frac{2}{3}t+20"]}}
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.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["g"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2(g-10)=2g+15","2g+15=2(g-10)"]}}
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 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)
b
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
c
To find the number of solutions of the equations from Parts A and B, let's solve them.
Equation from Part A
We will start with 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.
Equation from Part B
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.
Conclusion
The following statements are true.
- The equation from Part A has infinitely many solutions.
- The equation from Part B has no solutions.
Share
Report error
Translate
Pop Quiz
Determining the Number of Solutions to an Equation
Share
Report error
Translate
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 $280.
a
Write an equation in terms of w that represents the situation.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["w"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2(w+65)-50=280","280=2(w+65)-50","2w+130-50=280","280=2w+130-50","2w+80=280","280=2w+80","2w=280-80","280-80=2w","2w=200","200=2w"]}}
b
Solve the equation to find the initial amount of money Tiffaniqua had during the project.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"$","formTextAfter":null,"answer":{"text":["100"]}}
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.
Share
Report error
Translate
Writing and Solving Multi-Step Equations
Exercises
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["r"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["-4(r-3)-6=15","15=-4(r-3)-6"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["6+3(x-5)=8","8=6+3(x-5)"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["v"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["7(v+5)+2v=12","12=7(v+5)+2v"]}}
security
report
code
security
report
code
We have been given a statement and we want to translate it into an equation.
Subtract 6 from the product of - 4 and the difference of r and 3. The result is 15.
Every equation has an equals sign and values or expressions on either side of it. Key phrases, such as is,
is equal to,
and equals
tell us about the placement of the equals sign. Let's look for a keyword in the given sentence and replace it with the equals sign.
Subtract6 from the product of- 4and the difference ofr and3. The result is15.
⇕
Subtract6 from the product of- 4and the difference ofr and3 = 15
On the left-hand side we have three keywords — subtract,
product,
and difference.
The word difference
tells us that subtraction will be used in our equation.
Subtract6 from the product of- 4and the difference of r and 3
⇕
Subtract6 from the product of- 4and r - 3
The next keyword, product,
tells us that multiplication is also used.
Subtract6 from the product of - 4 and r-3
⇕
Subtract6 from - 4 * (r - 3)
Finally, the keyword subtract
tells us that subtraction will be used again.
Subtract 6 from -4(r-3)
⇕
-4(r-3) - 6
On the right-hand side, we have the number 15. If we put these sides together, we have a complete equation!
-4(r-3)-6= 15
We want to translate the given sentence into an equation.
The sum of 6 and the product of 3 and the difference of x and 5 is equal to 8.
Like every equation, our equation will have an equals sign and values or expressions on either side of it. Key phrases, like is,
is equal to,
and equals
tell us where the equals sign should be placed.
The sum of6and the product of3and the difference ofxand5 is equal to 8.
⇕
The sum of6and the product of3and the difference ofxand5 = 8
On the left-hand side, we have three keywords — sum,
product,
and difference.
The keyword difference
means that we will use subtraction in our equation.
The sum of6andthe product of3and the difference of x and 5
⇕
The sum of6andthe product of3and x - 5
The next keyword, product,
means that we will use multiplication in our equation.
The sum of6and the product of 3 and x - 5
⇕
The sum of6and 3 * (x - 5)
Finally, the keyword sum
tells us to use addition.
The sum of 6 and 3(x - 5)
⇕
6 + 3(x - 5)
On the right-hand side we have the number 8. Putting the two sides together, we have a complete equation.
6+3(x-5)= 8
Let's translate the last sentence into an equation.
The sum of the product of 7 and the sum of v and 5 and the product of 2 and v equals 12.
First, let's look for keywords to tell us where to place the equals sign.
The sum of the product of7and the sum ofvand5and the product of2andv equals 12.
⇕
The sum of the product of7and the sum ofvand5and the product of2andv = 12.
There are two keywords on the left-hand side of the equation. They are sum
and product,
and both appear twice. Sum
means that we will use addition in our equation.
The sum of the product of7and the sum of v and 5and the product of2 andv
⇕
The sum of the product of7and v + 5 and the product of2 andv
The next keyword is product,
so we will use multiplication in our equation.
The sum of the product of 7 and v + 5 and the product of 2 and v
⇕
The sum of 7 * (v + 5) and 2 * v
The final keyword sum
means that we will use addition again.
The sum of 7(v+5) and 2v
⇕
7(v+5) + 2v
We have rewritten the left-hand side of the equation. Finally, let's put 12 on the right-hand side of the equals sign and finish our equation.
7(v+5)+2v= 12
security
report
code
E
Hint & Answer
S
Solution
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["w"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["2+3(w-10)=5w+7","5w+7=2+3(w-10)"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["s"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["7(s-9)=s+3","s+3=7(s-9)"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["u"],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["17+6(u-9)=-5(u+2)+12","-5(u+2)+12=17+6(u-9)"]}}
security
report
code
security
report
code
We have been given a sentence and we want to translate it into an equation.
The sum of 2 and the product of 3 and the difference of w and 10 is the sum of the product of 5 and w and 7.
Every equation has an equals sign and values or expressions on either side of it. Key phrases such as is,
is equal to,
and equals
tell us about the placement of the equals sign. Let's look for these keywords in the given sentence and replace them with the equals sign.
The sum of2 and the product of3and the difference ofw and10 isthe sum of the product of5andwand7.
⇕
The sum of2 and the product of3and the difference ofw and10 = the sum of the product of5andwand7.
On the left-hand side we have three keywords — sum,
product,
and difference.
The keyword difference
tells us that subtraction will be used in our equation.
The sum of2 and the product of3and the difference of w and 10
⇕
The sum of2 and the product of3and w - 10
The next keyword, product,
tells us that multiplication is also used.
The sum of2 and the product of 3 and w - 10
⇕
The sum of2 and 3 * (w - 10)
Finally, the keyword sum
tells us that addition will be used.
The sum of 2 and 3(w - 10)
⇕
2 + 3(w-10)
On the right-hand side, we have the keywords sum
and product.
Remember that product
means to use multiplication.
the sum of the product of 5 and wand7
⇕
the sum of 5 * wand7
Sum
indicates addition.
the sum of 5w and 7
⇕
5w + 7
Finally, let's put these sides together to complete our equation.
2+3(w-10)=5w+7
We want to translate the given sentence into an equation.
The product of 7 and the difference of s and 9 is equal to the sum of s and 3.
Like every equation, our equation will have an equals sign and values or expressions on either side of it. Key phrases such as is,
is equal to,
and equals
tell us where the equals sign should be placed. Let's look for these words to find where to place our equals sign.
The product of7and the difference ofsand9 is equal tothe sum ofsand 3.
⇕
The product of7and the difference ofsand9 = the sum ofsand 3.
We have two keywords on the left-hand side of the equation, product
and difference.
The keyword difference
means that we will use subtraction in our equation.
The product of7and the difference of s and 9
⇕
The product of7and s - 9
The next keyword, product,
means that we will use multiplication in our equation.
The product of 7 and s - 9
⇕
7 * (s-9)
The only keyword on the right-hand side is sum,
which indicates addition.
the sum of s and 3
⇕
s + 3
Let's put the two sides together to complete our equation.
7(s-9)=s+3
Let's translate the final sentence into an equation.
The sum of 17 and the product of 6 and the difference of u and 9 equals the sum of the product of - 5 and the sum of u and 2 and 12.
Let's start by finding the key phrase that tells us where to place the equals sign.
The sum of17and the product of6and the difference ofuand9 equalsthe sum of the product of-5and the sum of u and2and 12.
⇕
The sum of17and the product of6and the difference ofuand9 = the sum of the product of-5and the sum of u and2and 12.
On the left-hand side, there are three keywords — sum,
product,
and difference.
The keyword difference
tells us to use subtraction in our equation.
The sum of17and the product of6and the difference of u and 9
⇕
The sum of17and the product of6and u - 9
Product
means that we will use multiplication in our equation.
The sum of17and the product of 6 and u - 9
⇕
The sum of17and 6 * (u-9)
The final keyword on this side is sum,
so we will use addition.
The sum of 17 and 6(u-9)
⇕
17 + 6(u-9)
On the right-hand side, we have the key words sum
and product.
The keyword sum
means we will use addition. This word appears twice on this side of the equation, so but let's skip the first one for now.
the sum of the product of-5and the sum of u and 2and 12
⇕
the sum of the product of-5and u + 2and 12
The next keyword, product,
tells us to use multiplication.
the sum of the product of -5 and u + 2and 12
⇕
the sum of -5 * (u+2)and 12
Now we will replace the first use of sum.
the sum of -5(u+2) and 12
⇕
-5(u+2) + 12
If we put the two sides together, we have a complete equation.
17+6(u-9)=-5(u+2)+12
security
report
code
E
Hint & Answer
S
Solution
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"y =","formTextAfter":null,"answer":{"text":["6"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"s =","formTextAfter":null,"answer":{"text":["-8"]}}
security
report
code
security
report
code
We have to isolate the variable y on one side to solve the given equation. -6(y-7)+4=10 In this equation, 7 is subtracted from y, and the difference is then multiplied by - 6. Finally, 4 is added to the product. To solve this equation, we use the inverse operations of the operations applied to the variable in reverse order. This means first subtracting 4 from both sides of the equation using the Subtraction Property of Equality.
We can undo the multiplication next by dividing both sides of the equation by - 6. This does not change the solution because of the Division Property of Equality.
Finally, we undo the subtraction using the Addition Property of Equality.
Our calculations show that the solution to the given equation is y=6. To check that this solution is correct, let's substitute y=6 into the original equation and simplify.
The left-hand side and right-hand side are equal, so y=6 is the correct solution.
For this equation, we need to isolate s on one side. -34 = 9(s+4)+2 First, let's simplify the equation using the Distributive Property and combining like terms.
In this equation, the variable s is multiplied by 9 and 38 is added to the result. We can solve the equation by undoing the operations applied to the variable in reverse order. First, we use the Subtraction Property of Equality to subtract 38 from both sides of the equation.
Next, the multiplication can be undone by dividing both sides of the equation by 9. The Division Property of Equality ensures that the solution remains the same.
The solution to the given equation is s=-8. Finally, let's substitute s=-8 into the original equation and check if our answer is correct.
Substituting s=- 8 resulted in a true statement, so we can conclude that s=- 8 is the correct solution!
security
report
code
E
Hint & Answer
S
Solution
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"r =","formTextAfter":null,"answer":{"text":["0"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"x =","formTextAfter":null,"answer":{"text":["3"]}}
security
report
code
security
report
code
We have to isolate the variable r on one side to solve the given equation. -2r-7=r-7 In this equation, there are variable terms on both sides of the equation. We will start by collecting all the variable terms on one side. Let's use the Subtraction Property of Equality to subtract r from both sides of the equation.
In this equation, the variable is multiplied by - 3 and then 7 is subtracted from the product. Let's use the inverse operations of the operations applied to the variable in reverse order. This means first adding 7 to both sides of the equation using the Addition Property of Equality.
Next, we can undo the multiplication by dividing both sides of the equation by - 3. This does not change the solution because of the Division Property of Equality.
Our calculations show that the solution to the given equation is r=0. To check that this solution is correct, let's substitute r=0 into the original equation and simplify.
The left-hand side and right-hand side are equal, so r=0 is the correct solution.
Let's solve the given equation. -4(x-3)=2x-6 First, let's simplify the equation using the Distributive Property and combining like terms.
There are variable terms on both sides of the equals sign, so let's collect all the variable terms on one side. We can do this using the Subtraction Property of Equality.
Next, we continue solving the equation by undoing the operations applied to the variable in reverse order. Let's use the Subtraction and Division Properties of Equality.
The solution to the given equation is x=3. Finally, let's substitute x=3 into the original equation and check if our answer is correct.
Substituting x=3 resulted in a true statement, so we can conclude that x=3 is the correct solution!
security
report
code
E
Hint & Answer
S
Solution
{"type":"choice","form":{"alts":["One Solution","Infinitely Many Solutions","No Solutions"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
{"type":"choice","form":{"alts":["One Solution","Infinitely Many Solutions","No Solutions"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
{"type":"choice","form":{"alts":["One Solution","Infinitely Many Solutions","No Solutions"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
security
report
code
security
report
code
We are asked to determine how many solutions the given equation has. 9(x+3)-2=9x+25 We will try to solve the equation. If we succeed, this means that the equation has one solution. If we arrive at an identity, the equation has infinitely many solutions, and if we arrive at a false statement, the equation has no solutions. First, let's simplify the equation using the Distributive Property.
There are variable terms on both sides of the equation, so we need to collect them all on one side. Let's use the Subtraction Property of Equality to subtract 9x from both sides of the equation.
This results in a true statement. This means that every number is a solution to the given equation, so there are infinitely many solutions.
We want to find out how many solutions the given equation has. 9(s+7)-6=7s+3 Let's try to solve the equation. We will begin by using the Distributive Property to start simplifying the left-hand side of the equation.
Notice that there are variable terms on both sides of the equation again. Let's the Subtraction Property of Equality to gather them all on one side.
Now we will continue solving the equation by undoing the operations applied to the variable in reverse order. Let's use the Subtraction and Division Properties of Equality.
The solution to the given equation is s=-27. The equation has exactly one solution!
Let's determine how many solutions the following equation has. 5(s-1)-7=5s-7 Remember, if the equation simplifies into a true statement, the equation has infinitely many solutions. If it simplifies into a false statement, there are no solutions. Otherwise, there is one solution to the equation. Let's start simplifying the equation by using the Distributive Property.
Let's collect all the variable terms on one side. We will use the Subtraction Property of Equality to subtract 5s from both sides of the equation.
This results in a false statement, meaning that the given equation has no solutions.
security
report
code
E
Hint & Answer
S
Solution
Writing and Solving Multi-Step Equations
Recommended exercises
To get personalized recommended exercises, please login first.
Loading content
Edit Lesson
≥
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
Loading content