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🔶 1. INTRODUCTION

A linear equation in two variables is an equation of the form:ax+by+c=0ax + by + c = 0ax+by+c=0

Where:

• a,b,ca, b, ca,b,c are real numbers
• xxx and yyy are variables
• a and b cannot both be zero
• The highest power of variables is 1

Examples:

• 2x+3y=72x + 3y = 72x+3y=7
• 3x−y+5=03x – y + 5 = 03x−y+5=0

This equation represents a straight line on a graph.

A pair of linear equations means two lines drawn together:a1x+b1y+c1=0a_1x + b_1y + c_1 = 0a1​x+b1​y+c1​=0 a2x+b2y+c2=0a_2x + b_2y + c_2 = 0a2​x+b2​y+c2​=0

The solution of the pair is the point where the two lines meet.

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🔶 2. GEOMETRIC INTERPRETATION

Every linear equation represents a line.
So, two equations represent two lines, and their solution depends on how they behave:

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1️⃣ Case 1: Intersecting Lines → Unique Solution

If two lines cut each other at one point:

Condition:a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}a2​a1​​=b2​b1​​

✔ Exactly one solution
✔ Lines are not parallel
✔ System is consistent

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2️⃣ Case 2: Parallel Lines → No Solution

If the lines never meet:

Condition:a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}a2​a1​​=b2​b1​​=c2​c1​​

✔ No intersection
✔ No solution exists
✔ System is inconsistent

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3️⃣ Case 3: Coincident Lines → Infinitely Many Solutions

If both lines lie on top of each other:

Condition:a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}a2​a1​​=b2​b1​​=c2​c1​​

✔ All points of one line lie on the other
✔ Infinitely many solutions
✔ System is consistent & dependent

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🔶 3. SOLVING LINEAR EQUATIONS — METHODS

There are 4 major methods, each with detailed steps.

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✨ METHOD 1: Graphical Method

Steps to solve graphically:

1. Convert equations to the form y=mx+cy = mx + cy=mx+c (Slope-intercept form)
2. Take two values of x, find corresponding y.
3. Plot points on graph.
4. Draw lines.
5. Their intersection point gives the solution.

Graphical Results:

• Intersecting lines → one solution
• Parallel lines → no solution
• Coincident lines → infinite solutions

Graphical method gives approximate solutions.

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✨ METHOD 2: Substitution Method

Steps:

1. Solve one equation for one variable
   (example: x=7−yx = 7 – yx=7−y)
2. Substitute in the other equation.
3. Solve for the remaining variable.
4. Put value back to find the other variable.

Advantages:

• Best when one variable is already isolated (like x=5x = 5x=5).

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✨ METHOD 3: Elimination Method (Most Used in Exams)

Steps:

1. Multiply equations to make coefficients of either x or y equal.
2. Add or subtract equations to eliminate that variable.
3. Solve for the remaining variable.
4. Substitute in any original equation.

Important Tips:

• If signs are same, subtract.
• If signs are opposite, add.
  (For faster elimination)

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✨ METHOD 4: Cross-Multiplication Method

Use when equations are written in standard form:a1x+b1y+c1=0a2x+b2y+c2=0a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0a1​x+b1​y+c1​=0a2​x+b2​y+c2​=0

Formula:

xb1c2−b2c1=yc1a2−c2a1=1a1b2−a2b1\frac{x}{b_1c_2 – b_2c_1} = \frac{y}{c_1a_2 – c_2a_1} = \frac{1}{a_1b_2 – a_2b_1}b1​c2​−b2​c1​x​=c1​a2​−c2​a1​y​=a1​b2​−a2​b1​1​

If denominator = 0 → no unique solution.

This is the fastest algebraic method.

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🔶 4. CONDITIONS FOR SOLVABILITY (VERY IMPORTANT)

Given:a1x+b1y+c1=0a2x+b2y+c2=0a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0a1​x+b1​y+c1​=0a2​x+b2​y+c2​=0

Case 1: Unique Solution

a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}a2​a1​​=b2​b1​​

Case 2: No Solution

a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}a2​a1​​=b2​b1​​=c2​c1​​

Case 3: Infinitely Many Solutions

a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}a2​a1​​=b2​b1​​=c2​c1​​

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🔶 5. SPECIAL CASES

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🔸1. Both equations in intercept form

xa1+yb1=1\frac{x}{a_1} + \frac{y}{b_1} = 1a1​x​+b1​y​=1 xa2+yb2=1\frac{x}{a_2} + \frac{y}{b_2} = 1a2​x​+b2​y​=1

Intersection point can be found by equating intercepts.

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🔸2. Equations with fractions

Multiply by LCM to remove fractions.

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🔸3. Word Problems Category

Word problems usually fall under:

• Age Problems
• Speed-Distance-Time
• Number/Fraction Problems
• Price/Cost Problems
• Geometry Based Problems
• Mixture/Ratio Problems

Understanding how to make equations is key.

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🔶 6. SOLVED EXAMPLES (DETAILED)

Example 1: (Elimination)

Solve:3x+2y=113x + 2y = 113x+2y=11 2x−y=42x – y = 42x−y=4

Multiply second equation by 2:4x−2y=84x – 2y = 84x−2y=8

Add with first:7x=19⇒x=1977x = 19 \Rightarrow x = \frac{19}{7}7x=19⇒x=719​

Substitute back:3(197)+2y=113\left(\frac{19}{7}\right) + 2y = 113(719​)+2y=11

Solve for y:577+2y=11→2y=207→y=107\frac{57}{7} + 2y = 11 → 2y = \frac{20}{7} → y = \frac{10}{7}757​+2y=11→2y=720​→y=710​

✔ Solution:(197, 107)\left(\frac{19}{7},\ \frac{10}{7}\right)(719​, 710​)

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Example 2: (Substitution)

Solve:x+y=7x−y=1x + y = 7 x – y = 1x+y=7x−y=1

From first:x=7−yx = 7 – yx=7−y

Substitute:7−y−y=1⇒−2y=−6⇒y=37 – y – y = 1 \Rightarrow -2y = -6 \Rightarrow y = 37−y−y=1⇒−2y=−6⇒y=3

Then x=4x = 4x=4.

✔ Solution: (4,3)(4, 3)(4,3)

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Example 3: (Cross Multiplication)

2x+3y−13=02x + 3y – 13 = 02x+3y−13=0 3x−y−5=03x – y – 5 = 03x−y−5=0

Apply formula:x3(−5)−(−1)(−13)=y(−13)(3)−(−5)(2)=12(−1)−(3)(3)\frac{x}{3(-5) – (-1)(-13)} = \frac{y}{(-13)(3) – (-5)(2)} = \frac{1}{2(-1) – (3)(3)}3(−5)−(−1)(−13)x​=(−13)(3)−(−5)(2)y​=2(−1)−(3)(3)1​

Solve step-by-step → final values.

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🔶 7. PRACTICE QUESTIONS (HIGH-YIELD)

If you want practice sets, I can generate:

• 20 questions easy
• 20 questions standard
• 20 questions HOTS
• Board pattern sample paper

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🔶 8. SUMMARY (EXAM-READY)

✔ Linear equation form:

ax+by+c=0ax + by + c = 0ax+by+c=0

✔ Conditions for solutions:

Condition	Meaning
a1/a2≠b1/b2a_1/a_2 ≠ b_1/b_2a1​/a2​=b1​/b2​	Unique solution
a1/a2=b1/b2≠c1/c2a_1/a_2 = b_1/b_2 ≠ c_1/c_2a1​/a2​=b1​/b2​=c1​/c2​	No solution
a1/a2=b1/b2=c1/c2a_1/a_2 = b_1/b_2 = c_1/c_2a1​/a2​=b1​/b2​=c1​/c2​	Infinite solutions

✔ Methods:

• Substitution
• Elimination
• Graphical
• Cross multiplication