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.
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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
In some equations, more than two operations are applied to the variable. These equations often require the use of properties of operations, like the Distributive Property, to solve. Sometimes, the variable appears on both sides of the equation. In these cases, inverse operations are used to transfer the variable terms all to one side. This lesson focuses on solving and writing multi-step equations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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 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.
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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Discussion
Solving Multi-Step Equations
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.
Equation (I):Equation (II):23(x−2)+5=14y+2y−4=11−2y
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.
Solving Equation (I)
Equation (I) contains a fraction and parentheses.23(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. In this case, distribute 23 to each term inside the parentheses. Then, combine like terms on the left-hand side.
23(x−2)+5=14
Distr
Distribute 23
23x−23⋅2+5=14
FracMultDenomToNumber
2a⋅2=a
23x−3+5=14
AddTerms
Add terms
23x+2=14
2
Isolate the Variable
expand_more Apply the Properties of Equality to isolate the variable on one side of the equation. In this case, first apply the Subtraction Property of Equality.
The coefficient of the variable is a fraction. The Multiplication Property of Equality can be used to multiply both sides of the equation by the reciprocal of the coefficient.
As such, x=8 is the solution of Equation (I).
23x=12
MultEqn
LHS⋅32=RHS⋅32
23x⋅32=12⋅32
CommutativePropMult
Commutative Property of Multiplication
23⋅32⋅x=12⋅32
MultFracByInverse
ba⋅ab=1
1⋅x=12⋅32
OneMult
1⋅a=a
x=12⋅32
MoveLeftFacToNum
a⋅cb=ca⋅b
x=312⋅2
Multiply
Multiply
x=324
CalcQuot
Calculate quotient
x=8
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.
In conclusion, y=3 is the solution of Equation (II).
5y−4=11
AddEqn
LHS+4=RHS+4
5y=11+4
AddTerms
Add terms
5y=15
DivEqn
LHS/5=RHS/5
y=515
CalcQuot
Calculate quotient
y=3
Example
Finding the Initial Value of an Investment
Tiffaniqua and her friend Kevin are talking about the stock market project. Kevin is telling her how his investment is doing after two weeks.
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b Solve the equation to find the amount Kevin invested in the stock.
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Hint
a The value of Kevin's investment after two weeks is 3 times its value after the first week.
b Use the Distributive Property and inverse operations to solve the equation.
Solution
a Let p be the starting value of Kevin's investment. Kevin said that during the first week, the value of the stock he bought fell by $15. This means that the value of the investment after the first week is the difference between p and 15.
Value After First Week:p−15
The value of the stock tripled the next week. In other words, during the second week, the value of Kevin's investment is three times the value at the first week.
Value After First Week:Value After Second Week:p−153(p−15)
If the value of the investment grows by $35 during the last week, it will be $35 more than the value after the second week.
Value After First Week:Value After Second Week:Value After Third Week:p−153(p−15)3(p−15)+35
The final value of Kevin's investment would be $140, which will complete the equation.
3(p−15)+35=140
b Solve the equation written in Part A to find the amount of money Kevin invested in the stock.
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
33p=3150
CrossCommonFac
Cross out common factors
33p=3150
SimpQuot
Simplify quotient
p=3150
CalcQuot
Calculate quotient
p=50
Example
Comparing Two Investments
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!
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b Solve the equation to find the amount Tearrik and Ramsha invested in the stocks.
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Hint
a Express the final value of each investment using m.
b Collect all the variable terms on one side of the equation.
Solution
a Let m be the starting value of each investment. The value of Tearrik's investment fell by $10 during the first week, then doubled during the second week. This means that the value of Tearrik's investment after two weeks can be expressed as 2(m−10).
Value of Tearrik’s Investment:2(m−10)
The value of Ramsha's investment grew by a third during the first week, so its value after the first week was 1m+31m=34m. During the second week, the value grew by $10. Then, the value of Ramsha's investment after two weeks is 34m+10.
Value of Ramsha’s Investment:34m+10
Since the final values of Tearrik's and Ramsha's investments are equal, an equation can be written by setting these two expressions equal to each other.
2(m−10)=34m+10
b Solve the equation written in Part A to find the amount of money that Tearrik and Ramsha invested in their respective stocks.
2(m−10)=34m+10
First, use the Distributive Property.
The resulting equation has variable terms on both sides. To solve the equation, all the variable terms are first collected on one side of the equation. Use the Subtraction Property of Equality to subtract 34m from both sides.
2m−20=34m+10
SubEqn
LHS−34m=RHS−34m
2m−20−34m=34m+10−34m
CommutativePropAdd
Commutative Property of Addition
2m−34m−20=34m−34m+10
Rewrite
Rewrite 2m as 36m
36m−34m−20=34m−34m+10
SubTerms
Subtract terms
32m−20=10
32m−20=10
AddEqn
LHS+20=RHS+20
32m−20+20=10+20
AddTerms
Add terms
32m=30
MultEqn
LHS⋅23=RHS⋅23
32m⋅23=30⋅23
CommutativePropMult
Commutative Property of Multiplication
32⋅23⋅m=30⋅23
MultFracByInverse
ba⋅ab=1
m=30⋅23
MoveLeftFacToNum
a⋅cb=ca⋅b
m=230⋅3
Multiply
Multiply
m=290
CalcQuot
Calculate quotient
m=45
Pop Quiz
Solving Different Equations
Solve the given equation for the indicated variable. If necessary, round the answer to two decimal places.
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. Now it is time to 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 instead, the equation has no solutions. In this case, it can be said that 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.
Example
Solving Equations With No Solutions or Infinitely Many Solutions
a Tiffaniqua and Zosia both invested the same amount of money in one stock each. After two weeks, they compare how their investments are doing.
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b Ali is tracking two different stocks. He compared how much each stock would be worth after two weeks if he invested g dollars into each at the beginning of the project.
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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]}
Hint
a Represent the unknown value using the variable. Write two different expressions that represent the ladies' investments.
b Write two expressions that represent the investments. Use the variable to represent the unknown value.
Solution
a Let t be the amount of money that Tiffaniqua and Zosia invested in their stocks. First, express the final value of Tiffaniqua's investment in terms of t. The investment lost a third of its value in the first week, so the value at this point is 32t. In the second week, the value grew by $20. The value of the investment after two weeks is 32t+20.
Value of Tiffaniqua’s Investment:32t+20
The value of Zosia's investment grew by $30 during the first week, so its value was t+30 at that point. Then the value fell by a third during the second week. This means that the value of Zosia's investment after two weeks is 32(t+30).
Value of Zosia’s Investment:32(t+30)
The values of Tiffaniqua's and Zosia's investments after two weeks are equal. This makes it possible to write an equation that models the given situation.
32t+20=32(t+30)
b Write an expression that describes the final value of each investment. The first stock would lose $10 in value during the first week, so its value would be g−10. The value would then double during the second week, so the final value would be 2(g−10).
First Stock:2(g−10)
The second stock doubled its value during the first week, so the value would be 2g at that point. In the second week, the value would grow by $15, so the final value is 2g+15.
Second Stock:2g+15
Ali wants to know how much money invested into both stocks would result in the final values being equal. The best way to find this information is to set the 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 solution.
32t+20=32(t+30)
Distr
Distribute 32
32t+20=32t+32⋅30
MoveRightFacToNum
ca⋅b=ca⋅b
32t+20=32t+32⋅30
Multiply
Multiply
32t+20=32t+360
CalcQuot
Calculate quotient
32t+20=32t+20
SubEqn
LHS−32t=RHS−32t
32t+20−32t=32t+20−32t
SubTerms
Subtract terms
20=20
- The equation from Part A has infinitely many solutions.
- The equation from Part B has no solutions.
Pop Quiz
Determining the Number of Solutions to an Equation
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 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.
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 To write an equation that models a real-life situation, a variable is used to represent an unknown quantity. In this situation, the unknown quantity is the amount of money that Tiffaniqua invested initially. This quantity can be represented by w.
Starting Amount of Money:w
Next, the final value of Tiffaniqua's stock can be expressed 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.
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
Start by analyzing the expression. Here, 65 is added to w and the result is then multiplied by 2. Finally, 50 is subtracted from the product. These operations can be undone using inverse operations. The first operation to be undone is the subtraction. Use the Addition Property of Inequality to undo the subtraction.
Then, the multiplication is undone using the Division Property of Equality.
2(w+65)=330
DivEqn
LHS/2=RHS/2
22(w+65)=2330
CrossCommonFac
Cross out common factors
22(w+65)=2330
SimpQuot
Simplify quotient
w+65=2330
CalcQuot
Calculate quotient
w+65=165
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 and then combining like terms.
In the resulting equation, the variable w is multiplied by 2 and then 80 is added to the result. These operations can be undone using inverse operations, starting with the addition. Use the Subtraction Property of Inequality to undo the addition.
Finally, the multiplication is undone using the Division Property of Equality.
The solution to the equation is w=100, the same result as before.
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
Writing and Solving Multi-Step Equations
Exercises
Write the given verbal expressions as equations.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"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":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["7(v+5)+2v=12","12=7(v+5)+2v"]}}
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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
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Write the given verbal expressions as equations.
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{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["s"],"constants":[]},"rendered":false},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"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":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["17+6(u-9)=-5(u+2)+12","-5(u+2)+12=17+6(u-9)"]}}
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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
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Solve the given equations. Check your answers.
-6(y−7)+4=10
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-34=9(s+4)+2
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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!
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Solve the given equations. Check your answers.
-2r−7=r−7
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-4(x−3)=2x−6
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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!
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Does each equation have one, infinitely many, or no solutions?
9(x+3)−2=9x+25
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9(s+7)−6=7s+3
{"type":"choice","form":{"alts":["One Solution","Infinitely Many Solutions","No Solutions"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
5(s−1)−7=5s−7
{"type":"choice","form":{"alts":["One Solution","Infinitely Many Solutions","No Solutions"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}
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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.
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Writing and Solving Multi-Step Equations
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