PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Concept Before Solving

• A pair of linear equations in two variables represents two straight lines.
• The solution of the system is the point where both lines intersect.
• Algebraically: Solve equations simultaneously.
• Graphically: Plot both equations → intersection point = solution.

Solution Roadmap (Exam Strategy)

1. Assign variables clearly.
2. Convert statements into equations.
3. Solve using substitution/elimination.
4. Verify by substituting values.
5. Interpret final answer in context.

Solution (i)

Step 1: Let variables

Let number of girls = \(x\)
Let number of boys = \(y\)

Step 2: Form equations

Girls are 4 more than boys:

$$\begin{align}x = y + 4 \tag{1}\end{align}$$

Total students = 10:

$$\begin{align}x + y = 10 \tag{2}\end{align}$$

Step 3: Substitute equation (1) into (2)

$$\begin{aligned} x + y &= 10 \\ (y + 4) + y &= 10 \\ y + 4 + y &= 10 \\ 2y + 4 &= 10 \\ 2y &= 10 - 4 \\ 2y &= 6 \\ y &= \frac{6}{2} \\ y &= 3 \end{aligned}$$

Step 4: Find value of \(x\)

$$\begin{aligned} x &= y + 4 \\ x &= 3 + 4 \\ x &= 7 \end{aligned}$$

Final Answer:
Number of girls = 7
Number of boys = 3

Verification

$$7 + 3 = 10 \quad \text{✔ correct}$$

Solution (ii)

Step 1: Let variables

Cost of one pencil = \(x\)
Cost of one pen = \(y\)

Step 2: Form equations

$$\begin{align5x + 7y = 50 \tag{1}\end{align}$$ $$\begin{align7x + 5y = 46 \tag{2}\end{align}$$

Step 3: Elimination method

Multiply (1) by 7 and (2) by 5: $$\begin{align} 35x + 49y &= 350 \tag{3}\\ 35x + 25y &= 230 \tag{4} \end{align}$$

Subtract (4) from (3):

$$\begin{aligned} (35x + 49y) - (35x + 25y) &= 350 - 230 \\ 35x + 49y - 35x - 25y &= 120 \\ 24y &= 120 \\ y &= \frac{120}{24} \\ y &= 5 \end{aligned}$$

Step 4: Substitute \(y = 5\) in equation (1)

$$\begin{aligned} 5x + 7y &= 50 \\ 5x + 7(5) &= 50 \\ 5x + 35 &= 50 \\ 5x &= 50 - 35 \\ 5x &= 15 \\ x &= \frac{15}{5} \\ x &= 3 \end{aligned}$$

Final Answer:
Cost of one pencil = ₹3
Cost of one pen = ₹5

Verification

$$5(3) + 7(5) = 15 + 35 = 50 \quad \text{✔}$$ $$7(3) + 5(5) = 21 + 25 = 46 \quad \text{✔}$$

Why This Question is Important

• Directly tests formation of equations from real-life situations.
• Very common in CBSE Board Exams (3–4 marks guaranteed type).
• Strengthens algebraic modelling (used in higher mathematics).
• Forms base for coordinate geometry and graph interpretation.

Competitive Exam Relevance

• Frequently asked in NTSE, Olympiads, and foundation exams.
• Improves speed in solving linear systems (important for JEE basics).
• Helps in word-problem translation — a key aptitude skill.

Concept Before Solving

• For a pair of linear equations:
  \(a_1x + b_1y + c_1 = 0\)
  \(a_2x + b_2y + c_2 = 0\)
• Compare ratios:

\[ \frac{a_1}{a_2}, \quad \frac{b_1}{b_2}, \quad \frac{c_1}{c_2} \]

• If \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) → Lines intersect → Unique solution
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → Parallel → No solution
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → Coincident → Infinite solutions

Solution Strategy

1. Write coefficients clearly.
2. Find all three ratios carefully.
3. Compare ratios properly (avoid sign mistakes).
4. State conclusion with solution type.

Board Exam Importance

• Very frequently asked 2–3 mark direct question.
• Concept-based → easy scoring if ratios are handled correctly.
• Common mistake: ignoring signs in ratios.

Competitive Exam Relevance

• Builds foundation for coordinate geometry.
• Important for graphical interpretation in higher maths.
• Used in elimination logic in aptitude and algebra problems.

Concept Before Solving

For a pair of linear equations:

\(a_1x + b_1y + c_1 = 0\)
\(a_2x + b_2y + c_2 = 0\)

Compare ratios:

• If \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) → Consistent (unique solution)
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → Inconsistent (no solution)
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → Consistent (infinite solutions)

Solution Strategy

1. Convert equations into standard form.
2. Extract coefficients carefully.
3. Handle fractions properly (very important).
4. Compare ratios and conclude.

Board Exam Importance

• Very common conceptual question (2–4 marks).
• Tests ratio logic and sign accuracy.
• Fractions-based questions are frequently asked.

Competitive Exam Relevance

• Important for linear system classification.
• Helps in graphical reasoning problems.
• Strengthens algebraic manipulation speed.

Concept Before Solving

• If two equations represent same line → infinite solutions.
• If lines intersect → unique solution.
• If lines are parallel → no solution.
• Graphically, solution = point of intersection.

Solution Strategy

1. Convert into standard form if needed.
2. Compare ratios.
3. If consistent and unique → solve algebraically.
4. Verify solution.

Solution (iii)

$$\begin{aligned} 2x + y - 6 &= 0\\ 4x - 2y - 4 &= 0 \end{aligned}$$

Step 1: Check consistency

$$\begin{aligned} \frac{a_1}{a_2} &= \frac{2}{4} = \frac{1}{2}\\ \frac{b_1}{b_2} &= \frac{1}{-2} = -\frac{1}{2} \end{aligned}$$ \[\frac{1}{2} \ne -\frac{1}{2}\]

Conclusion: Intersecting → unique solution

Step 2: Solve algebraically

From (1): \[y = 6 - 2x\] Substitute into (2): $$\begin{aligned} 4x - 2(6 - 2x) - 4 &= 0\\ 4x - 12 + 4x - 4 &= 0\\ 8x - 16 &= 0\\ 8x &= 16\\ x &= 2 \end{aligned}$$ $$\begin{aligned} y &= 6 - 2(2)\\ y &= 6 - 4\\ y &= 2 \end{aligned}$$

Solution point: (2, 2)

Verification:

\[2(2)+2=6 \quad ✔\] \[4(2)-2(2)-4=0 \quad ✔\]

Board Exam Importance

• Highly important 3–4 mark question combining concept + solving.
• Tests both classification and algebraic solving.
• Graph-based questions frequently asked.

Competitive Exam Relevance

• Important for graphical interpretation problems.
• Builds strong foundation for coordinate geometry.
• Used in elimination-based aptitude problems.

Concept Before Solving

• Perimeter of a rectangle = \(2(\text{length} + \text{width})\)
• Half of the perimeter = \(\text{length} + \text{width}\)
• Translate word statements carefully into equations.

Solution Strategy

1. Assume variables clearly.
2. Convert given conditions into equations.
3. Solve using substitution method.
4. Verify final answer.

Board Exam Importance

• Common word problem (3 marks).
• Tests translation of geometry into algebra.
• Easy scoring if equation formation is correct.

Competitive Exam Relevance

• Important for aptitude-based algebra problems.
• Used in mensuration + algebra combined questions.
• Builds modelling skills for real-life scenarios.

Concept Before Solving

• For two lines:
  \(a_1x + b_1y + c_1 = 0\)
  \(a_2x + b_2y + c_2 = 0\)
• Intersecting: \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\)
• Parallel: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\)
• Coincident: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Given Equation

Here: \(a_1 = 2,\ b_1 = 3,\ c_1 = -8\)

Board Exam Importance

• Tests conceptual clarity of ratio conditions.
• Common 2–3 mark question.
• Students often lose marks due to wrong ratio condition.

Competitive Exam Relevance

• Important for understanding slope and line relations.
• Used in coordinate geometry and system classification.
• Strengthens conceptual algebra.

Concept Before Solving

• Each linear equation represents a straight line.
• Triangle is formed by intersection of:
  • Line 1
  • Line 2
  • x-axis (y = 0)
• Vertices = points of intersection.

Solution

Given equations:

\[\begin{align}x - y + 1 = 0 \tag{1}\end{align}\] \[\begin{align}3x + 2y - 12 = 0 \tag{2}\end{align}\]

Step 1: Convert to slope form

$$\begin{aligned} (1):\ y &= x + 1 \\ (2):\ 2y &= 12 - 3x \\ y &= 6 - \frac{3}{2}x \end{aligned}$$

Step 2: Find intersection with x-axis (y = 0)

For (1): \[x + 1 = 0 \Rightarrow x = -1\] Point = (-1, 0)

For (2): \[3x - 12 = 0 \Rightarrow x = 4\] Point = (4, 0)

Step 3: Find intersection of two lines

$$\begin{aligned} x + 1 &= 6 - \frac{3}{2}x \\ x + \frac{3}{2}x &= 6 - 1 \\ \frac{5}{2}x &= 5 \\ x &= 2 \end{aligned}$$ \[y = x + 1 = 2 + 1 = 3\] Point = (2, 3)

Step 4: Vertices of triangle

• (-1, 0)
• (4, 0)
• (2, 3)

Final Answer:
Triangle vertices = (-1,0), (4,0), (2,3)

Board Exam Importance

• Very important 4-mark graphical question.
• Tests graph plotting + intersection concept.
• Common mistake: skipping intersection calculation.

Competitive Exam Relevance

• Important for coordinate geometry basics.
• Used in area and graph-based reasoning problems.
• Strengthens visualization skills.