Pair of Linear Equations in Two Variables

Every linear equation in two variables represents a straight line on the Cartesian plane. Therefore, a pair of linear equations represents two lines whose relative position determines the number of solutions.

The solution of the pair corresponds to the point(s) of intersection of these lines.

Geometric Interpretation with Algebraic Conditions

Important Observations

• A solution graphically means common point(s) of both lines.
• Graph method is useful for visual understanding but may lack precision.
• At least two points per line are required for accurate graph plotting.

CBSE Exam Insight

• Questions often ask to identify type of solution using graphs.
• Algebraic conditions must be linked with graphical interpretation.
• Case study questions combine real-life graphs + interpretation.

For a pair of linear equations:

\[ \begin{aligned} a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0 \end{aligned} \]

The nature of solutions depends on the ratio of coefficients. These conditions help determine whether the system is consistent, inconsistent, or dependent without solving it.

Determinant (Advanced Insight)

These conditions can also be understood using determinants:

\[ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1 \]

• If D ≠ 0 → Unique solution (lines intersect)
• If D = 0 → Lines are parallel or coincident

Conceptual Understanding

• Ratios compare slopes of lines
• Equal slopes → parallel or same line
• Different slopes → intersection guaranteed

Quick Decision Flow (Visual Logic)

CBSE Exam Insights

• Direct questions: identify nature using ratios
• Case-based questions combine algebra + graph interpretation
• Frequently used in 2–3 mark conceptual problems

Common Mistakes

• Comparing ratios incorrectly (especially signs)
• Ignoring condition on constant terms \(c_1, c_2\)
• Writing ratios in different order (must match positions)

Problem: Graphically determine whether the following pair of linear equations has no solution, unique solution, or infinitely many solutions:

\[ \begin{align} 5x - 8y &= 0 \tag{1}\\ 3x - \frac{24}{5}y + \frac{3}{5} &= 0 \tag{2} \end{align} \]

Step 1: Convert to Standard Form

Multiply Equation (2) by 5 to eliminate fractions:

\[ 15x - 24y + 3 = 0 \tag{2'} \]

Step 2: Compare Coefficients

For Equation (1): \(a_1 = 5,\ b_1 = -8,\ c_1 = 0\)
For Equation (2′): \(a_2 = 15,\ b_2 = -24,\ c_2 = 3\)

\[ \begin{aligned} \frac{a_1}{a_2} = \frac{5}{15} = \frac{1}{3}, \\\\ \frac{b_1}{b_2} = \frac{-8}{-24} = \frac{1}{3}, \\\\ \frac{c_1}{c_2} = \frac{0}{3} = 0 \end{aligned} \]

Step 3: Apply Algebraic Condition

Since:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \]

The lines are parallel.

Step 4: Graphical Interpretation

Parallel lines never intersect, so the system has no solution.

Final Conclusion

The given pair of equations represents parallel lines. Therefore, the system has no solution.

Verification Insight

Since both equations have the same slope but different constants, they represent distinct parallel lines, confirming inconsistency.

CBSE Exam Tip

• Always eliminate fractions before comparing ratios
• Check all three ratios carefully
• Graph confirms algebraic result (double verification = full marks)

Problem: Champa went to a sale to purchase pants and skirts. She said:

• The number of skirts is two less than twice the number of pants.
• The number of skirts is four less than four times the number of pants.

Find the number of pants and skirts she purchased.

Step 1: Formulation of Equations

Let number of pants = x, number of skirts = y

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

Step 2: Solve Algebraically

Equate both equations:

\[ 2x - 2 = 4x - 4 \]

Simplify:

\[ 2 = 2x \Rightarrow x = 1 \]

Substitute in (1):

\[ y = 2(1) - 2 = 0 \]

Step 3: Graphical Interpretation

The solution corresponds to the point of intersection of the lines.

Final Answer

Champa purchased 1 pant and 0 skirts.

Verification

• From (1): \(y = 2(1) - 2 = 0\) ✓
• From (2): \(y = 4(1) - 4 = 0\) ✓

Concept Insight

• Word problems translate into linear equations using keywords like "twice", "less than"
• Intersection point gives real-world solution
• Graph confirms algebraic answer

CBSE Exam Focus

• Frequently asked in 3–4 mark case-based questions
• Proper equation formation carries marks
• Final answer must include units/context

A pair of linear equations can be solved algebraically using systematic methods that provide accurate and exact solutions, unlike graphical methods.

1. Substitution Method

In this method, one equation is solved for one variable and substituted into the other equation.

• Step 1: Solve one equation for x or y
• Step 2: Substitute into the second equation
• Step 3: Solve for one variable
• Step 4: Substitute back to find the other variable

Example:
\(x + y = 5,\quad x - y = 1\)

\(x = 5 - y\) → substitute → \(5 - y - y = 1\) → \(y = 2,\ x = 3\)

2. Elimination Method

This method eliminates one variable by making coefficients equal.

• Step 1: Make coefficients of one variable equal
• Step 2: Add/Subtract equations to eliminate one variable
• Step 3: Solve resulting equation
• Step 4: Substitute to find second variable

Special Cases

• True statement (e.g., 0 = 0) → Infinitely many solutions
• False statement (e.g., 2 = 5) → No solution

Example:
\(2x + y = 7,\quad 2x - y = 1\)

Add → \(4x = 8 \Rightarrow x = 2\), then \(y = 3\)

3. Cross Multiplication Method

For equations:

\[ \begin{aligned} a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0 \end{aligned} \]

The solution is given by:

\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]

Note: This method is efficient but requires careful calculation of determinants.

Visual Solver Flow

Method Selection Strategy

• Use substitution when one equation is simple
• Use elimination when coefficients align easily
• Use cross multiplication for direct formula-based solving

CBSE Exam Insights

• Elimination method is most frequently tested
• Step marking is strict → show all steps clearly
• Verification step improves answer quality

Common Errors to Avoid

• Sign errors during elimination
• Incorrect substitution
• Wrong determinant calculation in cross multiplication

Solve the following pair of linear equations using the substitution method:

\[ \begin{align} 7x - 15y &= 2 \tag{1}\\ x + 2y &= 3 \tag{2} \end{align} \]

Step 1: Express One Variable

From equation (2):

\[ \begin{aligned} x + 2y &= 3 \\\\\Rightarrow y &= \frac{3 - x}{2} \end{aligned} \]

Step 2: Substitute into Equation (1)

\[ 7x - 15\left(\frac{3 - x}{2}\right) = 2 \]

\[ 14x - 15(3 - x) = 4 \]

\[ \begin{aligned} 14x - 45 + 15x &= 4\\\\ \Rightarrow 29x &= 49\\\\ \Rightarrow x &= \frac{49}{29} \end{aligned} \]

Step 3: Substitute Back

Substitute value of \(x\) in equation (2):

\[ \frac{49}{29} + 2y = 3 \]

\[ \begin{aligned} 2y &= 3 - \frac{49}{29}\\ &= \frac{87 - 49}{29}\\ &= \frac{38}{29}\\ \Rightarrow y &= \frac{19}{29} \end{aligned} \]

Step 4: Graphical Interpretation

The solution represents the intersection point of the two lines.

Final Answer

\[ x = \frac{49}{29}, \quad y = \frac{19}{29} \]

Verification

• Substitute in (1): \(7x - 15y = 2\) ✓
• Substitute in (2): \(x + 2y = 3\) ✓

Concept Insight

• Substitution is efficient when one equation is easy to rearrange
• Fractions are common—handle carefully to avoid errors
• Always verify the final answer in both equations

CBSE Exam Tips

• Show substitution step clearly to secure method marks
• Avoid skipping intermediate steps
• Final answer must be in simplest form

The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each saves ₹2000 per month, find their monthly incomes.

Step 1: Define Variables

Let common income factor = x
Let common expenditure factor = y

Then:

• Incomes → 9x and 7x
• Expenditures → 4y and 3y

Step 2: Form Equations

Savings = Income − Expenditure = ₹2000

\[ 9x - 4y = 2000 \tag{1} \] \[ 7x - 3y = 2000 \tag{2} \]

Step 3: Elimination Method

Multiply equation (1) by 3 and equation (2) by 4:

\[ 27x - 12y = 6000 \tag{3} \] \[ 28x - 12y = 8000 \tag{4} \]

Subtract (3) from (4):

\[ \begin{aligned} (28x - 27x) &= 8000 - 6000\\\\ \Rightarrow x &= 2000 \end{aligned} \]

Step 4: Find Incomes

\[ 9x = 9 \times 2000 = 18000 \] \[ 7x = 7 \times 2000 = 14000 \]

Final Answer

Monthly incomes are:
First person = ₹18,000
Second person = ₹14,000

Substitute x = 2000 in (1):

\[ \begin{aligned} 9(2000) - 4y &= 2000\\\\ \Rightarrow 18000 - 4y &= 2000\\\\ \Rightarrow 4y &= 16000\\\\ \Rightarrow y &= 4000 \end{aligned} \]

Now check:

• First person: ₹18000 − ₹16000 = ₹2000 ✓
• Second person: ₹14000 − ₹12000 = ₹2000 ✓

Graphical Interpretation

The equations represent two straight lines whose intersection gives the solution.

Concept Insight

• Ratio problems often use common multipliers (x, y)
• Elimination is fastest when coefficients align
• Always verify using real-world meaning (income − expenditure)

CBSE Exam Focus

• Common 3–4 mark word problem
• Proper equation formation is crucial
• Units (₹) must be included in final answer

The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers are possible?

Step 1: Define Variables

Let the unit digit = x
Let the tens digit = y

Then:

Original number = \(10y + x\)
Reversed number = \(10x + y\)

Step 2: Form Equation from Sum

\[ \begin{align} (10y + x) + (10x + y) &= 66\\\\ \Rightarrow 11x + 11y &= 66\\\\ \Rightarrow x + y &= 6 \tag{1} \end{align} \]

Step 3: Form Equation from Difference

Digits differ by 2 → two cases:

• \(x - y = 2\)
• \(y - x = 2\)

Case 1: \(x - y = 2\)

\[ \begin{aligned} x + y &= 6 \\\\\ x - y &= 2 \end{aligned} \]

\[ \begin{aligned} 2x &= 8 \\\\\Rightarrow x &= 4,\\\\ y &= 2 \end{aligned} \]

Number = 24

Case 2: \(y - x = 2\)

\[ \begin{aligned} x + y &= 6 \\\\ y - x &= 2 \end{aligned} \]

\[ \begin{aligned} 2y &= 8 \\\\\Rightarrow y &= 4,\\\\ x &= 2 \end{aligned} \]

Number = 42

Final Answer

The required numbers are 24 and 42.
Total number of such numbers = 2

Verification

• 24 + 42 = 66 ✓
• |Digits difference| = 2 ✓

Graphical Interpretation

The equations \(x + y = 6\) and \(x - y = 2\) represent straight lines intersecting at (4,2). Similarly, the second case intersects at (2,4).

Concept Insight

• Two-digit numbers → \(10 \times \text{tens} + \text{units}\)
• “Digits differ” → always consider both cases
• Reversing digits swaps positions

CBSE Exam Focus

• Very common 3–4 mark question
• Missing second case leads to loss of marks
• Always write final answer clearly with count

• A pair of linear equations in two variables can be represented and solved using:
  1. Graphical Method
  2. Algebraic Methods
• Graphical Interpretation:
  1. Intersecting lines → Unique solution → Consistent & Independent
  2. Coincident lines → Infinitely many solutions → Consistent & Dependent
  3. Parallel lines → No solution → Inconsistent
• Algebraic Methods of Solving:
  1. Substitution Method
  2. Elimination Method
  3. Cross Multiplication Method (shortcut technique)
• Algebraic Conditions: For equations \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\):
  1. \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) → Unique solution
  2. \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → No solution
  3. \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\) → Infinite solutions
• Determinant Insight (Advanced but Powerful):
  \(D = a_1b_2 - a_2b_1\)
  • \(D \ne 0\) → Unique solution
  • \(D = 0\) → Parallel or coincident lines
• Key Terminology:
  • Consistent: At least one solution exists
  • Inconsistent: No solution
  • Dependent: Infinite solutions (same line)
  • Independent: Unique solution
• Minimum Graph Requirement: At least two points are needed to draw each line accurately.
• Many real-life problems may initially appear non-linear but can be transformed into linear equations using substitutions or simplifications.

Visual Summary

CBSE Exam Smart Tips

• Always compare ratios carefully (sign errors are common)
• Verify solutions in both equations for full marks
• Graph + algebra combination questions are frequently asked
• Missing “nature of solution” statement can reduce marks

1. Homogeneous Pair of Linear Equations

A system of the form:

\[ a_1x + b_1y = 0,\quad a_2x + b_2y = 0 \]

• Always has at least one solution: (0,0)
• If ratios differ → unique solution (trivial)
• If ratios equal → infinite solutions

2. Consistency Test Without Solving

You can determine the nature of solutions without solving equations:

• Compare ratios directly → fastest method in exams
• Saves time in MCQs and case study questions

3. Slope Interpretation (Hidden Concept)

Convert equations into slope-intercept form:

\[ y = mx + c \]

• Equal slopes → parallel or coincident lines
• Different slopes → lines intersect

4. Geometric Meaning of Solution

• Solution = point satisfying both equations simultaneously
• Intersection point = common solution

5. Special Case Detection Shortcut

• If one equation is a multiple of another → infinite solutions
• If only constants differ → no solution

6. Reduction to Linear Form

Some problems are not directly linear but can be converted:

• Substitution (e.g., \(x = 1/u\))
• Simplification of fractions or ratios

7. Real-Life Interpretation Constraints

• Solutions must be realistic (e.g., no negative ages)
• Context decides valid solutions

8. Graph Accuracy Limitations

• Graph gives approximate solutions
• Algebra gives exact values

9. Error Detection Strategy

• Substitute final answer in both equations
• Check sign mistakes in elimination
• Verify ratios carefully

10. Exam Trap Alerts

• Forgetting second case in digit problems
• Ignoring constant ratio \(c_1/c_2\)
• Writing incomplete final answer (missing units/context)

11. Time-Saving Strategy (Topper Trick)

• Use elimination when coefficients match quickly
• Use substitution when one equation is simple
• Use cross multiplication for fast direct answer

12. Concept Linkage (Higher Classes)

• Forms basis for matrices and determinants (Class XI)
• Used in coordinate geometry and optimization