Algebra Basics

Learn to work with variables, solve equations, and use formulas: the foundation for financial planning, business analysis, and problem-solving.

What is Algebra?

Algebra is arithmetic with unknowns. Instead of always working with specific numbers, we use letters (variables) to represent unknown or changing values.

Why it matters:

• Solve for unknowns (How much do I need to save monthly?)
• Create formulas (Calculate total cost for any quantity)
• Model relationships (How does revenue change with price?)

Variables and Expressions

Understanding Variables

A variable is a letter that represents a number.

Common conventions:

• x, y, z: unknown values to solve for
• n: number of items
• t: time
• P: principal (money)
• r: rate (interest, growth)

Example: If x = 5, then:

• x + 3 = 5 + 3 = 8
• 2x = 2 × 5 = 10
• x² = 5 × 5 = 25

Expressions vs Equations

Expression: A mathematical phrase without an equals sign

• 3x + 5
• 2(x − 4)
• x² + 2x + 1

Equation: A mathematical statement that two expressions are equal

• 3x + 5 = 20
• 2(x − 4) = 10
• x² = 25

Solving Simple Equations

The Golden Rule

Whatever you do to one side, do to the other side.

The equation stays balanced when you perform the same operation on both sides.

One-Step Equations

Addition/Subtraction

Example: x + 7 = 15

• Subtract 7 from both sides
• x = 15 − 7
• x = 8

Life Example: Unknown expense

• Total expenses: $350
• Known expenses: $275
• Unknown: x + 275 = 350
• x = $75

Multiplication/Division

Example: 5x = 45

• Divide both sides by 5
• x = 45 ÷ 5
• x = 9

Business Example: Unit cost

• Total cost: $450
• Number of units: 30
• Cost per unit: 30x = 450
• x = $15

Two-Step Equations

Example: 3x + 4 = 19

Steps:

1. Subtract 4 from both sides: 3x = 15
2. Divide both sides by 3: x = 5

Check: 3(5) + 4 = 15 + 4 = 19 ✓

Life Example: Gym membership

• Monthly fee plus signup: mx + 50 = 230
• Total paid: $230
• Signup fee: $50
• For 6 months: 6m + 50 = 230
• 6m = 180
• m = $30/month

Multi-Step Equations

Example: 2(x + 3) = 16

Steps:

1. Distribute: 2x + 6 = 16
2. Subtract 6: 2x = 10
3. Divide by 2: x = 5

Business Example: Pricing with markup

• Cost: c
• Markup: 50% (multiply by 1.5)
• Discount: $20 off
• Final price: $130
• Equation: 1.5c − 20 = 130
• 1.5c = 150
• c = $100

Working with Formulas

Using Existing Formulas

Many real-world problems have standard formulas. Learn to identify which values you have and which you need to find.

Distance Formula: D = r × t

• D = distance
• r = rate (speed)
• t = time

Example: How long to drive 300 miles at 60 mph?

• 300 = 60 × t
• t = 300 ÷ 60
• t = 5 hours

Common Formulas

Rearranging Formulas (Solving for Different Variables)

Sometimes you need to solve for a different variable in a formula.

Example: Distance formula D = rt, solve for time

Steps:

1. Start with: D = rt
2. Divide both sides by r
3. Result: t = D/r

Application:

• Drive 180 miles at 45 mph
• t = 180/45 = 4 hours

Example: Simple interest I = Prt, solve for Principal

Steps:

1. Start with: I = Prt
2. Divide both sides by rt
3. Result: P = I/(rt)

Application:

• Want to earn $500 interest in 2 years at 5% rate
• P = 500/(0.05 × 2) = 500/0.1 = $5,000 needed

Word Problems

The Process

1. Read carefully: understand what's being asked
2. Identify unknowns: assign variables
3. Write equation: translate words to math
4. Solve: use algebra techniques
5. Check: does the answer make sense?

Example 1: Cost Problem

Problem: "A phone plan costs $30/month plus $0.10 per text. How many texts can you send if your budget is $45?"

Solution:

• Let t = number of texts
• Fixed cost: $30
• Variable cost: $0.10 per text = 0.10t
• Total: 30 + 0.10t = 45
• Solve: 0.10t = 15, so t = 150 texts

Example 2: Age Problem

Problem: "John is 3 times as old as his daughter. In 10 years, he'll be twice as old as she is. How old are they now?"

Solution:

• Let d = daughter's age now
• John's age now: 3d
• In 10 years: Daughter is d + 10, John is 3d + 10
• Equation: 3d + 10 = 2(d + 10)
• Simplify: 3d + 10 = 2d + 20
• Solve: d = 10
• Daughter: 10 years old, John: 30 years old

Example 3: Business Problem

Problem: "A business has fixed costs of $5,000/month and variable costs of $25 per unit. If they sell units at $45 each, how many must they sell to break even?"

Solution:

• Let x = number of units
• Total costs: 5000 + 25x
• Total revenue: 45x
• Break even when costs = revenue: 5000 + 25x = 45x
• Solve: 5000 = 20x, so x = 250 units

Systems of Equations (Two Unknowns)

Sometimes you need to find two unknowns using two equations.

Substitution Method

Problem: "Adult tickets cost $12, child tickets cost $8. A family bought 5 tickets for $52. How many of each?"

Solution:

• Let a = adult tickets, c = child tickets
• Equation 1: a + c = 5 (total tickets)
• Equation 2: 12a + 8c = 52 (total cost)

Steps:

1. From equation 1: a = 5 − c
2. Substitute into equation 2: 12(5 − c) + 8c = 52
3. Simplify: 60 − 12c + 8c = 52
4. Solve: 60 − 4c = 52, so 4c = 8, thus c = 2
5. Find a: a = 5 − 2 = 3
6. Answer: 3 adult tickets, 2 child tickets

Elimination Method

Problem: Same as above

Solution:

• Equation 1: a + c = 5
• Equation 2: 12a + 8c = 52

Steps:

1. Multiply equation 1 by -8: −8a − 8c = −40
2. Add to equation 2: 12a + 8c = 52
3. Result: 4a = 12, so a = 3
4. Substitute back: 3 + c = 5, so c = 2
5. Answer: 3 adult tickets, 2 child tickets

Inequalities

Instead of equals, we can have greater than (>), less than (<), or their "or equal to" versions (≥, ≤).

Example: "You need at least $500 saved. You have $200 and save $30/week. How many weeks?"

Solution:

• Let w = weeks
• Inequality: 200 + 30w ≥ 500
• Solve: 30w ≥ 300
• w ≥ 10 weeks (need at least 10 weeks)

Exponents in Algebra

Basic Rules

Financial Application: Compound Growth

Formula: A = P(1 + r)^t

• A = final amount
• P = principal
• r = rate (as decimal)
• t = time periods

Example: Investment doubles

• $10,000 invested at 7% annually
• After 10 years: A = 10000(1.07)^10
• A = 10000(1.967) ≈ $19,670

Practice Problems

Solving Equations

1. x + 12 = 35
2. 7x = 91
3. 4x − 8 = 28
4. 3(x + 5) = 33

Word Problems

5. A car rental costs $50/day plus $0.25/mile. Total bill is $87.50 for one day. How many miles driven?

6. Two numbers sum to 50. One is 8 more than the other. Find both numbers.

7. A product costs $120 after a 20% markup. What was the original cost?

Formulas

8. Using D = rt, find speed if distance is 240 miles and time is 3.5 hours.

9. Using I = Prt, find the interest rate if $2,000 earns $300 interest in 3 years.

10. Temperature is 25°C. What is it in Fahrenheit? Use F = 9/5C + 32

Solutions

1. x = 23 (subtract 12 from both sides)
2. x = 13 (divide both sides by 7)
3. x = 9 (add 8, then divide by 4)
4. x = 6 (distribute: 3x + 15 = 33, so 3x = 18)
5. 150 miles (50 + 0.25m = 87.50, so 0.25m = 37.50, thus m = 150)
6. 21 and 29 (x + (x + 8) = 50, so 2x = 42, thus x = 21)
7. $100 (1.20c = 120, so c = 100)
8. 68.57 mph (r = 240/3.5)
9. 5% (r = 300/(2000 × 3) = 300/6000 = 0.05)
10. 77°F (F = 9/5(25) + 32 = 45 + 32)

Key Takeaways

✓ Variables represent unknowns: use letters to stand for values you don't know yet
✓ Balance equations: whatever you do to one side, do to the other
✓ Work systematically: isolate the variable step by step
✓ Check your answers: plug back into original equation
✓ Translate words to math: word problems follow predictable patterns

Real-World Applications

• Budgeting: Solve for unknown expenses or required savings
• Business: Break-even analysis, pricing decisions
• Investments: Calculate required returns or time to goals
• Project Planning: Determine resources needed
• Negotiations: Find acceptable ranges for deals

Next Steps

Move to Chapter 03: Ratios & Proportions to learn how to compare quantities and solve scaling problems. These skills are essential for recipes, unit conversions, and business comparisons.