📘 Chapter 1: Linear Equations & Algebraic Fundamentals
Building blocks for solving real-world problems with confidence
📖 Chapter Summary
✨ Core concepts covered: In this chapter, we explore the foundation of algebra —
linear equations in one variable. You'll learn how to isolate variables
using inverse operations (addition, subtraction, multiplication, division). The key principle:
whatever you do to one side of an equation, you must do to the other to maintain balance.
We also review the distributive property: a(b + c) = ab + ac, and how to combine like terms.
By the end, you'll be able to solve equations of the form ax + b = c and
a(x + b) = c. Mastering these skills prepares you for more complex modeling,
inequalities, and functions.
🧩 Worked Example
📌 Example: Solve for x
$ 3(x - 4) + 5 = 2x + 1 $
Step-by-step solution:
Step 1: Apply the distributive property:
$ 3x - 12 + 5 = 2x + 1 $ → $ 3x - 7 = 2x + 1 $
Step 2: Isolate the variable terms on one side. Subtract $2x$ from both sides: $ 3x - 2x - 7 = 1 $ → $ x - 7 = 1 $
Step 3: Add 7 to both sides: $ x = 8 $
Step 4: Check the solution: LHS: $ 3(8-4)+5 = 3(4)+5 = 12+5=17 $ | RHS: $ 2(8)+1 = 16+1=17 $ ✓
✅ Final answer: $ x = 8 $
Step 2: Isolate the variable terms on one side. Subtract $2x$ from both sides: $ 3x - 2x - 7 = 1 $ → $ x - 7 = 1 $
Step 3: Add 7 to both sides: $ x = 8 $
Step 4: Check the solution: LHS: $ 3(8-4)+5 = 3(4)+5 = 12+5=17 $ | RHS: $ 2(8)+1 = 16+1=17 $ ✓
✅ Final answer: $ x = 8 $
💡 Always verify by substituting back into the original equation.
✏️ Practice Question
❓ Question: Solve for $ y $ in the equation below:
$ 4(y + 2) - 3 = 2y + 7 $
📐 General Strategy / Formula:
For linear equations of the form $ a(y + b) + c = dy + e $:
• Step A: Use distributive property: $ a \cdot y + a \cdot b + c = dy + e $
• Step B: Combine constant terms on LHS.
• Step C: Collect variable terms on one side, constants on the opposite side.
• Step D: Divide by the coefficient of $ y $ (if not 1).
🧠 Isolation principle: "undo" operations in reverse order of operations (PEMDAS backward).
For linear equations of the form $ a(y + b) + c = dy + e $:
• Step A: Use distributive property: $ a \cdot y + a \cdot b + c = dy + e $
• Step B: Combine constant terms on LHS.
• Step C: Collect variable terms on one side, constants on the opposite side.
• Step D: Divide by the coefficient of $ y $ (if not 1).
🧠 Isolation principle: "undo" operations in reverse order of operations (PEMDAS backward).
🔄 Similar Example (step-by-step):
Equation: $ 2(x + 3) - 4 = x + 5 $
• Distribute: $ 2x + 6 - 4 = x + 5 $ → $ 2x + 2 = x + 5 $
• Subtract $ x $: $ x + 2 = 5 $
• Subtract 2: $ x = 3 $
✅ Solution: $ x = 3 $.
🧩 Use exactly the same steps for the practice question.
Equation: $ 2(x + 3) - 4 = x + 5 $
• Distribute: $ 2x + 6 - 4 = x + 5 $ → $ 2x + 2 = x + 5 $
• Subtract $ x $: $ x + 2 = 5 $
• Subtract 2: $ x = 3 $
✅ Solution: $ x = 3 $.
🧩 Use exactly the same steps for the practice question.
✅ Full Solution for $ 4(y + 2) - 3 = 2y + 7 $:
1. Apply distributive property:
$ 4 \cdot y + 4 \cdot 2 - 3 = 2y + 7 $ → $ 4y + 8 - 3 = 2y + 7 $
2. Combine constants on left: $ 8 - 3 = 5 $ → $ 4y + 5 = 2y + 7 $
3. Move variable terms to one side: Subtract $ 2y $ from both sides:
$ 4y - 2y + 5 = 7 $ → $ 2y + 5 = 7 $
4. Isolate the term with $ y $: Subtract 5 from both sides:
$ 2y = 2 $
5. Divide both sides by 2: $ y = 1 $
6. Check: LHS: $ 4(1+2) - 3 = 4(3)-3 = 12-3=9 $ ; RHS: $ 2(1)+7 = 2+7=9 $ ✓
🎯 Final Answer: $ y = 1 $
1. Apply distributive property:
$ 4 \cdot y + 4 \cdot 2 - 3 = 2y + 7 $ → $ 4y + 8 - 3 = 2y + 7 $
2. Combine constants on left: $ 8 - 3 = 5 $ → $ 4y + 5 = 2y + 7 $
3. Move variable terms to one side: Subtract $ 2y $ from both sides:
$ 4y - 2y + 5 = 7 $ → $ 2y + 5 = 7 $
4. Isolate the term with $ y $: Subtract 5 from both sides:
$ 2y = 2 $
5. Divide both sides by 2: $ y = 1 $
6. Check: LHS: $ 4(1+2) - 3 = 4(3)-3 = 12-3=9 $ ; RHS: $ 2(1)+7 = 2+7=9 $ ✓
🎯 Final Answer: $ y = 1 $
💡 Interactive tip: Click any button to toggle extra help. Try solving yourself before revealing the solution!
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