Linear Equations (One & Two Variables)
Linear Equations — एक और दो Variables
Linear Equations (One & Two Variables)
- Algebra Basics
- Linear Equations (One & Two Variables)
Solve equations with one variable, and pairs of equations with two variables using elimination or substitution.
🎯 Learning Objective
Solve equations with one variable, and pairs of equations with two variables using elimination or substitution.
💡 Concept
- A linear equation has variables only to the power 1 (no x², no xy).
- To solve for one variable: move constants to one side, variable terms to the other, then divide.
- Whatever operation you do to one side, do exactly the same to the other side.
- Two variables need two equations. ELIMINATION: add/subtract to cancel one variable.
- SUBSTITUTION: express one variable from an equation and plug it into the other.
🧮 Key Formulas
ax + b = c → x = (c − b) / a
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Two variables: eliminate one, solve, back-substitute
✏️ Easy Example
Q. Solve: 2x + 5 = 15
- 2x = 15 − 5 = 10
- x = 10 / 2
Answer: x = 5
🇮🇳 Real-Life Example
Working out how many ₹20 platform tickets you can buy with ₹100 after a ₹40 snack is just solving 20x + 40 = 100 — everyday budgeting is linear algebra.
📝 Exam-Level Example
Q. Solve the pair: 2x + 3y = 13 and 3x − y = 3.
- From the second equation: y = 3x − 3
- Substitute into the first: 2x + 3(3x − 3) = 13
- 2x + 9x − 9 = 13 → 11x = 22 → x = 2
- y = 3(2) − 3 = 3
Answer: x = 2, y = 3
📝 Exam-Level Example
Q. The cost of 2 chairs and 3 tables is ₹1,400, and 3 chairs and 2 tables cost ₹1,100. Find the cost of each.
- 2c + 3t = 1400 … (i); 3c + 2t = 1100 … (ii)
- (i)×3: 6c + 9t = 4200; (ii)×2: 6c + 4t = 2200
- Subtract: 5t = 2000 → t = 400
- 2c + 3(400) = 1400 → 2c = 200 → c = 100
Answer: Chair = ₹100, Table = ₹400
🪄 Memory Trick
Make the coefficients of one variable equal, then add or subtract to eliminate it in a single step. Pick the variable that is easiest to match.
⚠️ Common Mistakes
- ❌ Changing the sign wrong when moving a term across the equals sign
- ❌ Multiplying only one side of an equation
- ❌ Substituting a value back into the wrong equation and not verifying
🏆 Exam Tips
- ✅ After solving, plug both values back into both equations to confirm
- ✅ Choose substitution when one variable already has coefficient 1
📌 Summary
- Isolate the variable: shift constants, then divide
- Do the same operation to both sides
- Two variables → elimination or substitution
- Always verify by back-substitution