SLC-S22W1/Variables and Expressions
Hello everyone
I hope you all are doing well and enjoying your life . Today I am excited to share my homework task for this week contest. In this post I will explain some basic concepts of algebra such as variable expression and their types. I will also show you how to solve algebraic expression and apply them to real life problem.
As a tuition teacher I teach math to my students every day and explaining variable and expressions is something I often do. While we have already covered many types of variables and expressions in class I want to discuss two additional types: discrete variable and continuous variable. These concepts are very practical and easy to understand especialy with example from daily life.
| Task 1:Understanding Discrete & Continuou Variables with Practical & Algebraic Examples |
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1. Discrete Variables and Expressions
Explanation:
A discrete variable represent things that can be counte and only take specific separate value. These value are whole numbers and cannot have decimal or fractions. I often explain this to my students by using example like counting the number of student in a classroom.
Practical Example:
Imagine we are calculating the number of book on a shelf. you can have 5 10 or 15 book but you cannot have 10.5 book (assuming books are not split in half). The number of books is a discrete variable because it is countable and doesn’t include fractions.
Algebraic Example:
If xrepresents the number of book and each book cost $3 the total cost can be expresed as:
= 3x
Here x is a discrete variable because it only represent whole numbers. I often use such examples in class when teaching my students about multiplying whole number with variables.
2. Continuous Variables and Expressions
Explanation:
A continuous variable is different because it can take any value within a range including decimal and fraction. Continuous variables are commonly used when we measure thing like height weight or time.
Practical Example:
When I teach about measuring distances or time I tell my students that continuous variables can represent things like the time it takes to complete homework. For example student may finish their homework in 1.5 hours or 1.75 hours. Time is a continuous variable because it can take any value not just whole number.
Algebraic Example:
Suppose you are driving a car at a speed of 60 km/hour and represented the time in hours. The distance traveled can be expresed as:
Distance = 60t
Here t is a continuous variable because time can be any values like 2.5 or 3.75 hours. This kind of example help my students understand how continuous variables are used in real-life calculations.
How I Relate It to Teaching:
When teaching my student I often use relatable examples like counting the number of pencil (discrete) or measuring the time taken for a task (continuous). This makes it easier for them to understand the difference. Discrete variables help with understanding counting problems while continuous variables teach them how to work with measurement and fractios.
By connecting these mathematical idea to everyday situations my student learn not just theory but also how to apply these concepts in real life scenarios.
| Task 2: How I Evaluate an Algebraic Expresion When Variable Values Are Given |
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I frequently help my student understand how to evaluate algebraic expressions. Evaluating these expressions mean substituting the given values of the variable into the expression and solving it step by steps. Here my way of explaining and solving such problem in a clear and detailed manner
Step by Step Evaluation Proces
Example 1:
Expression: 2x + 3y - 5
Given Values: x = 4) y = 2
Step 1: Substitute the given value into the expression.
Replace x with 4 and with 2:
(2(4) + 3(2) - 5Step 2: Perform multiplication first.
Multiply ( 2 times 4 = 8 ) and ( 3 time 2 = 6):
( 8 + 6 - 5 )Step 3: Perform addition and subtraction in order.
First add ( 8 + 6 = 14 ). Then subtract ( 5 ):
( 14 - 5 = 9 )
Final Answer: The value of the expression is ( 9 ).
Example 2:
Expression: 3a^2 - 4b + 7
Given Valus: a = 2 b = 5
Step 1: Substitute the give values into the expresion.
Replace a with 2 and b with 5:
3(2^2) - 4(5) + 7Step 2: Solve the exponent first.
( 2^2 = 4 :
3(4) - 4(5) + 7Step 3: Perform multiplication.
Multiply 3 time 4 = 12 & 4 time 5 = 20:
12 - 20 + 7Step 4: Perform addition and subtraction in order.
First subtract 12 - 20 = -8 . Then add 7 :
-8 + 7 = -1
Final Answer: The value of the expression is -1 .
General Example from Daily Life
Let say a student is working on problem to calculate the cost of buying pen and notebook.
Expression: 5p + 3n
Given Values: p = 10 cost of a pen n = 15 cost of a notebook
Substitute p = 10 and n = 15
(5(10) + 3(15)Perform multiplication:
5 times 10 = 50 and 3 times 15 = 45
50 + 45Perform addition:
50 + 45 = 95
Final Result : The total cost is 95 .
| Task 3 |
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Simplify this expresion: 3(2x - 1) + 2(x + 4) - 5
Evaluate this expresion: (x^2 + 2x - 3) / (x + 1) when x = 2
Solve the following equations: 2x + 5 = 3(x - 2) + 1
| Task 4 part 1 Daily Profit Calculation for a Bakery |
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Step 1: Define the variables
Let:
- x = Number of white bread loaves sold
- The bakery sells 30 more whole wheat loave than white bread loave.
So the number of Whole wheat loave = x + 30.
The total number of loaves solds per day is 250. We can writes this as:
x + (x + 30) = 250
Step 2: Solve for x
Simplify the equation:
2x + 30 = 250
Subtract 30 from both sides:
2x = 220
Divide by 2
x = 110
So the bakery sell:
- 110 white bread loaves
- 140 whole wheat loaves since 110 + 30 = 140
Step 3: Write the profit expression
The bakery makes:
- $0.50 profit per white bread loaf.
- $0.75 profit per whole wheat loaf.
The total profit is:
= (0.50 times x) + (0.75 times (x + 30)
Step 4: Substitute x = 110
Now replace x with ( 110
Total Profit = (0.50 times 110) + (0.75 times (110 + 30)
Simplify step by step:
- 0.50 times 110 = 55
- 110 + 30 = 140 so ( 0.75 times 140 = 105 ).
Add them together:
Total Profit = 55 + 105 = 160
Final Answer
The total daily profit of the bakery is $160.
The expression for profit is:
Total Profit = (0.50 times x) + (0.75 times (x + 30)
Where x = 110 represents the number of white bread loaves sold.
| Task part 2: Simplifying the Expression for Total Cost of Renting a Car |
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- Regular Cost: 2x + 15 , where x is the number of hours the car is rented.
- Special Package: 3x - 2 which applies when the car is rented for more than 4 hours.
Total Cost for Renting the Car
We are task with writing single expresion for the total cost and simplifying it base on the give condition.
Step 1: Understandng the Situation
- For a regular rental: The cost per day is calculated as 2x + 15 .
- For customers renting for more than 4 hours: They are offered a discounted package of 3x - 2 .
Now, to write an expression for the total cost of renting the car:
- If x > 4 (car rented for more than 4 hours), the total cost = 3x - 2.
- If x ≤ 4, the total cost = 2x + 15.
Step 2: Expression for the Total Cost
We can combine the two cases into a single mathematical expression using a condition
Total Cost={ 2x+15 if x≤4}
{3x−2 if x>4}
This piecewise function represents the total cost of renting the car for ( x ) hours.
Step 3: Simplification of Each Case
Let me show the simplification of each case:
For Regular Rental x ≤ 4:
- The given expression 2x + 15 is already in its simplest form.
For Package Rental x > 4:
- The given expression 3x - 2 is also already simplified.
Total Cost={ 2x+15 if x≤4} {3x−2 if x>4}
This expression clearly shows how the cost changes based on the number of hours