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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
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
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)
The way three or more factors are grouped when multiplied does not affect the value of the product.
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.
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.
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
ca⋅b=ca⋅b
Multiply
Calculate quotient
Add terms
C=25
ca⋅b=ca⋅b
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.
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.
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 23
2a⋅2=a
Add terms
LHS⋅32=RHS⋅32
Commutative Property of Multiplication
ba⋅ab=1
1⋅a=a
a⋅cb=ca⋅b
Multiply
Calculate quotient
LHS+2y=RHS+2y
Commutative Property of Addition
Add terms
LHS+4=RHS+4
Add terms
LHS/5=RHS/5
Calculate quotient
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.
Distribute 8
Multiply
LHS+4.8=RHS+4.8
LHS/8=RHS/8
Calculate quotient
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
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 |
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 |
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
To make the tower of a cardboard castle, Zosia needs to cut out a triangle and a rectangle, both with the same area, whose dimensions are shown in the figure.
The area of a triangle is half the product of its base and its height. Let's start by writing an expression that represents the area of the triangle that Zosia needs. Triangle's Area [0.15cm] 1/2x* 40 = 40x/2 We need to know the value of x in order to calculate the area of the triangle. To find it, let's also write an expression representing the area of the rectangle that Zosia needs. We will use the fact that the area of a rectangle is equal to the length multiplied by the width. Rectangle's Area 12.5(x+18) Since the triangle and the rectangle have the same area, we can equate the expressions so that we get an equation in terms of x. 40x/2 = 12.5(x+18) We can find the value of x by solving the equation. To start, let's multiply both sides of the equation by 2 to get rid of the fractions.
Next, we can distribute 25 to clear all the parentheses.
Finally, we will group the variable terms on the left-hand side. We can then apply inverse operations to isolate x.
The value of x is 30, which means that the base of the triangle is 30 millimeters long. Knowing this, we can find the area of the triangle.
The area of the triangle is 600 square millimeters. Since the triangle and the rectangle have the same area, we could also substitute x=30 into the expression representing the area of the rectangle — we would get the same answer, though.
To determine the next even number, we need to find the value of the four consecutive even numbers. Let's start by calling the smallest of the four even numbers x. \begin{gathered} \bm{1}^\text{st} \to {\color{#0000FF}{x}} \end{gathered} Since the four numbers are even and consecutive, the second even number is two units greater than x. This means that we can write it as x+2. \begin{aligned} \bm{1}^\text{st} &\to {\color{#0000FF}{x}} \\ \bm{2}^\text{nd} &\to {\color{#009600}{x+2}} \end{aligned} The third even number is two units greater than the second one. Then, the third number equals x+2+2, which is the same as x+4. \begin{aligned} \bm{1}^\text{st} &\to {\color{#0000FF}{x}} \\ \bm{2}^\text{nd} &\to {\color{#009600}{x+2}} \\ \bm{3}^\text{rd} &\to {\color{#FD9000}{x+4}} \end{aligned} Similarly, the fourth even number is two units greater than the third one. Therefore, the fourth number equals x+4+2, or x+6. \begin{aligned} \bm{1}^\text{st} &\to {\color{#0000FF}{x}} \\ \bm{2}^\text{nd} &\to {\color{#009600}{x+2}} \\ \bm{3}^\text{rd} &\to {\color{#FD9000}{x+4}} \\ \bm{4}^\text{th} &\to {\color{#A800DD}{x+6}} \end{aligned} Now that we have expressions for each of the four consecutive even numbers in terms of x, we can add them and equate the sum to 92. x + ( x+2) + ( x+4) + ( x+6) = 92 We can find the value of x by solving the equation. Note that the equation has no fractions and the variable terms are all grouped on the left-hand side. This means that we can simply combine like terms and isolate the variable through inverse operations to solve the equation.
The value of x is 20, so the smallest of the four consecutive even numbers is 20. Knowing this, we can determine the next three even numbers. \begin{aligned} \bm{1}^\text{st} &\to {\color{#0000FF}{20}} \\ \bm{2}^\text{nd} &\to {\color{#009600}{22}} \\ \bm{3}^\text{rd} &\to {\color{#FD9000}{24}} \\ \bm{4}^\text{th} &\to {\color{#A800DD}{26}} \end{aligned} We can conclude that the fifth even number is 26+2=28.
To determine the length of the longest side of the triangle, we need to know its three side lengths. Let's try to write expressions representing the side lengths in terms of one variable. To start, we can call x the length of BC. BC → x We are told that AB is 3 centimeters shorter than BC, which allows us to write the length of AB in terms of x. ABis3centimeters shorter thanBC ⇓ AB → x-3 Also, we know that AC is twice as long as AB. This means that the length of AC is 2 times the length of AB. ACis twice as long asAB ⇓ AC → 2(x-3) Now we have the side lengths written in terms of only one variable. The perimeter of a triangle equals the sum of the three side lengths. We know that the perimeter of △ ABC is 23 centimeters. Therefore, the sum of the three expressions is equal to 23. x + x-3 + 2(x-3) = 23 Next, let's solve the equation for x. We will start by distributing 2 to clear all the parentheses and continue by combining like terms.
Finally, we isolate the variable on the left-hand side by applying inverse operations.
The value of x is 8. This means that BC is 8 centimeters long. Knowing this, the other two side lengths can be calculated. BC & → 8 cm AB & → 8-3 = 5 cm AC & → 2(8-3) = 10 cm As we can see, the longest side of △ ABC is AC, which is 10 centimeters long.
The following figure is made of four identical rectangles and one square whose area is half the area of a rectangle.
We are told that the figure is made of four identical rectangles and one square. Let x be the area of each rectangle. Since the area of the square is half the area of a rectangle, the area of the square is 12x. Rectangle's Area &→ x [0.3em] Square's Area &→ 1/2x There are four rectangles in the figure and the area of all of them together is 4x. We can find the total area of the figure by adding together the area of the four rectangles and the area of the square. We also know that this area is 54 square inches, so we will set this expression equal to 54. 4x + 1/2x = 54 Let's solve the equation to find the value of x. To start, we will multiply both sides of the equation by 2 to get rid of the fraction. Then, we will combine like terms and isolate the variable using inverse operations.
The area of each rectangle is 12 square inches. This means that the area of the square is 6 square inches.