NCERT Solutions Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables
Finding authentic, step-by-step NCERT Solutions Class 10 Mathematics Chapter 3 Pair of Linear Equations in Two Variables is essential for students aiming for top scores in CBSE Class 10 Board exams. Chapter 3 is a foundational pillar of high school algebra, bridging arithmetic relationships and graphical coordinate geometry. Whether you are solving algebraic systems using the substitution method, applying the elimination technique, or analyzing consistency conditions without plotting graphs, these comprehensive NCERT Mathematics Class 10 solutions provide rigorous, NCERT-aligned, exam-ready answers. With 3 core exercises and diverse real-world application models, mastering this chapter guarantees high accuracy across 1-mark objective questions, 3-mark analytical problems, and 5-mark long-answer case studies.
Chapter Overview
A linear equation in two variables represents a straight line on the Cartesian coordinate plane. When two such equations are considered together, they form a pair of linear equations in two variables (or a simultaneous linear system). In CBSE Class 10 Mathematics, Chapter 3 explores both geometric (graphical) and algebraic solutions to determine the coordinates $(x, y)$ that simultaneously satisfy both equations.
The standard algebraic form of a pair of linear equations in two variables $x$ and $y$ is expressed as:
- Equation 1: $a_1x + b_1y + c_1 = 0$
- Equation 2: $a_2x + b_2y + c_2 = 0$
where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers such that $a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$.
Core Topics Covered in Chapter 3
- Graphical Representation and Geometric Interpretation: Plotting linear pairs on graph paper to find points of intersection, parallel lines, or coincident lines.
- Conditions of Consistency: Comparing ratio coefficients $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ to determine whether a system has a unique solution, infinitely many solutions, or no solution.
- Algebraic Method 1 — Substitution Method: Expressing one variable in terms of the other from one equation and substituting it into the second equation.
- Algebraic Method 2 — Elimination Method: Multiplying one or both equations by suitable non-zero constants to equalize coefficients of one variable and eliminate it by addition or subtraction.
- Real-World Word Problems: Formulating linear pairs for problems involving age, geometry, fractions, speed-distance (upstream/downstream), and fixed vs. unit costs.
In the CBSE Class 10 Board examination, Chapter 3 holds a weightage of 6 to 8 marks under the 20-mark Algebra unit. Thorough conceptual clarity and systematic presentation of steps are critical to securing full marks.
Important Formulas and Theorems
Before solving exercise questions, students must memorize the condition table for comparing coefficient ratios and understand the algebraic algorithms.
Consistency and Graphical Conditions Table
| Ratio Comparison | Graphical Representation | Algebraic Interpretation | Consistency of System |
|---|---|---|---|
| $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ | Intersecting Lines | Exactly one (Unique) solution | Consistent |
| $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ | Coincident Lines (Overlapping) | Infinitely many solutions | Consistent (Dependent) |
| $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ | Parallel Lines (Never meet) | No solution | Inconsistent |
Key Algebraic Algorithms
- Substitution Method Algorithm:
- Step 1: Pick one equation and express $y$ in terms of $x$ (i.e., $y = f(x)$) or $x$ in terms of $y$.
- Step 2: Substitute this value into the other equation to get a single linear equation in one variable.
- Step 3: Solve for that variable.
- Step 4: Substitute the obtained value back into either original equation to calculate the second variable.
- Elimination Method Algorithm:
- Step 1: Multiply both equations by appropriate non-zero constants so that the absolute value of the coefficient of one variable (either $x$ or $y$) becomes identical in both equations.
- Step 2: Add or subtract the two equations to eliminate that variable.
- Step 3: Solve the resulting single-variable equation.
- Step 4: Substitute this root back into any original equation to solve for the other variable.
- Relative Speed in Upstream & Downstream Motion:
- Let the speed of the boat in still water be $u\text{ km/h}$ and the speed of the water current/stream be $v\text{ km/h}$ ($u > v$).
- Speed Downstream (with the flow) $= (u + v)\text{ km/h}$
- Speed Upstream (against the flow) $= (u - v)\text{ km/h}$
- $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$
Exercise Solutions
Here are detailed, step-by-step solutions to essential NCERT exercise questions formatted according to official CBSE marking guidelines.
Exercise 3.1 — Question: Graphical and Consistency Verification
Problem: Check graphically whether the pair of equations $x + 3y = 6$ and $2x - 3y = 12$ is consistent. If so, solve them graphically.
Given:
Equation (1): $x + 3y = 6 \implies x = 6 - 3y$
Equation (2): $2x - 3y = 12 \implies 2x = 12 + 3y \implies x = \frac{12 + 3y}{2}$
Step 1: Determine consistency using coefficient ratios
Writing in standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
- $a_1 = 1,\; b_1 = 3,\; c_1 = -6$
- $a_2 = 2,\; b_2 = -3,\; c_2 = -12$
Ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{-3} = -1$.
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point. Hence, the system is consistent.
Step 2: Find coordinate points for Equation (1): $x = 6 - 3y$
- When $y = 0 \implies x = 6 - 3(0) = 6 \implies \text{Point } A(6, 0)$
- When $y = 2 \implies x = 6 - 3(2) = 0 \implies \text{Point } B(0, 2)$
- When $y = 1 \implies x = 6 - 3(1) = 3 \implies \text{Point } C(3, 1)$
Step 3: Find coordinate points for Equation (2): $x = \frac{12 + 3y}{2}$
- When $y = 0 \implies x = \frac{12 + 0}{2} = 6 \implies \text{Point } P(6, 0)$
- When $y = -2 \implies x = \frac{12 - 6}{2} = 3 \implies \text{Point } Q(3, -2)$
- When $y = -4 \implies x = \frac{12 - 12}{2} = 0 \implies \text{Point } R(0, -4)$
Step 4: Graphical Intersection Point
Plotting these lines on coordinate axes shows that both lines intersect precisely at the common point $(6, 0)$.
Final Answer:
$x = 6,\; y = 0$
Exercise 3.2 — Question: Substitution Method (Fraction Problem)
Problem: A fraction becomes $\frac{9}{11}$, if $2$ is added to both the numerator and the denominator. If $3$ is added to both the numerator and the denominator, it becomes $\frac{5}{6}$. Find the fraction.
Step 1: Define variables and formulate algebraic equations
Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.
Condition 1: Adding $2$ to both numerator and denominator:
$$\frac{x + 2}{y + 2} = \frac{9}{11}$$
$$11(x + 2) = 9(y + 2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4 \quad \text{--- (Equation 1)}$$
Condition 2: Adding $3$ to both numerator and denominator:
$$\frac{x + 3}{y + 3} = \frac{5}{6}$$
$$6(x + 3) = 5(y + 3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3 \quad \text{--- (Equation 2)}$$
Step 2: Apply Substitution Method
From Equation (1), express $x$ in terms of $y$:
$$11x = 9y - 4 \implies x = \frac{9y - 4}{11} \quad \text{--- (Equation 3)}$$
Substitute Equation (3) into Equation (2):
$$6\left(\frac{9y - 4}{11}\right) - 5y = -3$$
$$\frac{54y - 24}{11} - 5y = -3$$
Multiply the entire equation by $11$ to clear the denominator:
$$54y - 24 - 55y = -33$$
$$-y - 24 = -33 \implies -y = -33 + 24 \implies -y = -9 \implies y = 9$$
Step 3: Calculate the value of $x$
Substitute $y = 9$ into Equation (3):
$$x = \frac{9(9) - 4}{11} = \frac{81 - 4}{11} = \frac{77}{11} = 7$$
Final Answer:
The required fraction is $\frac{7}{9}$.
Exercise 3.2 — Question: Substitution Method (Age Problem)
Problem: Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
Step 1: Variable Declaration
Let Jacob's present age $= x\text{ years}$ and his son's present age $= y\text{ years}$.
Step 2: Formulate Equations
Five years hence (future):
Jacob's age $= x + 5$, Son's age $= y + 5$
$$x + 5 = 3(y + 5) \implies x + 5 = 3y + 15 \implies x - 3y = 10 \quad \text{--- (Equation 1)}$$
Five years ago (past):
Jacob's age $= x - 5$, Son's age $= y - 5$
$$x - 5 = 7(y - 5) \implies x - 5 = 7y - 35 \implies x - 7y = -30 \quad \text{--- (Equation 2)}$$
Step 3: Solve by Substitution
From Equation (1), $x = 3y + 10$.
Substitute this expression into Equation (2):
$$(3y + 10) - 7y = -30$$
$$-4y + 10 = -30 \implies -4y = -40 \implies y = 10$$
Substitute $y = 10$ back into Equation (1):
$$x = 3(10) + 10 = 30 + 10 = 40$$
Final Answer:
Jacob's present age is $40\text{ years}$ and his son's present age is $10\text{ years}$.
Exercise 3.3 — Question: Elimination Method (Two-Digit Number Problem)
Problem: The sum of a two-digit number and the number obtained by reversing the digits is $66$. If the digits of the number differ by $2$, find the number. How many such numbers are there?
Step 1: Set up the digits
Let the tens digit be $x$ and the units digit be $y$.
- Original Number $= 10x + y$
- Reversed Number $= 10y + x$
Step 2: Formulate equations
$$\text{Original Number} + \text{Reversed Number} = 66$$
$$(10x + y) + (10y + x) = 66 \implies 11x + 11y = 66 \implies x + y = 6 \quad \text{--- (Equation 1)}$$
Given that the difference between the digits is $2$, two cases arise:
Case I: $x - y = 2 \quad \text{--- (Equation 2a)}$
Adding Equation (1) and Equation (2a):
$$(x + y) + (x - y) = 6 + 2 \implies 2x = 8 \implies x = 4$$
Substituting $x = 4$ into Equation (1):
$$4 + y = 6 \implies y = 2$$
Thus, the number is $10(4) + 2 = 42$.
Case II: $y - x = 2 \implies -x + y = 2 \quad \text{--- (Equation 2b)}$
Adding Equation (1) and Equation (2b):
$$(x + y) + (-x + y) = 6 + 2 \implies 2y = 8 \implies y = 4$$
Substituting $y = 4$ into Equation (1):
$$x + 4 = 6 \implies x = 2$$
Thus, the number is $10(2) + 4 = 24$.
Final Answer:
There are two such numbers: $42$ and $24$.
Exercise 3.3 — Question: Elimination Method (Fixed and Running Cost)
Problem: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\text{ km}$, the charge paid is $\text{Rs } 105$ and for a journey of $15\text{ km}$, the charge paid is $\text{Rs } 155$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25\text{ km}$?
Step 1: Variable Definition
Let the fixed charge be $\text{Rs } x$ and the running charge per kilometer be $\text{Rs } y$.
Step 2: Formulate the Linear System
$$x + 10y = 105 \quad \text{--- (Equation 1)}$$
$$x + 15y = 155 \quad \text{--- (Equation 2)}$$
Step 3: Solve by Elimination Method
Subtract Equation (1) from Equation (2):
$$(x + 15y) - (x + 10y) = 155 - 105$$
$$5y = 50 \implies y = \frac{50}{5} = 10$$
Substitute $y = 10$ into Equation (1):
$$x + 10(10) = 105 \implies x + 100 = 105 \implies x = 105 - 100 = 5$$
Step 4: Calculate Cost for $25\text{ km}$ Journey
$$\text{Total Fare} = x + 25y = 5 + 25(10) = 5 + 250 = \text{Rs } 255$$
Final Answer:
Fixed charge $= \text{Rs } 5$, Charge per km $= \text{Rs } 10$, and the fare for $25\text{ km} = \text{Rs } 255$.
High-Order Thinking Skills (HOTS) — Upstream and Downstream Motion
Problem: A boat covers $30\text{ km}$ upstream and $44\text{ km}$ downstream in $10\text{ hours}$. In $13\text{ hours}$, it can cover $40\text{ km}$ upstream and $55\text{ km}$ downstream. Determine the speed of the stream and that of the boat in still water.
Step 1: Set Up Motion Equations
Let the speed of the boat in still water be $x\text{ km/h}$ and speed of the stream be $y\text{ km/h}$.
- Speed upstream $= (x - y)\text{ km/h}$
- Speed downstream $= (x + y)\text{ km/h}$
Using $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
$$\frac{30}{x - y} + \frac{44}{x + y} = 10 \quad \text{--- (Equation 1)}$$
$$\frac{40}{x - y} + \frac{55}{x + y} = 13 \quad \text{--- (Equation 2)}$$
Step 2: Reducible Substitution
Let $\frac{1}{x - y} = u$ and $\frac{1}{x + y} = v$. The equations become:
$$30u + 44v = 10 \quad \text{--- (Equation 3)}$$
$$40u + 55v = 13 \quad \text{--- (Equation 4)}$$
Step 3: Solve for $u$ and $v$ using Elimination
Multiply Equation (3) by $4$ and Equation (4) by $3$ to equate coefficients of $u$:
$$120u + 176v = 40 \quad \text{--- (Equation 5)}$$
$$120u + 165v = 39 \quad \text{--- (Equation 6)}$$
Subtract Equation (6) from Equation (5):
$$11v = 1 \implies v = \frac{1}{11}$$
Substitute $v = \frac{1}{11}$ into Equation (3):
$$30u + 44\left(\frac{1}{11}\right) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{6}{30} = \frac{1}{5}$$
Step 4: Solve for $x$ and $y$
$$\frac{1}{x - y} = \frac{1}{5} \implies x - y = 5 \quad \text{--- (Equation 7)}$$
$$\frac{1}{x + y} = \frac{1}{11} \implies x + y = 11 \quad \text{--- (Equation 8)}$$
Adding Equations (7) and (8):
$$2x = 16 \implies x = 8\text{ km/h}$$
Subtracting Equation (7) from Equation (8):
$$2y = 6 \implies y = 3\text{ km/h}$$
Final Answer:
Speed of boat in still water $= 8\text{ km/h}$; Speed of stream $= 3\text{ km/h}$.
Common Mistakes and Tips
Analyzing past CBSE board exam evaluator reports highlights recurring student pitfalls in Chapter 3:
- Inconsistent Standard Form when Comparing Ratios: When testing consistency, ensure constants $c_1$ and $c_2$ are on the same side of the equality sign. For instance, comparing $2x + 3y = 7$ with $4x + 6y + 14 = 0$ directly without shifting $14$ leads to incorrect sign ratios for $\frac{c_1}{c_2}$.
- Sign Reversal in Elimination Subtraction: When subtracting one equation from another (e.g., $(3x - 5y) - (2x - 7y)$), students often forget to invert the negative signs, writing $-5y - 7y$ instead of $-5y + 7y = +2y$. Always enclose the second expression in brackets before expanding.
- Neglecting Dual Cases in Digit Difference Problems: In problems where "digits differ by $2$", $(x - y = 2)$ is only one possibility. You must also evaluate $(y - x = 2)$ unless the problem explicitly states which digit is larger.
- Forgetting to State Physical Units: In applied word problems involving speed ($\text{km/h}$), money ($\text{Rs}$), and time ($\text{hours}$/$\text{years}$), leaving answers as bare numbers causes deductions in final step marks.
- Quick Verification Check: Always plug your final values of $x$ and $y$ back into both original equations during revision. If both equations balance to true identities, your solution is guaranteed to be 100% accurate.
Board Exam Relevance
In the CBSE Class 10 Board Exam question paper blueprint, questions from Chapter 3 are distributed across multiple sections:
- Section A (1 Mark - MCQs / Assertion-Reason): Ratio comparison questions finding the value of unknown constant $k$ for which a linear system has infinitely many solutions, a unique solution, or no solution (e.g., parallel lines condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$).
- Section B & C (2 & 3 Marks - Short Answer Questions): Direct evaluation of simultaneous equations via substitution or elimination, fraction formulations, or perimeter/area linear formulations.
- Section D & E (4 & 5 Marks - Long Answer & Case Study Based Questions): High-weightage real-life word problems such as upstream/downstream watercraft movement, fixed plus distance-based transport fare systems, age relations, and commercial billing models.
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