Chapter 3 — PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Concept: Substitution Method

In a pair of linear equations in two variables, the substitution method involves:

• Expressing one variable in terms of another from one equation
• Substituting that expression into the second equation
• Solving the resulting equation
• Back-substituting to find the remaining variable

When to Use?

• When one equation is already simple (e.g., x + y = ...)
• When isolating a variable is easy
• Very useful in board exams for step marking

Solution Roadmap

1. Select the simpler equation
2. Make one variable subject
3. Substitute into the other equation
4. Solve for one variable
5. Back-substitute to find second variable
6. Verify solution

Graphical Insight

Intersection point represents the solution of the system.

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1. Given:

   \[ x + y = 14 \tag{1} \] \[ x - y = 4 \tag{2} \]

   Step 1: Express \( y \) from (2)

   \[ x - y = 4 \] \[ -y = 4 - x \] \[ y = x - 4 \tag{3} \]

   Step 2: Substitute into (1)

   \[ x + (x - 4) = 14 \]

   Step 3: Solve

   \[ 2x - 4 = 14 \] \[ 2x = 18 \] \[ x = 9 \]

   Step 4: Back substitute

   \[ y = 9 - 4 = 5 \]

   Solution: \( x = 9,\ y = 5 \)

2. Given:

   \[ s - t = 3 \tag{1} \] \[ \frac{s}{3} + \frac{t}{2} = 6 \tag{2} \]

   Step 1: Express \( s \)

   \[ s = t + 3 \tag{3} \]

   Step 2: Substitute

   \[ \frac{t+3}{3} + \frac{t}{2} = 6 \]

   Step 3: Multiply by LCM = 6

   \[ 2(t+3) + 3t = 36 \]

   Step 4: Simplify

   \[ 2t + 6 + 3t = 36 \] \[ 5t + 6 = 36 \] \[ 5t = 30 \] \[ t = 6 \]

   Step 5: Back substitute

   \[ s = 6 + 3 = 9 \]

   Solution: \( s = 9,\ t = 6 \)

3. Given:

   \[ 3x - y = 3 \] \[ 9x - 3y = 9 \]

   Step 1: Express \( y \)

   \[ y = 3x - 3 \]

   Step 2: Substitute

   \[ 9x - 3(3x - 3) = 9 \]

   Step 3: Simplify

   \[ 9x - 9x + 9 = 9 \] \[ 9 = 9 \]

   Infinitely many solutions (coincident lines)

   \[ y = 3x - 3 \]

4. Given:

   \[ 0.2x + 0.3y = 1.3 \] \[ 0.4x + 0.5y = 2.3 \]

   Step 1: Remove decimals

   \[ 2x + 3y = 13 \] \[ 4x + 5y = 23 \]

   Step 2: Express \( x \)

   \[ x = \frac{13 - 3y}{2} \]

   Step 3: Substitute

   \[ 4\left(\frac{13 - 3y}{2}\right) + 5y = 23 \]

   Step 4: Solve

   \[ 2(13 - 3y) + 5y = 23 \] \[ 26 - 6y + 5y = 23 \] \[ 26 - y = 23 \] \[ y = 3 \]

   Step 5: Back substitute

   \[ x = 2 \]

   Solution: \( x = 2,\ y = 3 \)

5. Given:

   \[ \sqrt{2}x + \sqrt{3}y = 0 \] \[ \sqrt{3}x + \sqrt{8}y = 0 \]

   Step 1: Express \( x \)

   \[ x = -\sqrt{\frac{3}{2}}y \]

   Step 2: Substitute

   \[ \sqrt{3}\left(-\sqrt{\frac{3}{2}}y\right) + \sqrt{8}y = 0 \]

   Step 3: Simplify

   \[ -\frac{3}{\sqrt{2}}y + 2\sqrt{2}y = 0 \] \[ y\left(\frac{1}{\sqrt{2}}\right)=0 \] \[ y = 0 \]

   Step 4:

   \[ x = 0 \]

   Solution: \( x = 0,\ y = 0 \)

6. Given:

   \[ \frac{3}{2}x - \frac{5}{3}y = -2 \] \[ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \]

   Step 1: Clear denominators

   \[ 9x - 10y = -12 \] \[ 2x + 3y = 13 \]

   Step 2: Express \( x \)

   \[ x = \frac{10y - 12}{9} \]

   Step 3: Substitute

   \[ 2\left(\frac{10y - 12}{9}\right) + 3y = 13 \]

   Step 4:

   \[ 20y - 24 + 27y = 117 \] \[ 47y = 141 \] \[ y = 3 \]

   Step 5:

   \[ x = 2 \]

   Solution: \( x = 2,\ y = 3 \)

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Exam Significance

• Very high probability question in CBSE Board (3–4 marks)
• Tests algebraic manipulation + conceptual clarity
• Forms base for coordinate geometry & graphs
• Frequently used in entrance exams like JEE Foundation, NTSE

Common Mistakes

• Wrong substitution (sign errors)
• Skipping steps (losing marks in board exams)
• Not clearing fractions properly
• Ignoring special cases (no solution / infinite solution)

Concept Applied

This problem combines two key ideas:

• Solving a pair of linear equations
• Using the obtained solution in another linear relation

Solution Strategy

1. Solve the system using substitution
2. Find ordered pair \( (x, y) \)
3. Substitute into \( y = mx + 3 \)
4. Solve for slope \( m \)

Graph Insight

Intersection point gives the solution used to determine slope \( m \).

Solution

Given equations:

\[ 2x + 3y = 11 \tag{1} \] \[ 2x - 4y = -24 \tag{2} \]

Step 1: Express \(2x\) from equation (1)

\[ 2x + 3y = 11 \] \[ 2x = 11 - 3y \tag{3} \]

Step 2: Substitute into equation (2)

\[ 2x - 4y = -24 \] Replace \(2x\) using (3): \[ (11 - 3y) - 4y = -24 \]

Step 3: Simplify carefully

\[ 11 - 3y - 4y = -24 \] \[ 11 - 7y = -24 \]

Step 4: Transpose and solve

\[ -7y = -24 - 11 \] \[ -7y = -35 \] \[ y = \frac{-35}{-7} \] \[ y = 5 \]

Step 5: Substitute \( y = 5 \) into (3)

\[ 2x = 11 - 3(5) \] \[ 2x = 11 - 15 \] \[ 2x = -4 \] \[ x = \frac{-4}{2} \] \[ x = -2 \]

Solution of system: \( x = -2,\ y = 5 \)

Step 6: Use relation \( y = mx + 3 \)

Substitute \( x = -2,\ y = 5 \): \[ 5 = m(-2) + 3 \]

Step 7: Solve for \( m \)

\[ 5 = -2m + 3 \] \[ 5 - 3 = -2m \] \[ 2 = -2m \] \[ m = \frac{2}{-2} \] \[ m = -1 \]

Required value: \( m = -1 \)

Final Answer

\( x = -2,\ y = 5,\ m = -1 \)

Exam Significance

• Very common CBSE 3-mark question (combined concept)
• Tests substitution + application in linear form
• Important for slope interpretation in coordinate geometry
• Frequently seen in NTSE & foundation-level competitive exams

Common Mistakes

• Sign error while substituting \(2x\)
• Wrong handling of negative division
• Forgetting to substitute into second part (finding \(m\))

Concept: Converting Word Problems into Linear Equations

Many real-life situations can be translated into linear equations by:

• Assigning variables to unknown quantities
• Forming equations based on given conditions
• Solving using substitution method

Solution Roadmap

1. Define variables clearly
2. Translate statements into equations
3. Use substitution method
4. Solve step-by-step
5. Interpret answer in context

Modeling Insight

Real-world problems reduce to intersection of two linear relations.

Solutions

1. Let numbers be: larger \( x \), smaller \( y \)

   \[ x = 3y \tag{1} \] \[ x - y = 26 \tag{2} \] Substitute (1) into (2): \[ 3y - y = 26 \] \[ 2y = 26 \] \[ y = 13 \] Back substitute: \[ x = 3(13) = 39 \]

   Numbers: 39 and 13

2. Let angles be: larger \( x \), smaller \( y \)

   \[ x + y = 180 \tag{1} \] \[ x = y + 18 \tag{2} \] Substitute: \[ y + 18 + y = 180 \] \[ 2y + 18 = 180 \] \[ 2y = 162 \] \[ y = 81 \] \[ x = 99 \]

   Angles: 99°, 81°

3. Let: cost of bat = \( x \), cost of ball = \( y \)

   \[ 7x + 6y = 3800 \tag{1} \] \[ 3x + 5y = 1750 \tag{2} \] From (1): \[ x = \frac{3800 - 6y}{7} \] Substitute into (2): \[ 3\left(\frac{3800 - 6y}{7}\right) + 5y = 1750 \] Multiply by 7: \[ 11400 - 18y + 35y = 12250 \] \[ 17y = 850 \] \[ y = 50 \] Substitute: \[ 3x + 250 = 1750 \] \[ 3x = 1500 \] \[ x = 500 \]

   Bat = ₹500, Ball = ₹50

4. Let: fixed charge = \( x \), per km = \( y \)

   \[ x + 10y = 105 \tag{1} \] \[ x + 15y = 155 \tag{2} \] From (1): \[ x = 105 - 10y \] Substitute: \[ 105 - 10y + 15y = 155 \] \[ 105 + 5y = 155 \] \[ 5y = 50 \] \[ y = 10 \] \[ x = 5 \] For 25 km: \[ x + 25y = 5 + 250 = 255 \]

   Fixed ₹5, Rate ₹10/km, 25 km = ₹255

5. Let fraction = \( \frac{x}{y} \)

   \[ \frac{x+2}{y+2} = \frac{9}{11} \] \[ 11(x+2) = 9(y+2) \] \[ 11x + 22 = 9y + 18 \] \[ 11x = 9y - 4 \tag{1} \] \[ \frac{x+3}{y+3} = \frac{5}{6} \] \[ 6(x+3) = 5(y+3) \] \[ 6x + 18 = 5y + 15 \] \[ 6x - 5y = -3 \tag{2} \] Substitute (1): \[ 6\left(\frac{9y - 4}{11}\right) - 5y = -3 \] Multiply by 11: \[ 54y - 24 - 55y = -33 \] \[ -y - 24 = -33 \] \[ y = 9 \] \[ x = 7 \]

   Fraction = \( \frac{7}{9} \)

6. Let: Jacob = \( x \), Son = \( y \)

   5 years ago: \[ x - 5 = 7(y - 5) \] \[ x = 7y - 30 \tag{1} \] 5 years hence: \[ x + 5 = 3(y + 5) \] \[ x + 5 = 3y + 15 \] Substitute: \[ 7y - 30 + 5 = 3y + 15 \] \[ 7y - 25 = 3y + 15 \] \[ 4y = 40 \] \[ y = 10 \] \[ x = 40 \]

   Jacob = 40 years, Son = 10 years

Exam Significance

• Very important CBSE 4–5 marks case-based question
• Tests mathematical modelling + algebra
• Frequently repeated pattern in board exams
• Highly relevant for NTSE, Olympiads, and foundation exams

Common Mistakes

• Wrong variable assumption
• Incorrect translation of statements
• Skipping algebra steps
• Not interpreting final answer properly