3. Solving Multi-Step Equations in One Variable
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Chapter 1
3.
Solving Multi-Step Equations in One Variable
In algebra, multi step equations hold significant importance. These equations often necessitate multiple operations to derive the solution. The lesson elaborates on the nuances of tackling such equations, emphasizing the pivotal role of properties of addition and multiplication. For instance, the commutative property indicates that the sequence in which numbers are added or multiplied doesn't alter the outcome. The distributive property, on the other hand, assists in breaking down expressions with parentheses. Everyday scenarios, such as determining the total cost of items or calculating discounts, frequently involve these multi step equations. By grasping the techniques showcased, individuals can proficiently address algebraic challenges and implement them in real-world situations.
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Lesson Settings & Tools
| | 13 Theory slides |
| | 13 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Multi-Step Equations in One Variable
Slide of 13
When it comes to modeling daily life situations using equations, it is common to end with equations that require more than one step to be solved. For such cases, it is convenient to know which properties can be applied to solve the obtained equations. To learn about it, different types of equations will be set and solved along this lesson.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Writing Equation for Number of Tickets Bought
Mark wants to buy tickets online to play paintball with his friends next weekend. On the website, he finds the following price list.
a Mark and his friends decide to choose Course 1 and pay $103. Let t be the number of tickets bought. Write an equation in terms of t that models this situation.
b Mark and his friends plan to go again next month but on Course 2. They were told that when the group has more than eight people, the processing fee of five tickets is free. Write an equation modeling the situation in which the group gets the discount and pays $230.50.
c How many tickets did Mark buy for his first visit? How many tickets will Mark buy next month?
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Challenge
Matching Equivalent Expressions
Two numerical expressions are equivalent if one of the expressions is obtained from the other by applying properties of addition and multiplication. Knowing this, match each expression in the left column with the equivalent expression.
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Discussion
Properties of Addition and Multiplication
Apart from the Properties of Equality, the properties of addition and multiplication are also heavily used when rewriting and simplifying expressions or equations. Thus, it is convenient to know these properties. The commutative property of each operation will be discussed first.
Rule
Commutative Property of Addition
The order in which two or more terms are added does not affect the value of the sum. In other words, the addends can be written in any order.
a+ b= b+ a
For example, adding 3 to 6 produces the same result as adding 6 to 3. In both cases, the sum is 9. This property also applies to the sum of more than two terms. 4+ 5+ 1 &= 1+ 5+ 4 &⇓ 10 &= 10 ✓
Since the Commutative Property of Addition is an axiom, it does not need a proof.Rule
Commutative Property of Multiplication
The order in which two or more factors are multiplied does not affect the value of the product. That is, the multiplicands can be written in any order.
a* b = b* a
For example, multiplying 5 by 4 produces the same result as multiplying 4 by 5. In both cases, the product is 20. This property also applies to the product of more than two terms. 3* 2* 6 &= 2* 6 * 3 &⇓ 36 &= 36 ✓
Since the Commutative Property of Multiplication is an axiom, it does not need a proof.As seen, the commutative property says that the order in which two or more terms are added or multiplied will not change the resulting sum or product. Next, the associative property of addition and multiplication will be presented.
Rule
Associative Property of Addition
The way three or more terms are grouped when added does not affect the value of the sum.
( a+ b)+ c= a+( b+ c)
For example, consider the sum 3+9+4. Grouping 3+9 and adding it to 4 produces the same result as grouping 9+4 and adding it to 3. ( 3+ 9)+ 4&= 3+( 9+ 4) &⇓ 12 + 4 &= 3 + 13 &⇓ 16 &= 16 ✓
Since the Associative Property of Addition is an axiom, it does not need a proof.Rule
Associative Property of Multiplication
The way three or more factors are grouped when multiplied does not affect the value of the product.
( a* b)* c= a*( b* c)
For example, consider the product 2* 4* 6. Grouping 2* 4 and multiplying it by 6 produces the same result as grouping 4* 6 and multiplying it by 2. ( 2* 4)* 6&= 2*( 4* 6) &⇓ 8* 6&= 2* 24 &⇓ 48 &= 48 ✓
Since the Associative Property of Multiplication is an axiom, it does not need a proof.
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Pop Quiz
Recognizing Equivalent Expressions
Two expressions are equivalent when one of them is obtained by applying properties of addition and multiplication to the other. For each pair of given expressions, determine whether they are equivalent. In the affirmative case, name the property that transforms one expression into the other.
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Discussion
Substituting Expressions
When solving equations, there are instances in which some parts of an expression can be substituted with an equivalent expression. As an example, consider the following equation. 5t + (9-3) = 30 Since 9-3 equals 6, the second term on the left-hand side can be substituted with 6. 5t + (9-3)^6 = 30 ⇓ 5t + 6 = 30 The previous simplification is able thanks to the Substitution Property of Equality, which also guarantees that the resulting equation and the initial are equivalent.
Rule
Substitution Property of Equality
If two real numbers are equal, then one can be substituted for another in any expression.
If a=b, then a can be substituted for b in any expression.
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Example
Temperature Conversion
This morning in science class, Tadeo and Magdalena learned how to convert degrees Celsius to Fahrenheit using the following formula. Celsius to Fahrenheit: F = 9/5C+32 Later, while eating lunch in the schoolyard, they saw that the thermometer indicated a temperature of 77^(∘)F. They wondered what the temperature would be in Celsius. To make the conversion, they used the formula they learned in the morning. However, they got different results.
Who is correct?
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Hint
Apply the Substitution Property of Equality to determine who got the correct answer. Start by substituting 77 for F into the conversion formula. Then, substitute the values Tadeo and Magdalena found for C. If a true statement is obtained, the corresponding value is a solution; otherwise, it is not.
Solution
Tadeo and Magdalena want to convert 77^(∘)F into degrees Celsius. Therefore, by the Substitution Property of Equality, 77 can be substituted for F in the conversion formula.
77 = 9/5C+32
Again, by the Substitution Property of Equality, the values found by Tadeo and Magdalena will be substituted for C in the above equation one at a time.
Tadeo:& 23^(∘)C Magdalena: & 25^(∘)C
After substituting each value and simplifying, if a true statement is obtained, the substituted value is a solution of the equation. Otherwise, it is not. First, C= 23 will be substituted.
77 = 9/5C+32
▼
Substitute 23 for C and simplify
Substitute
C= 23
77 ? = 9/5( 23)+32
MoveRightFacToNum
a/c* b = a* b/c
77 ? = 9* 23/5+32
Multiply
Multiply
77 ? = 207/5+32
CalcQuot
Calculate quotient
77 ? = 41.4+32
AddTerms
Add terms
77 = 73.4 *
A false statement was obtained, which suggests that Tadeo's answer is not correct. Next, substitute Magdalena's answer.
77 = 9/5C+32
▼
Substitute 25 for C and simplify
Substitute
C= 25
77 ? = 9/5( 25)+32
MoveRightFacToNum
a/c* b = a* b/c
77 ? = 9* 25/5+32
Multiply
Multiply
77 ? = 225/5+32
CalcQuot
Calculate quotient
77 ? = 45+32
AddTerms
Add terms
77 = 77 ✓
A true statement was obtained, which implies that Magdalena is correct in saying that 77^(∘)F is the same as 25^(∘)C.
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Discussion
Distributive Property of Multiplication
In addition to the commutative and associative properties, multiplication has another useful property that helps in the process of simplifying expressions with parentheses. This property is called the Distributive Property.
Rule
Distributive Property
Multiplying a number by the sum of two or more addends produces the same result as multiplying the number by each addend individually and then adding all the products together.
a( b ± c) &= a* b ± a* c [1ex] ( b ± c) a &= a* b ± a* c
Note that the factor outside the parentheses is multiplied, or distributed, to every term inside. The Distributive Property is used to simplify expressions with parentheses.
Note that, by the Symmetric Property of Equality, the Distributive Property can also be used to take out a common factor in two or more addends. 9x+ 9* 2 = 9(x+2)
In some cases, the right-hand side expression is more useful than the one on the left.
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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.
rl Equation (I):& 3/2(x-2)+5 = 14 [0.8em] Equation (II):& y+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. 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. In this case, 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. 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.
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
1* x = 12*2/3
OneMult
1* a=a
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.
y = 15/5
CalcQuot
Calculate quotient
y = 3
In conclusion, y=3 is the solution of Equation (II).
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Example
Finding the Usual Price Based on the Discounted Price
At Ali's fruit store, strawberries are on sale! Today they are $0.60 cheaper per pound than usual. Magdalena stopped at the store on her way home from school and bought eight pounds of strawberries. She paid a total of $14.40.
a Let x be the usual price of one pound of strawberries. Write an equation, in terms of x, modeling Magdalena's purchase.
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b If Tadeo bought six pounds of strawberries two weeks ago, how much did he pay?
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Hint
a Today, the special price per pound of strawberries is x-0.60 dollars. Multiply this price by the number of pounds bought by Magdalena and equate it to the total amount she paid.
b Because Tadeo bought strawberries two weeks ago, he paid the usual price for them. Solve the equation set in Part A to find the usual price. Then, multiply it by the number of pounds bought by Tadeo.
Solution
a Let x be the usual price of one pound of strawberries. According to Ali's offer, the special price is $0.60 cheaper per pound than the usual price. Therefore, the special price equals the usual price minus 0.60.
Special price = Usual price - 0.60 ⇓ Special price = x - 0.60 With this special price, Magdalena paid $14.40 for eight pounds of strawberries. Consequently, 8 multiplied by the special price is equal to the total amount paid. Using this information, an equation in terms of x can be established. ( c Number of pounds ) ( c Special price ) = c Total paid ⇓ 8( x - 0.60) = 14.40 This equation can be used to determine the usual price of a pound of strawberries.
b Since Tadeo bought the strawberries two weeks ago, he paid the usual price. Therefore, start by finding the usual price by solving the equation set in Part A.
8(x-0.60) = 14.40 First, apply the Distributive Property to clear the parentheses. Then, apply the Properties of Equality to isolate the x variable.
8(x-0.60) = 14.40
Distr
Distribute 8
8x-8* 0.60 = 14.40
Multiply
Multiply
8x-4.8 = 14.40
AddEqn
LHS+4.8=RHS+4.8
8x = 19.20
DivEqn
.LHS /8.=.RHS /8.
x = 19.20/8
CalcQuot
Calculate quotient
x = 2.40
This means that the usual price of one pound of strawberries is $ 2.40. To find how much Tadeo paid, multiply 6 — the number of pounds he bought — by the usual price.
Consequently, last week Tadeo paid $14.40 for six pounds of strawberries. Note that Tadeo paid the same as Magdalena, but he bought 2 pounds fewer.
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Example
Solving an Equation With Variables on Both Sides
After dinner, Magdalena works on a 3D model of a castle using cardboard for a history project. She wants the walls of the ramparts to be 7 inches wide. Additionally, she wants the value of the perimeter of the wall, in inches, to be the same as the value of its area, in square inches.
How high should the wall be?
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Hint
The area of the wall is 7x minus the area of the two small rectangles at the top. Equate the area expression with the perimeter of the wall. Then, solve the equation for x.
Solution
Since Magdalena wants the wall to have the same perimeter and area, regardless of the units of measure, start by finding the perimeter of the wall. To do so, add all the side lengths of the wall. Notice that the shorter sides of the small rectangles have a length of 1 inch. Knowing this, the length of the top of the wall can be found.
The right-hand side of the wall has the same length as the left-hand side of the wall — that is, both have a length of x inches. Adding these two lengths to the base length, 7 inches, and to the length of the top of the wall, the perimeter of the wall will be obtained.
P = x+7+x+Top Length
▼
Substitute 11 for Top Length and simplify
Substitute
Top Length= 11
P = x+7+x+ 11
CommutativePropAdd
Commutative Property of Addition
P = x+x+7+11
AssociativePropAdd
Associative Property of Addition
P = (x+x)+(7+11)
AddTerms
Add terms
P = 2x+18
To find the wall's area, note that it is made by cutting two small rectangles from the bigger rectangle, whose dimensions are 7 by x.
Consequently, the wall's area is equal to the area of the big rectangle minus the area of the two small rectangles. The big rectangle area is 7x, and the area of each small rectangle is 2 square inches.
A = A_(Big) - 2* A_(Small)
SubstituteII
A_(Big)= 7x, A_(Small)= 2
A = 7x - 2* 2
Multiply
Multiply
A = 7x - 4
Since the perimeter and area of the wall must be the same, equate P and A. P &= A &⇓ 2x+18 &= 7x - 4 To find the wall's height, solve the equation for x. Since the equation has variables on both sides, start by collecting them on one side of the equation.
2x+18 = 7x - 4
SubEqn
LHS-2x=RHS-2x
18 = 7x - 4 - 2x
CommutativePropAdd
Commutative Property of Addition
18 = 7x - 2x - 4
SubTerms
Subtract terms
18 = 5x - 4
AddEqn
LHS+4=RHS+4
18 + 4 = 5x
AddTerms
Add terms
22 = 5x
DivEqn
.LHS /5.=.RHS /5.
22/5 = x
CalcQuot
Calculate quotient
4.4 = x
RearrangeEqn
Rearrange equation
x = 4.4
Therefore, the wall should be 4.4 inches high.
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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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Closure
Finding the Number of Tickets Bought
At the beginning of the lesson, Mark was buying some paintball tickets. He found the following prices online, where the $4 service charge is per transaction, not per ticket.
a Write an equation modeling the situation in which Mark and his friends chose Course 1, got no discount, and paid $103.
b Write an equation modeling the situation in which Mark and his friends chose Course 2, got the discount, and paid $230.50.
c How many tickets did Mark buy in the first situation? How many tickets did Mark buy in the second?
Answer
a Example Equation: 15t+1.5t+4=103
b Example Equation: 18t+1.5(t-5)+4=230.5
c Mark bought 6 tickets in the first situation and 12 in the second.
Hint
a Find the amount paid only for the tickets, without fees. Next, find the amount paid in fees. Then, add these two costs to the service charge. Finally, equate the resulting expression with the actual money paid.
b Find the amount paid only for the tickets, without fees. Next, find the amount paid in fees. Remember, five of the tickets paid no fee. Then, add these two costs to the service charge. Finally, equate the resulting expression with the actual money paid.
c Solve the corresponding equation for t. Use the Distributive Property, the Associative Property of Addition, the Substitution Property of Equality, and the Properties of Equality.
Solution
a Start by listing the prices corresponding to Course 1.
| Course 1 | |
|---|---|
| Ticket price | $15.00 |
| Processing fee per ticket | $1.50 |
| Service charge | $4.00 |
Let t be the number of purchased tickets. Then, the amount paid only for the tickets without any fees will be the product of 15 and t. 15t Since the group did not get a discount, they paid the processing fee for all the tickets bought. To find the amount paid only in fees, multiply the price of the processing fee by t. 1.5t They also had to pay a service charge, which is $4 no matter the number of tickets bought. The sum of the three costs together will give an expression representing the total cost of purchasing t tickets for Course 1. 15t+1.5t+4 Finally, it is said that they paid a total of $103. Therefore, equate the previous expression to 103.
15t+1.5t+4 = 103
This equation models the situation in which Mark and his friends chose Course 1, got no discount, and paid $103.
b This time, the group decided to play on Course 2. Recall the prices corresponding to Course 2.
| Course 2 | |
|---|---|
| Ticket price | $18.00 |
| Processing fee per ticket | $1.50 |
| Service charge | $4.00 |
Let t be the number of purchased tickets. Since the price per ticket is $18.00, the amount paid only for the tickets — without any fees — will be the product of the price per ticket and t. 18t Since the group received a discount this time, they did not pay the fee for five of the tickets bought. Thus, to find the amount paid in fees, multiply the processing fee by (t-5). 1.5(t-5) They also paid a service charge, which is $4 no matter the number of tickets bought. By adding the three costs, an expression representing the money paid for the purchase of t tickets for Course 2 will be found. 18t+1.5(t-5)+4 Finally, it is said that they paid a total of $230.50. Therefore, equate the previous expression to 230.5.
18t+1.5(t-5)+4 = 230.5
This equation models the situation in which Mark and his friends chose Course 2, got a discount, and paid $230.50.
c To find the number of tickets bought by Mark the first time, the equation set in Part A will be solved for t.
15t+1.5t+4 = 103
▼
Solve for t
AssociativePropAdd
Associative Property of Addition
(15t+1.5t)+4 = 103
AddTerms
Add terms
16.5t+4 = 103
SubEqn
LHS-4=RHS-4
16.5t = 99
DivEqn
.LHS /16.5.=.RHS /16.5.
t = 99/16.5
CalcQuot
Calculate quotient
t = 6
Consequently, the first time Mark bought 6 tickets. This confirms why he did not get the discount. To find the number of tickets he bought the second time, the equation set in Part B will be solved for t.
18t+1.5(t-5)+4 = 230.5
▼
Solve for t
Distr
Distribute 1.5
18t+1.5t-1.5* 5+4 = 230.5
Multiply
Multiply
18t+1.5t-7.5+4 = 230.5
CommutativePropAdd
Commutative Property of Addition
18t+1.5t+4-7.5 = 230.5
AssociativePropAdd
Associative Property of Addition
(18t+1.5t)+(4-7.5) = 230.5
AddSubTerms
Add and subtract terms
19.5t+(-3.5) = 230.5
AddNeg
a+(- b)=a-b
19.5t-3.5 = 230.5
AddEqn
LHS+3.5=RHS+3.5
19.5t= 234
DivEqn
.LHS /19.5.=.RHS /19.5.
t = 234/19.5
CalcQuot
Calculate quotient
t = 12
In conclusion, Mark bought 12 tickets the second time.
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Solving Multi-Step Equations in One Variable
Exercise 3.1
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In the given equation, only x is a variable. Therefore, we can treat the rest of the terms as numbers. The first thing we note is that there are no fractions and the variable appears on both sides of the equation. k x + 1 = 2 x + 4 Let's start by collecting all the terms that contain x on the left-hand side and the rest of the terms on the right-hand side.
To isolate x, we factor it out on the left-hand side.
From this, the solution to the equation is 3k-2.
Let's start by writing the solution found in the previous part.
x = 3/k-2
Since the solution is a fraction and fractions are undefined when the denominator is zero, we can say that the equation has no solution when the denominator is zero.
k - 2 = 0
By solving this equation for k, we get k=2. This implies the fraction is undefined for k=2. In other words, the equation has no solution when k=2. To check our answer, let's substitute 2 for k into the given equation and try to solve it for x.
We got a false statement, which confirms that when k=2, the equation has no solution. Note that this means that the equation has a solution for any value of k other than 2.
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Hint & Answer
S
Solution
Scientists weighed three diamond stones labeled as S, T, and U and recorded the following information.
| Stone | Information |
|---|---|
| S | Weighs 16 grams more than double the weight of stone T. |
| U | Weighs 40 grams less than three times the weight of stone T. |
The average weight of the three stones is 102 grams. How much does the heaviest stone weigh?
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Note that the weights of stones S and U are given in terms of the weight of stone T. Therefore, let's start by representing the weight of stone T with the variable x. T → x From the given information, the stone S weighs 16 grams more than double the weight of stone T. This means that the stone S weights 16 grams more than 2x, which is the same as saying that stone S weighs 2x+16 grams. T & → x S & → 2x+16 Also, we are told that stone U weighs 40 grams less than three times the weight of stone T. Then, stone U weighs 40 grams less than 3x. In other words, the weight of stone T is 3x-40 grams. T & → x S & → 2x+16 U & → 3x-40 Now we have the three weights in terms of x. From the given information, the average weight of the three stones is 102 grams. This means that if we add the weights of the three stones and divide the sum by 3, the result is equal to 102. x + (2x+16) + (3x-40)/3 = 102 The value of x can be found by solving the obtained equation. To start, we multiply both sides by 3 to get rid of the fraction.
Next, we will combine like terms on the left-hand side. Then we will isolate the variable by using inverse operations.
The value of x is 55, which means that the stone T weighs 55 grams. With this information, we can find the weights of the other two stones. T & → 55 g S & → 2(55)+16 = 126 g U & → 3(55)-40 = 125 g We can see that the heaviest stone is stone S, whose weight is 126 grams.
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Hint & Answer
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Solution
Solving Multi-Step Equations in One Variable
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>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
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