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| 13 Theory slides |
| 13 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
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.
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.
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.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.
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.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.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.
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.
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.
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.
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.
C= 23
a/c* b = a* b/c
Multiply
Calculate quotient
Add terms
C= 25
a/c* b = a* b/c
Multiply
Calculate quotient
Add terms
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.
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, 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.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.
Distribute 3/2
a/2* 2 = a
Add terms
LHS * 2/3=RHS* 2/3
Commutative Property of Multiplication
a/b* b/a=1
1* a=a
a*b/c= a* b/c
Multiply
Calculate quotient
LHS+2y=RHS+2y
Commutative Property of Addition
Add terms
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.
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.
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.
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.
Top Length= 11
Commutative Property of Addition
Associative Property of Addition
Add terms
A_(Big)= 7x, A_(Small)= 2
Multiply
LHS-2x=RHS-2x
Commutative Property of Addition
Subtract terms
LHS+4=RHS+4
Add terms
.LHS /5.=.RHS /5.
Calculate quotient
Rearrange equation
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.
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.
Associative Property of Addition
Add terms
LHS-4=RHS-4
.LHS /16.5.=.RHS /16.5.
Calculate quotient
Distribute 1.5
Multiply
Commutative Property of Addition
Associative Property of Addition
Add and subtract terms
a+(- b)=a-b
LHS+3.5=RHS+3.5
.LHS /19.5.=.RHS /19.5.
Calculate quotient
In the equation kx + 1 = 2x+4, k is a constant and x is a variable.
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.
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. |
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.