1. Foundation of Algebra

What is Algebra? (Introduction & Real-Life Applications)

    1. Foundation for Higher Mathematics: Algebra is a building block for advanced mathematical concepts such as Geometry, Trigonometry, and Calculus. Mastery of algebraic principles is essential for progress in these subjects.

    2. Critical Thinking and Problem-Solving Skills: Engaging with algebra develops critical thinking and enhances problem-solving abilities. It challenges individuals to analyze situations, formulate strategies, and arrive at logical solutions.

    3. Real-World Applications: Algebra is frequently used in various real-world scenarios, including business, science, and technology. From calculating profits and analyzing data to programming and engineering, algebraic concepts are crucial in functioning effectively in many fields.

    In summary, algebra is fundamental for academic advancement and practical problem-solving and decision-making in everyday life.

    • Finance & Budgeting: Calculating savings, loan interest, and expenses using formulas.

    • Business & Marketing: Profit calculations, analyzing trends, pricing strategies.

    • Computer Science & Coding: Writing algorithms and logic in programming.

    • Sports Analytics: Predicting performance, tracking stats, and optimizing strategies.

    • Engineering & Construction: Designing buildings and calculating materials needed.

    • Medicine & Science: Drug dosages, medical research, physics equations.

Practice Questions

1. Budgeting & Expenses:
Jay wants to save up for a new camera that costs $800. He already has $200 saved and plans to save $50 per week until he reaches his goal.

a. Write an equation that represents how many weeks (w) it will take for Jay to afford the camera.
b. Solve for (w) to determine how many weeks it will take.

Understanding the Situation:

  • Jay already has $200 saved.

  • He adds $50 per week to his savings.

  • He needs a total of $800 to buy the camera.

  • We need to find out how many weeks (w) it will take him to reach his goal.

Breaking It Down:

  1. Start with what he has: Jay starts with $200.

  2. Add what he saves each week: Every week, he saves $50, so after w weeks, he will have saved 50w dollars.

  3. Set the equation: His total savings after w weeks must equal $800.

200 + 50w = 800

Step 1: Write an equation

Let w represent the number of weeks he needs to save:

200 + 50w = 800

Step 2: Solve for w

Subtract 200 from both sides:

50w = 600

Divide both sides by 50:

w = 12

Answer: Jay needs to save for 12 weeks to afford the camera.

2. Sports Performance:
A basketball player scores an average of 4 points per quarter. In a game where he plays all 4 quarters, he also makes an additional 6 points from free throws.

a. Write an equation representing his total points (P) in the game.
b. If he wants to score at least 25 points, how many extra points does he need to score?

Understanding the Situation:

The player scores 4 points per quarter.

  • He plays 4 quarters in a complete game.

  • He also scores an additional 6 points from free throws.

  • We need to find his total points scored (P).

Breaking It Down:

  1. Identify the repeating action: Every quarter, he scores 4 points.

    (If he plays q quarters, his total points from quarters is 4q)

  2. Account for the extra points: He scores six free-throw points regardless of quarters played.

  3. Set up the equation: His total points scored is the sum of points per quarter and the free throw points.

P = 4q + 6

Step 1: Write an equation

Let P represent the total points scored in the game:

P = 4q + 6

Since the player plays all 4 quarters, substitute q = 4:

P = 4(4) + 6

Step 2: Solve for P

Multiply:

P= 16 + 6

Add:

P = 22

If the player wants to score at least 25 points, we must find how many more points are needed:

25 - 22 = 3

Answer: The player scored 22 points and needs three more points to reach 25.

3. Ride-Share Costs:
A ride-sharing company charges $2.50 per mile plus a $5 base fee for every trip.

a. Write an equation representing the total cost (C) of a ride (m) miles long.
b. If someone takes a 10-mile ride, how much will they have to pay?

Understanding the Situation:

The ride costs $2.50 per mile.

  • There is a $5 base fee (a fixed cost you pay no matter how far you go).

  • We need to find the total cost (C) for a trip of mmm miles.

Breaking It Down:

  1. Identify the fixed cost: No matter how many miles you ride, you always pay $5.

  2. Identify the cost per mile: Since each mile adds $2.50, write it as “2.50m.”

  3. Set up the equation: The total cost is the sum of the fixed base fee and the cost per mile multiplied by the miles traveled.

    C = 2.50m + 5

Step 1: Write an equation

Let C represent the total cost and m be the number of miles driven:

C = 2.50m + 5

Step 2: Find the cost for a 10-mile ride

Substituting m = 10:

C = 2.50(10) + 5

Multiply:

C = 25 + 5

Add:

C = 30

Answer: The total cost for a 10-mile ride is $30.