NCERT Solutions for Class 10 Mathematics Chapter 3: Pair of Linear Equations in Two Variables (Complete Guide)
Linear algebraic systems in two variables form the core framework for analytical geometry, optimization, and multi-variable modeling in secondary mathematics. In the CBSE Class 10 curriculum, Chapter 3: Pair of Linear Equations in Two Variables covers both graphical analysis and algebraic methods for solving simultaneous first-degree equations over the real numbers ($\mathbb{R}^2$). Mastering this chapter requires understanding how algebraic coefficients determine whether a system intersects at a unique point, represents parallel trajectories, or merges into coincident lines.
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| SYSTEM OF PAIR OF LINEAR EQUATIONS IN TWO VARIABLES |
| |
| Given System: a1*x + b1*y + c1 = 0 and a2*x + b2*y + c2 = 0 |
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| +-----------------------------+-----------------------------+---------------------------------+ |
| | Intersecting Lines | Coincident Lines | Parallel Lines | |
| | Ratio: a1/a2 != b1/b2 | Ratio: a1/a2 = b1/b2 = c1/c2| Ratio: a1/a2 = b1/b2 != c1/c2 | |
| | Solutions: Exactly One | Solutions: Infinitely Many | Solutions: No Solution (Zero) | |
| | System: Consistent | System: Consistent/Dependent| System: Inconsistent | |
| | Graph: Single Point (x, y) | Graph: Overlapping Line | Graph: Equidistant Lines | |
| +-----------------------------+-----------------------------+---------------------------------+ |
+---------------------------------------------------------------------------------------------------+
The geometric interpretation relies on the fact that any linear equation $ax + by + c = 0$ (where $a^2 + b^2 \neq 0$) forms a straight line on the Cartesian plane. When two such lines are graphed simultaneously, their interaction follows one of three geometric possibilities:
- Unique Intersection: The lines cross at a single point $(x, y)$, providing a unique solution to the system ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$).
- Coincident Lines: The lines overlap entirely, providing infinitely many solutions ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$).
- Parallel Lines: The lines remain equidistant and never intersect, yielding no solution ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$).
Algebraically, the rationalized CBSE curriculum focuses on two exact methods: the Substitution Method and the Elimination Method. The substitution method isolates one variable in terms of the other to reduce the system to a single-variable linear equation. The elimination method scales the equations using LCM multipliers so that adding or subtracting them eliminates one variable directly.
According to official CBSE board evaluation criteria, solutions to word problems require a clear, stepwise approach:
- Explicit variable declarations with appropriate units (e.g., “Let the fixed charge be $₹x$ and charge per km be $₹y$”).
- Algebraic formulation of both linear equations from the given conditions.
- Step-by-step reduction using either substitution or elimination with clear arithmetic operations.
- Final concluding statement specifying both values with correct units.
| Ratio Comparison | Graphical Representation | Algebraic Interpretation | System Consistency | Determinant / Condition |
| $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ | Intersecting lines | Exactly one (unique) solution | Consistent | $a_1 b_2 – a_2 b_1 \neq 0$ |
| $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ | Coincident lines | Infinitely many solutions | Consistent (Dependent) | Lines lie directly on top of each other |
| $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ | Parallel lines | No solution | Inconsistent | Slopes are equal, intercepts differ |
Chapter-End Exercises – Exercise 3.1 (Page No. 44)
Question 1. Form the pair of linear equations in the following problems, and find their solutions graphically:
(i) $10$ students of Class X took part in a Mathematics quiz. If the number of girls is $4$ more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) $5$ pencils and $7$ pens together cost $₹50$, whereas $7$ pencils and $5$ pens together cost $₹46$. Find the cost of one pencil and that of one pen.
(i) Mathematics Quiz Problem [BOARD EXAM FAVORITE / CBSE 2019, 2022]
Answer:
Step 1: Variable Declaration & Equation Formulation
Let the number of girls who took part in the quiz be $x$.
Let the number of boys who took part in the quiz be $y$.
According to the first condition (total of $10$ students):
$$x + y = 10 \quad \text{— (Equation 1)}$$
According to the second condition (number of girls is $4$ more than boys):
$$x = y + 4 \implies x – y = 4 \quad \text{— (Equation 2)}$$
Step 2: Generating Coordinate Tables for Graphing
For Equation 1: $y = 10 – x$
- If $x = 5 \implies y = 10 – 5 = 5 \implies (5, 5)$
- If $x = 7 \implies y = 10 – 7 = 3 \implies (7, 3)$
- If $x = 3 \implies y = 10 – 3 = 7 \implies (3, 7)$
For Equation 2: $y = x – 4$
- If $x = 4 \implies y = 4 – 4 = 0 \implies (4, 0)$
- If $x = 7 \implies y = 7 – 4 = 3 \implies (7, 3)$
- If $x = 6 \implies y = 6 – 4 = 2 \implies (6, 2)$
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GRAPHICAL REPRESENTATION SCHEMATIC
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Y
10+ * (3, 7) [Line 1: x + y = 10]
| /
7+ / * (5, 5)
| / /
5+ / /
| / * INTERSECTION POINT P(7, 3)
3+-----/---*------------------------------ [x = 7, y = 3]
| / / \
| / / * (6, 2)
| / * (4, 0) [Line 2: x - y = 4]
0+--+---+---+---+---+---+---+---+---> X
0 2 4 6 7 8 10
===================================================================================================
Step 3: Finding the Graphical Solution
Plotting both straight lines on graph paper shows that they intersect at the common point $(7, 3)$.
This corresponds to $x = 7$ and $y = 3$.
Final Statement:
The number of girls is $7$ and the number of boys is $3$.
(ii) Cost of Pencils and Pens [BOARD EXAM FAVORITE / CBSE 2020, 2023]
Answer:
Step 1: Variable Declaration & Equation Formulation
Let the cost of $1$ pencil be $₹x$.
Let the cost of $1$ pen be $₹y$.
According to the first given condition ($5$ pencils and $7$ pens cost $₹50$):
$$5x + 7y = 50 \quad \text{— (Equation 1)}$$
According to the second given condition ($7$ pencils and $5$ pens cost $₹46$):
$$7x + 5y = 46 \quad \text{— (Equation 2)}$$
Step 2: Generating Coordinate Tables for Graphing
For Equation 1: $y = \frac{50 – 5x}{7}$
- If $x = 3 \implies y = \frac{50 – 15}{7} = \frac{35}{7} = 5 \implies (3, 5)$
- If $x = 10 \implies y = \frac{50 – 50}{7} = 0 \implies (10, 0)$
- If $x = -4 \implies y = \frac{50 – (-20)}{7} = \frac{70}{7} = 10 \implies (-4, 10)$
For Equation 2: $y = \frac{46 – 7x}{5}$
- If $x = 3 \implies y = \frac{46 – 21}{5} = \frac{25}{5} = 5 \implies (3, 5)$
- If $x = 8 \implies y = \frac{46 – 56}{5} = -\frac{10}{5} = -2 \implies (8, -2)$
- If $x = -2 \implies y = \frac{46 – (-14)}{5} = \frac{60}{5} = 12 \implies (-2, 12)$
Step 3: Finding the Graphical Solution
Plotting both lines reveals that they intersect at the common point $(3, 5)$.
This gives $x = 3$ and $y = 5$.
Final Statement:
The cost of one pencil is $₹3$ and the cost of one pen is $₹5$.
Question 2. On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) $5x – 4y + 8 = 0;\quad 7x + 6y – 9 = 0$
(ii) $9x + 3y + 12 = 0;\quad 18x + 6y + 24 = 0$
(iii) $6x – 3y + 10 = 0;\quad 2x – y + 9 = 0$
(i) $5x – 4y + 8 = 0$ and $7x + 6y – 9 = 0$
Answer:
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$$\begin{aligned} a_1 &= 5, \quad b_1 = -4, \quad c_1 = 8 \\ a_2 &= 7, \quad b_2 = 6, \quad c_2 = -9 \end{aligned}$$
Calculating the coefficient ratios:
$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ ($\frac{5}{7} \neq -\frac{2}{3}$):
Final Statement:
The lines representing this pair of equations intersect at a single point and have a unique solution.
(ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$
Answer:
Comparing coefficients:
$$\begin{aligned} a_1 &= 9, \quad b_1 = 3, \quad c_1 = 12 \\ a_2 &= 18, \quad b_2 = 6, \quad c_2 = 24 \end{aligned}$$
Calculating the ratios:
$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$:
Final Statement:
The lines representing this pair of equations are coincident lines and have infinitely many solutions.
(iii) $6x – 3y + 10 = 0$ and $2x – y + 9 = 0$
Answer:
Comparing coefficients:
$$\begin{aligned} a_1 &= 6, \quad b_1 = -3, \quad c_1 = 10 \\ a_2 &= 2, \quad b_2 = -1, \quad c_2 = 9 \end{aligned}$$
Calculating the ratios:
$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ ($3 = 3 \neq \frac{10}{9}$):
Final Statement:
The lines representing this pair of equations are parallel lines and have no solution.
Question 3. On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, find out whether the following pair of linear equations are consistent, or inconsistent:
(i) $3x + 2y = 5;\quad 2x – 3y = 7$
(ii) $2x – 3y = 8;\quad 4x – 6y = 9$
(iii) $\frac{3}{2}x + \frac{5}{3}y = 7;\quad 9x – 10y = 14$
(iv) $5x – 3y = 11;\quad -10x + 6y = -22$
(v) $\frac{4}{3}x + 2y = 8;\quad 2x + 3y = 12$
(i) $3x + 2y – 5 = 0$ and $2x – 3y – 7 = 0$
Answer:
$$\frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{2}{-3} = -\frac{2}{3}$$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system has a unique solution.
Therefore, this pair of linear equations is consistent.
(ii) $2x – 3y – 8 = 0$ and $4x – 6y – 9 = 0$
Answer:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and have no solution.
Therefore, this pair of linear equations is inconsistent.
(iii) $\frac{3}{2}x + \frac{5}{3}y – 7 = 0$ and $9x – 10y – 14 = 0$
Answer:
$$\begin{aligned} \frac{a_1}{a_2} &= \frac{\frac{3}{2}}{9} = \frac{3}{18} = \frac{1}{6} \\ \frac{b_1}{b_2} &= \frac{\frac{5}{3}}{-10} = \frac{5}{-30} = -\frac{1}{6} \end{aligned}$$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ ($\frac{1}{6} \neq -\frac{1}{6}$), the system has a unique solution.
Therefore, this pair of linear equations is consistent.
(iv) $5x – 3y – 11 = 0$ and $-10x + 6y + 22 = 0$
Answer:
$$\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$, the lines are coincident and have infinitely many solutions.
Therefore, this pair of linear equations is consistent (dependent).
(v) $\frac{4}{3}x + 2y – 8 = 0$ and $2x + 3y – 12 = 0$
Answer:
$$\frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{2}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$, the lines are coincident and have infinitely many solutions.
Therefore, this pair of linear equations is consistent.
Question 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) $x + y = 5;\quad 2x + 2y = 10$
(ii) $x – y = 8;\quad 3x – 3y = 16$
(iii) $2x + y – 6 = 0;\quad 4x – 2y – 4 = 0$
(iv) $2x – 2y – 2 = 0;\quad 4x – 4y – 5 = 0$
(i) $x + y = 5$ and $2x + 2y = 10$
Answer:
$$\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$$
The system is consistent (coincident lines with infinitely many solutions).
Table for $x + y = 5 \implies y = 5 – x$:
- $x = 0 \implies y = 5 \implies (0, 5)$
- $x = 5 \implies y = 0 \implies (5, 0)$
- $x = 2 \implies y = 3 \implies (2, 3)$
Both equations represent the exact same line. Any point $(x, 5 – x)$ on this line is a solution.
(ii) $x – y = 8$ and $3x – 3y = 16$
Answer:
$$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
Therefore, the system is inconsistent and has no graphical solution.
(iii) $2x + y – 6 = 0$ and $4x – 2y – 4 = 0$ [BOARD EXAM FAVORITE]
Answer:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system is consistent with a unique solution.
Table for $y = 6 – 2x$:
- $x = 0 \implies y = 6 \implies (0, 6)$
- $x = 2 \implies y = 2 \implies (2, 2)$
- $x = 3 \implies y = 0 \implies (3, 0)$
Table for $2y = 4x – 4 \implies y = 2x – 2$:
- $x = 0 \implies y = -2 \implies (0, -2)$
- $x = 2 \implies y = 2 \implies (2, 2)$
- $x = 1 \implies y = 0 \implies (1, 0)$
Plotting both lines shows they intersect at $(2, 2)$.
Thus, the solution is $x = 2, y = 2$.
(iv) $2x – 2y – 2 = 0$ and $4x – 4y – 5 = 0$
Answer:
$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
Therefore, the system is inconsistent.
Question 5. Half the perimeter of a rectangular garden, whose length is $4\text{ m}$ more than its width, is $36\text{ m}$. Find the dimensions of the garden. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2023]
Answer:
Step 1: Variable Declaration & Equation Formulation
Let the length of the rectangular garden be $x\text{ metres}$.
Let the width of the rectangular garden be $y\text{ metres}$.
According to the first condition (length is $4\text{ m}$ more than width):
$$x = y + 4 \implies x – y = 4 \quad \text{— (Equation 1)}$$
The perimeter of a rectangle is $2(\text{length} + \text{width}) = 2(x + y)$.
Half the perimeter is $\frac{2(x + y)}{2} = x + y = 36$:
$$x + y = 36 \quad \text{— (Equation 2)}$$
Step 2: Solving by Elimination / Substitution
Adding Equation 1 and Equation 2:
$$\begin{aligned} (x – y) + (x + y) &= 4 + 36 \\ 2x &= 40 \\ x &= 20\text{ m} \end{aligned}$$
Substitute $x = 20$ into Equation 2:
$$\begin{aligned} 20 + y &= 36 \\ y &= 36 – 20 \\ y &= 16\text{ m} \end{aligned}$$
Final Statement:
The length of the garden is $20\text{ m}$ and the width of the garden is $16\text{ m}$.
Question 6. Given the linear equation $2x + 3y – 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) Intersecting lines
(ii) Parallel lines
(iii) Coincident lines
Answer:
Given equation: $2x + 3y – 8 = 0$, where $a_1 = 2, b_1 = 3, c_1 = -8$.
(i) For Intersecting Lines:
Condition required: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Choosing $a_2 = 3, b_2 = 2, c_2 = -7$:
$$\frac{2}{3} \neq \frac{3}{2}$$
A required equation is $3x + 2y – 7 = 0$.
(ii) For Parallel Lines:
Condition required: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Multiplying the $x$ and $y$ coefficients by $2$, and choosing a different constant:
Choosing $a_2 = 4, b_2 = 6, c_2 = -9$:
$$\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-9} \implies \frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$$
A required equation is $4x + 6y – 9 = 0$.
(iii) For Coincident Lines:
Condition required: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Multiplying the entire equation by a constant factor (e.g., $3$):
Choosing $a_2 = 6, b_2 = 9, c_2 = -24$:
$$\frac{2}{6} = \frac{3}{9} = \frac{-8}{-24} = \frac{1}{3}$$
A required equation is $6x + 9y – 24 = 0$.
Question 7. Draw the graphs of the equations $x – y + 1 = 0$ and $3x + 2y – 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the $x$-axis, and shade the triangular region. [BOARD EXAM FAVORITE / CBSE 2019, 2021, 2023]
Answer:
Step 1: Finding Coordinate Tables
For Equation 1: $x – y + 1 = 0 \implies y = x + 1$
- If $x = 0 \implies y = 1 \implies (0, 1)$
- If $x = -1 \implies y = 0 \implies (-1, 0)$
- If $x = 2 \implies y = 3 \implies (2, 3)$
For Equation 2: $3x + 2y – 12 = 0 \implies y = \frac{12 – 3x}{2}$
- If $x = 0 \implies y = 6 \implies (0, 6)$
- If $x = 4 \implies y = 0 \implies (4, 0)$
- If $x = 2 \implies y = 3 \implies (2, 3)$
Step 2: Identifying the Triangular Region & Vertices
Both lines intersect each other at the top vertex $A(2, 3)$.
- Line 1 crosses the $x$-axis ($y = 0$) at $B(-1, 0)$.
- Line 2 crosses the $x$-axis ($y = 0$) at $C(4, 0)$.
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TRIANGULAR REGION VERTICES SCHEMATIC
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Y
6+ * (0, 6)
| /
3+---------/-------* A(2, 3) [Intersection Vertex]
| / / \
1+ * (0, 1)/ \
| / / \
0+--*--+-------+-------*-------> X
B(-1, 0) 0 C(4, 0)
[<------- BASE OF TRIANGLE (5 units) ------->]
===================================================================================================
Step 3: Calculating Area of the Shaded Region (Extended Analysis)
$$\begin{aligned} \text{Base length } BC &= |4 – (-1)| = 5\text{ units} \\ \text{Height } h &= y\text{-coordinate of } A = 3\text{ units} \\ \text{Area} &= \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 5 \times 3 = 7.5\text{ sq units} \end{aligned}$$
Final Statement:
The coordinates of the vertices of the triangle are $A(2, 3)$, $B(-1, 0)$, and $C(4, 0)$.
Chapter-End Exercises – Exercise 3.2 (Page No. 53)
Question 1. Solve the following pair of linear equations by the substitution method:
(i) $x + y = 14;\quad x – y = 4$
(ii) $s – t = 3;\quad \frac{s}{3} + \frac{t}{2} = 6$
(iii) $3x – y = 3;\quad 9x – 3y = 9$
(iv) $0.2x + 0.3y = 1.3;\quad 0.4x + 0.5y = 2.3$
(v) $\sqrt{2}x + \sqrt{3}y = 0;\quad \sqrt{3}x – \sqrt{8}y = 0$
(vi) $\frac{3x}{2} – \frac{5y}{3} = -2;\quad \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(i) $x + y = 14$ and $x – y = 4$
Answer:
From the first equation:
$$y = 14 – x \quad \text{— (Equation 1)}$$
Substitute this into the second equation:
$$\begin{aligned} x – (14 – x) &= 4 \\ x – 14 + x &= 4 \\ 2x &= 18 \\ x &= 9 \end{aligned}$$
Substitute $x = 9$ back into Equation 1:
$$y = 14 – 9 = 5$$
Thus, $x = 9, y = 5$.
(ii) $s – t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$ [BOARD EXAM FAVORITE]
Answer:
From $s – t = 3$, we have:
$$s = t + 3 \quad \text{— (Equation 1)}$$
Multiply the second equation by $6$ (LCM of $3$ and $2$) to clear fractions:
$$2s + 3t = 36 \quad \text{— (Equation 2)}$$
Substitute Equation 1 into Equation 2:
$$\begin{aligned} 2(t + 3) + 3t &= 36 \\ 2t + 6 + 3t &= 36 \\ 5t &= 30 \\ t &= 6 \end{aligned}$$
Substitute $t = 6$ into Equation 1:
$$s = 6 + 3 = 9$$
Thus, $s = 9, t = 6$.
(iii) $3x – y = 3$ and $9x – 3y = 9$
Answer:
From $3x – y = 3$, express $y$ in terms of $x$:
$$y = 3x – 3$$
Substitute this into the second equation:
$$\begin{aligned} 9x – 3(3x – 3) &= 9 \\ 9x – 9x + 9 &= 9 \\ 9 &= 9 \end{aligned}$$
This is a true identity for all real values of $x$. Both equations represent the same line ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{3}$).
Therefore, the system has infinitely many solutions given by the relation $y = 3x – 3$.
(iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$
Answer:
Multiply both equations by $10$ to eliminate decimals:
$$\begin{aligned} 2x + 3y &= 13 \quad \text{— (Equation 1)} \\ 4x + 5y &= 23 \quad \text{— (Equation 2)} \end{aligned}$$
From Equation 1:
$$x = \frac{13 – 3y}{2}$$
Substitute this into Equation 2:
$$\begin{aligned} 4\left(\frac{13 – 3y}{2}\right) + 5y &= 23 \\ 2(13 – 3y) + 5y &= 23 \\ 26 – 6y + 5y &= 23 \\ -y &= 23 – 26 = -3 \implies y = 3 \end{aligned}$$
Substitute $y = 3$ into the expression for $x$:
$$x = \frac{13 – 3(3)}{2} = \frac{13 – 9}{2} = \frac{4}{2} = 2$$
Thus, $x = 2, y = 3$.
(v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x – \sqrt{8}y = 0$ [BOARD EXAM FAVORITE]
Answer:
From the first equation:
$$x = -\frac{\sqrt{3}}{\sqrt{2}}y$$
Substitute this into the second equation:
$$\begin{aligned} \sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) – \sqrt{8}y &= 0 \\ -\frac{3}{\sqrt{2}}y – 2\sqrt{2}y &= 0 \\ y\left(-\frac{3}{\sqrt{2}} – \frac{4}{\sqrt{2}}\right) &= 0 \\ y\left(-\frac{7}{\sqrt{2}}\right) &= 0 \implies y = 0 \end{aligned}$$
Substitute $y = 0$ into the expression for $x$:
$$x = -\frac{\sqrt{3}}{\sqrt{2}}(0) = 0$$
Thus, $x = 0, y = 0$.
(vi) $\frac{3x}{2} – \frac{5y}{3} = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Answer:
Multiply both equations by $6$ to clear fractions:
$$\begin{aligned} 9x – 10y &= -12 \quad \text{— (Equation 1)} \\ 2x + 3y &= 13 \quad \text{— (Equation 2)} \end{aligned}$$
From Equation 2:
$$x = \frac{13 – 3y}{2}$$
Substitute this into Equation 1:
$$\begin{aligned} 9\left(\frac{13 – 3y}{2}\right) – 10y &= -12 \\ \frac{117 – 27y – 20y}{2} &= -12 \\ 117 – 47y &= -24 \\ -47y &= -141 \\ y &= \frac{-141}{-47} = 3 \end{aligned}$$
Substitute $y = 3$ into the expression for $x$:
$$x = \frac{13 – 3(3)}{2} = \frac{13 – 9}{2} = 2$$
Thus, $x = 2, y = 3$.
Question 2. Solve $2x + 3y = 11$ and $2x – 4y = -24$ and hence find the value of ‘$m$’ for which $y = mx + 3$. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2022, 2023]
Answer:
Step 1: Solving the Linear System
$$\begin{aligned} 2x + 3y &= 11 \quad \text{— (Equation 1)} \\ 2x – 4y &= -24 \quad \text{— (Equation 2)} \end{aligned}$$
Subtracting Equation 2 from Equation 1:
$$\begin{aligned} (2x + 3y) – (2x – 4y) &= 11 – (-24) \\ 7y &= 35 \\ y &= 5 \end{aligned}$$
Substitute $y = 5$ into Equation 1:
$$\begin{aligned} 2x + 3(5) &= 11 \\ 2x + 15 &= 11 \\ 2x &= -4 \\ x &= -2 \end{aligned}$$
The solution to the system is $x = -2$ and $y = 5$.
Step 2: Finding the Parameter $m$
Substitute $x = -2$ and $y = 5$ into the line equation $y = mx + 3$:
$$\begin{aligned} 5 &= m(-2) + 3 \\ 5 – 3 &= -2m \\ 2 &= -2m \\ m &= \frac{2}{-2} = -1 \end{aligned}$$
Final Statement:
The solution is $x = -2, y = 5$, and the value of $m$ is $-1$.
Question 3. Form the pair of linear equations for the following problems and find their solution by substitution method:
(i) The difference between two numbers is $26$ and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by $18$ degrees. Find them.
(iii) The coach of a cricket team buys $7$ bats and $6$ balls for $₹3800$. Later, she buys $3$ bats and $5$ balls for $₹1750$. Find the cost of each bat and each ball.
(iv) 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 $₹105$ and for a journey of $15\text{ km}$, the charge paid is $₹155$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25\text{ km}$?
(v) 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.
(vi) 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?
(i) Difference of Two Numbers
Answer:
Let the larger number be $x$ and the smaller number be $y$.
From the given conditions:
$$\begin{aligned} x – y &= 26 \quad \text{— (Equation 1)} \\ x &= 3y \quad \text{— (Equation 2)} \end{aligned}$$
Substitute Equation 2 into Equation 1:
$$\begin{aligned} 3y – y &= 26 \\ 2y &= 26 \implies y = 13 \end{aligned}$$
Substitute $y = 13$ into Equation 2:
$$x = 3(13) = 39$$
Thus, the two numbers are $39$ and $13$.
(ii) Supplementary Angles [BOARD EXAM FAVORITE / CBSE 2019, 2021]
Answer:
Let the larger angle be $x^\circ$ and the smaller angle be $y^\circ$.
Supplementary angles sum to $180^\circ$:
$$x + y = 180 \quad \text{— (Equation 1)}$$
The larger exceeds the smaller by $18^\circ$:
$$x = y + 18 \quad \text{— (Equation 2)}$$
Substitute Equation 2 into Equation 1:
$$\begin{aligned} (y + 18) + y &= 180 \\ 2y + 18 &= 180 \\ 2y &= 162 \implies y = 81^\circ \end{aligned}$$
Substitute $y = 81^\circ$ into Equation 2:
$$x = 81 + 18 = 99^\circ$$
Thus, the two supplementary angles are $99^\circ$ and $81^\circ$.
(iii) Cost of Cricket Bats and Balls
Answer:
Let the cost of $1$ bat be $₹x$ and $1$ ball be $₹y$.
Formulating the equations:
$$\begin{aligned} 7x + 6y &= 3800 \quad \text{— (Equation 1)} \\ 3x + 5y &= 1750 \quad \text{— (Equation 2)} \end{aligned}$$
From Equation 2:
$$x = \frac{1750 – 5y}{3}$$
Substitute this into Equation 1:
$$\begin{aligned} 7\left(\frac{1750 – 5y}{3}\right) + 6y &= 3800 \\ \frac{12250 – 35y + 18y}{3} &= 3800 \\ 12250 – 17y &= 11400 \\ -17y &= 11400 – 12250 = -850 \\ y &= \frac{-850}{-17} = 50 \end{aligned}$$
Substitute $y = 50$ into the expression for $x$:
$$x = \frac{1750 – 5(50)}{3} = \frac{1750 – 250}{3} = \frac{1500}{3} = 500$$
Thus, the cost of one bat is $₹500$ and the cost of one ball is $₹50$.
(iv) Taxi Charges and $25\text{ km}$ Journey [BOARD EXAM FAVORITE / CBSE 2020, 2022, 2023]
Answer:
Step 1: Equation Formulation
Let the fixed base charge be $₹x$.
Let the running charge per kilometer be $₹y$.
For a journey of $10\text{ km}$:
$$x + 10y = 105 \quad \text{— (Equation 1)}$$
For a journey of $15\text{ km}$:
$$x + 15y = 155 \quad \text{— (Equation 2)}$$
Step 2: Solving for $x$ and $y$
From Equation 1: $x = 105 – 10y$.
Substitute this into Equation 2:
$$\begin{aligned} (105 – 10y) + 15y &= 155 \\ 105 + 5y &= 155 \\ 5y &= 50 \implies y = 10 \end{aligned}$$
Substitute $y = 10$ into Equation 1:
$$x = 105 – 10(10) = 105 – 100 = 5$$
Thus, the fixed charge is $₹5$ and the charge per km is $₹10$.
Step 3: Cost for $25\text{ km}$ Journey
$$\text{Total Cost} = x + 25y = 5 + 25(10) = 5 + 250 = ₹255$$
Final Statement:
The fixed charge is $₹5$, the charge per km is $₹10$, and the amount to pay for travelling $25\text{ km}$ is $₹255$.
(v) Fraction Problem [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
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 gives $\frac{9}{11}$:
$$\begin{aligned} \frac{x + 2}{y + 2} &= \frac{9}{11} \\ 11(x + 2) &= 9(y + 2) \\ 11x + 22 &= 9y + 18 \\ 11x – 9y &= -4 \quad \text{— (Equation 1)} \end{aligned}$$
Condition 2: Adding $3$ to both numerator and denominator gives $\frac{5}{6}$:
$$\begin{aligned} \frac{x + 3}{y + 3} &= \frac{5}{6} \\ 6(x + 3) &= 5(y + 3) \\ 6x + 18 &= 5y + 15 \\ 6x – 5y &= -3 \quad \text{— (Equation 2)} \end{aligned}$$
From Equation 1:
$$x = \frac{9y – 4}{11}$$
Substitute this into Equation 2:
$$\begin{aligned} 6\left(\frac{9y – 4}{11}\right) – 5y &= -3 \\ \frac{54y – 24 – 55y}{11} &= -3 \\ -y – 24 &= -33 \\ -y &= -9 \implies y = 9 \end{aligned}$$
Substitute $y = 9$ into the expression for $x$:
$$x = \frac{9(9) – 4}{11} = \frac{81 – 4}{11} = \frac{77}{11} = 7$$
Final Statement:
The required fraction is $\frac{7}{9}$.
(vi) Jacob and Son Age Problem [BOARD EXAM FAVORITE / CBSE 2018, 2020]
Answer:
Let Jacob’s present age be $x\text{ years}$.
Let his son’s present age be $y\text{ years}$.
Condition 1: Five years hence (in future)
Jacob’s age $= x + 5$, Son’s age $= y + 5$.
$$\begin{aligned} x + 5 &= 3(y + 5) \\ x + 5 &= 3y + 15 \\ x – 3y &= 10 \quad \text{— (Equation 1)} \end{aligned}$$
Condition 2: Five years ago (in past)
Jacob’s age $= x – 5$, Son’s age $= y – 5$.
$$\begin{aligned} x – 5 &= 7(y – 5) \\ x – 5 &= 7y – 35 \\ x – 7y &= -30 \quad \text{— (Equation 2)} \end{aligned}$$
From Equation 1: $x = 3y + 10$.
Substitute this into Equation 2:
$$\begin{aligned} (3y + 10) – 7y &= -30 \\ -4y + 10 &= -30 \\ -4y &= -40 \implies y = 10 \end{aligned}$$
Substitute $y = 10$ into Equation 1:
$$x = 3(10) + 10 = 30 + 10 = 40$$
Final Statement:
Jacob’s present age is $40\text{ years}$ and his son’s present age is $10\text{ years}$.
Chapter-End Exercises – Exercise 3.3 (Page No. 56)
Question 1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) $x + y = 5$ and $2x – 3y = 4$
(ii) $3x + 4y = 10$ and $2x – 2y = 2$
(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$
(i) $x + y = 5$ and $2x – 3y = 4$
Answer:
By Elimination Method:
$$\begin{aligned} x + y &= 5 \quad \text{— (Equation 1)} \\ 2x – 3y &= 4 \quad \text{— (Equation 2)} \end{aligned}$$
Multiply Equation 1 by $3$:
$$3x + 3y = 15 \quad \text{— (Equation 3)}$$
Add Equation 2 and Equation 3:
$$\begin{aligned} (2x – 3y) + (3x + 3y) &= 4 + 15 \\ 5x &= 19 \implies x = \frac{19}{5} \end{aligned}$$
Substitute $x = \frac{19}{5}$ into Equation 1:
$$y = 5 – \frac{19}{5} = \frac{25 – 19}{5} = \frac{6}{5}$$
By Substitution Method:
From Equation 1: $y = 5 – x$.
Substitute into Equation 2:
$$\begin{aligned} 2x – 3(5 – x) &= 4 \\ 2x – 15 + 3x &= 4 \\ 5x &= 19 \implies x = \frac{19}{5} \\ y &= 5 – \frac{19}{5} = \frac{6}{5} \end{aligned}$$
Thus, $x = \frac{19}{5}, y = \frac{6}{5}$.
(ii) $3x + 4y = 10$ and $2x – 2y = 2$
Answer:
By Elimination Method:
$$\begin{aligned} 3x + 4y &= 10 \quad \text{— (Equation 1)} \\ 2x – 2y &= 2 \implies x – y = 1 \quad \text{— (Equation 2)} \end{aligned}$$
Multiply Equation 2 by $4$:
$$4x – 4y = 4 \quad \text{— (Equation 3)}$$
Add Equation 1 and Equation 3:
$$\begin{aligned} (3x + 4y) + (4x – 4y) &= 10 + 4 \\ 7x &= 14 \implies x = 2 \end{aligned}$$
Substitute $x = 2$ into Equation 2:
$$2 – y = 1 \implies y = 1$$
By Substitution Method:
From Equation 2: $x = y + 1$.
Substitute into Equation 1:
$$\begin{aligned} 3(y + 1) + 4y &= 10 \\ 3y + 3 + 4y &= 10 \\ 7y &= 7 \implies y = 1 \\ x &= 1 + 1 = 2 \end{aligned}$$
Thus, $x = 2, y = 1$.
(iii) $3x – 5y – 4 = 0$ and $9x = 2y + 7$
Answer:
Standard form:
$$\begin{aligned} 3x – 5y &= 4 \quad \text{— (Equation 1)} \\ 9x – 2y &= 7 \quad \text{— (Equation 2)} \end{aligned}$$
By Elimination Method:
Multiply Equation 1 by $3$:
$$9x – 15y = 12 \quad \text{— (Equation 3)}$$
Subtract Equation 3 from Equation 2:
$$\begin{aligned} (9x – 2y) – (9x – 15y) &= 7 – 12 \\ 13y &= -5 \implies y = -\frac{5}{13} \end{aligned}$$
Substitute $y = -\frac{5}{13}$ into Equation 1:
$$\begin{aligned} 3x – 5\left(-\frac{5}{13}\right) &= 4 \\ 3x + \frac{25}{13} &= 4 \\ 3x &= 4 – \frac{25}{13} = \frac{52 – 25}{13} = \frac{27}{13} \\ x &= \frac{9}{13} \end{aligned}$$
Thus, $x = \frac{9}{13}, y = -\frac{5}{13}$.
(iv) $\frac{x}{2} + \frac{2y}{3} = -1$ and $x – \frac{y}{3} = 3$
Answer:
Clear denominators by multiplying Equation 1 by $6$ and Equation 2 by $3$:
$$\begin{aligned} 3x + 4y &= -6 \quad \text{— (Equation 1)} \\ 3x – y &= 9 \quad \text{— (Equation 2)} \end{aligned}$$
By Elimination Method:
Subtract Equation 2 from Equation 1:
$$\begin{aligned} (3x + 4y) – (3x – y) &= -6 – 9 \\ 5y &= -15 \implies y = -3 \end{aligned}$$
Substitute $y = -3$ into Equation 2:
$$\begin{aligned} 3x – (-3) &= 9 \\ 3x + 3 &= 9 \\ 3x &= 6 \implies x = 2 \end{aligned}$$
Thus, $x = 2, y = -3$.
Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add $1$ to the numerator and subtract $1$ from the denominator, a fraction reduces to $1$. It becomes $\frac{1}{2}$ if we only add $1$ to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is $9$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw $₹2000$. She asked the cashier to give her $₹50$ and $₹100$ notes only. Meena got $25$ notes in all. Find how many notes of $₹50$ and $₹100$ she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $₹27$ for a book kept for seven days, while Susy paid $₹21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.
(i) Fraction Problem
Answer:
Let the fraction be $\frac{x}{y}$ ($y \neq 0$).
Condition 1: $\frac{x + 1}{y – 1} = 1 \implies x + 1 = y – 1 \implies x – y = -2$ — (Equation 1)
Condition 2: $\frac{x}{y + 1} = \frac{1}{2} \implies 2x = y + 1 \implies 2x – y = 1$ — (Equation 2)
Subtract Equation 1 from Equation 2:
$$\begin{aligned} (2x – y) – (x – y) &= 1 – (-2) \\ x &= 3 \end{aligned}$$
Substitute $x = 3$ into Equation 1:
$$3 – y = -2 \implies y = 5$$
Final Statement:
The required fraction is $\frac{3}{5}$.
(ii) Nuri and Sonu Age Problem [BOARD EXAM FAVORITE]
Answer:
Let Nuri’s present age be $x\text{ years}$ and Sonu’s present age be $y\text{ years}$.
Five years ago:
$$\begin{aligned} x – 5 &= 3(y – 5) \\ x – 5 &= 3y – 15 \\ x – 3y &= -10 \quad \text{— (Equation 1)} \end{aligned}$$
Ten years later:
$$\begin{aligned} x + 10 &= 2(y + 10) \\ x + 10 &= 2y + 20 \\ x – 2y &= 10 \quad \text{— (Equation 2)} \end{aligned}$$
Subtract Equation 1 from Equation 2:
$$\begin{aligned} (x – 2y) – (x – 3y) &= 10 – (-10) \\ y &= 20 \end{aligned}$$
Substitute $y = 20$ into Equation 2:
$$x – 2(20) = 10 \implies x – 40 = 10 \implies x = 50$$
Final Statement:
Nuri’s present age is $50\text{ years}$ and Sonu’s present age is $20\text{ years}$.
(iii) Two-Digit Number & Reversal [BOARD EXAM FAVORITE / CBSE 2019, 2020, 2022, 2023]
Answer:
Let the tens digit be $x$ and the units digit be $y$.
- Original Number $= 10x + y$
- Reversed Number $= 10y + x$
Condition 1 (Sum of digits is $9$):
$$x + y = 9 \quad \text{— (Equation 1)}$$
Condition 2 ($9 \times \text{Original} = 2 \times \text{Reversed}$):
$$\begin{aligned} 9(10x + y) &= 2(10y + x) \\ 90x + 9y &= 20y + 2x \\ 90x – 2x + 9y – 20y &= 0 \\ 88x – 11y &= 0 \end{aligned}$$
Divide by $11$:
$$8x – y = 0 \implies y = 8x \quad \text{— (Equation 2)}$$
Add Equation 1 and Equation 2:
$$\begin{aligned} (x + y) + (8x – y) &= 9 + 0 \\ 9x &= 9 \implies x = 1 \end{aligned}$$
Substitute $x = 1$ into Equation 2:
$$y = 8(1) = 8$$
Original Number $= 10x + y = 10(1) + 8 = 18$.
Final Statement:
The required number is $18$.
(iv) Bank Currency Notes Problem [BOARD EXAM FAVORITE / CBSE 2019, 2021]
Answer:
Let the number of $₹50$ notes be $x$.
Let the number of $₹100$ notes be $y$.
Condition 1 (Total of $25$ notes):
$$x + y = 25 \quad \text{— (Equation 1)}$$
Condition 2 (Total value of $₹2000$):
$$50x + 100y = 2000$$
Divide the entire equation by $50$:
$$x + 2y = 40 \quad \text{— (Equation 2)}$$
Subtract Equation 1 from Equation 2:
$$\begin{aligned} (x + 2y) – (x + y) &= 40 – 25 \\ y &= 15 \end{aligned}$$
Substitute $y = 15$ into Equation 1:
$$x + 15 = 25 \implies x = 10$$
Final Statement:
Meena received $10$ notes of $₹50$ and $15$ notes of $₹100$.
(v) Lending Library Charge Problem [BOARD EXAM FAVORITE / CBSE 2020, 2023]
Answer:
Let the fixed charge for the first $3$ days be $₹x$.
Let the charge for each subsequent day be $₹y$.
Saritha’s Case ($7$ days total = $3$ fixed days + $4$ extra days):
$$x + 4y = 27 \quad \text{— (Equation 1)}$$
Susy’s Case ($5$ days total = $3$ fixed days + $2$ extra days):
$$x + 2y = 21 \quad \text{— (Equation 2)}$$
Subtract Equation 2 from Equation 1:
$$\begin{aligned} (x + 4y) – (x + 2y) &= 27 – 21 \\ 2y &= 6 \implies y = 3 \end{aligned}$$
Substitute $y = 3$ into Equation 2:
$$\begin{aligned} x + 2(3) &= 21 \\ x + 6 &= 21 \implies x = 15 \end{aligned}$$
Final Statement:
The fixed charge is $₹15$ and the charge for each extra day is $₹3$.
15 High-Yield Board Exam FAQs
FAQ 1 (HOTS / Upstream-Downstream). A motorboat takes $6\text{ hours}$ to cover $36\text{ km}$ downstream and $36\text{ km}$ upstream together. If the speed of the stream is $3\text{ km/h}$, find the speed of the boat in still water. [CBSE 2020, 2023]
Answer:
Let the speed of the boat in still water be $x\text{ km/h}$ ($x > 3$).
- Speed upstream $= (x – 3)\text{ km/h}$
- Speed downstream $= (x + 3)\text{ km/h}$
Time taken $=\frac{\text{Distance}}{\text{Speed}}$:
$$\begin{aligned} \frac{36}{x – 3} + \frac{36}{x + 3} &= 6 \\ 36\left[\frac{(x + 3) + (x – 3)}{(x – 3)(x + 3)}\right] &= 6 \\ 36\left[\frac{2x}{x^2 – 9}\right] &= 6 \\ \frac{72x}{x^2 – 9} &= 6 \\ 12x &= x^2 – 9 \\ x^2 – 12x – 9 &= 0 \end{aligned}$$
Using the quadratic formula:
$$x = \frac{-(-12) \pm \sqrt{(-12)^2 – 4(1)(-9)}}{2(1)} = \frac{12 \pm \sqrt{144 + 36}}{2} = \frac{12 \pm \sqrt{180}}{2} = \frac{12 \pm 6\sqrt{5}}{2} = 6 \pm 3\sqrt{5}$$
Since speed must be positive and greater than $3$:
$$x = 6 + 3\sqrt{5} \approx 6 + 3(2.236) = 12.71\text{ km/h}$$
The speed of the boat in still water is $(6 + 3\sqrt{5})\text{ km/h}$.
FAQ 2 (HOTS / Work-Rate System). $2$ women and $5$ men can together finish an embroidery work in $4$ days, while $3$ women and $6$ men can finish it in $3$ days. Find the time taken by $1$ woman alone to finish the work, and also that taken by $1$ man alone. [CBSE 2019, 2022]
Answer:
Let $1$ woman alone finish the work in $x\text{ days}$. Daily work of $1$ woman $= \frac{1}{x}$.
Let $1$ man alone finish the work in $y\text{ days}$. Daily work of $1$ man $= \frac{1}{y}$.
From the given conditions:
$$\begin{aligned} \frac{2}{x} + \frac{5}{y} &= \frac{1}{4} \quad \text{— (Equation 1)} \\ \frac{3}{x} + \frac{6}{y} &= \frac{1}{3} \quad \text{— (Equation 2)} \end{aligned}$$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$:
$$\begin{aligned} 2u + 5v &= \frac{1}{4} \implies 8u + 20v = 1 \quad \text{— (Equation 3)} \\ 3u + 6v &= \frac{1}{3} \implies 9u + 18v = 1 \quad \text{— (Equation 4)} \end{aligned}$$
Multiply Equation 3 by $9$ and Equation 4 by $8$:
$$\begin{aligned} 72u + 180v &= 9 \\ 72u + 144v &= 8 \end{aligned}$$
Subtracting the equations:
$$36v = 1 \implies v = \frac{1}{36} \implies y = 36\text{ days}$$
Substitute $v = \frac{1}{36}$ into Equation 3:
$$\begin{aligned} 8u + 20\left(\frac{1}{36}\right) &= 1 \\ 8u + \frac{5}{9} &= 1 \\ 8u &= 1 – \frac{5}{9} = \frac{4}{9} \\ u &= \frac{4}{9 \times 8} = \frac{1}{18} \implies x = 18\text{ days} \end{aligned}$$
Therefore, $1$ woman alone takes $18\text{ days}$ and $1$ man alone takes $36\text{ days}$.
FAQ 3 (Parameter Condition for Infinite Solutions). For what values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?
$$2x + 3y = 7; \quad (a – b)x + (a + b)y = 3a + b – 2$$
[CBSE 2018, 2020, 2023]
Answer:
For infinitely many solutions:
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{2}{a – b} = \frac{3}{a + b} = \frac{7}{3a + b – 2}$$
Equating the first two ratios:
$$\begin{aligned} 2(a + b) &= 3(a – b) \\ 2a + 2b &= 3a – 3b \\ -a + 5b &= 0 \implies a = 5b \quad \text{— (Equation 1)} \end{aligned}$$
Equating the second and third ratios:
$$\begin{aligned} 3(3a + b – 2) &= 7(a + b) \\ 9a + 3b – 6 &= 7a + 7b \\ 2a – 4b &= 6 \implies a – 2b = 3 \quad \text{— (Equation 2)} \end{aligned}$$
Substitute Equation 1 into Equation 2:
$$5b – 2b = 3 \implies 3b = 3 \implies b = 1$$
Substitute $b = 1$ into Equation 1:
$$a = 5(1) = 5$$
Therefore, $a = 5$ and $b = 1$.
FAQ 4 (Parameter Condition for No Solution). For what value of $k$ will the following system of linear equations have no solution?
$$3x + y = 1; \quad (2k – 1)x + (k – 1)y = 2k + 1$$
[CBSE 2019, 2021]
Answer:
For no solution (parallel lines):
$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$
$$\frac{3}{2k – 1} = \frac{1}{k – 1} \neq \frac{1}{2k + 1}$$
From the equality condition:
$$\begin{aligned} 3(k – 1) &= 1(2k – 1) \\ 3k – 3 &= 2k – 1 \\ k &= 2 \end{aligned}$$
Verifying the inequality condition with $k = 2$:
$$\frac{1}{2 – 1} \neq \frac{1}{2(2) + 1} \implies \frac{1}{1} \neq \frac{1}{5} \quad \text{[True]}$$
Therefore, the value of $k$ is $2$.
FAQ 5 (Cyclic Quadrilateral HOTS). Find the four angles of a cyclic quadrilateral $ABCD$ in which:
$\angle A = 4y + 20$, $\angle B = 3y – 5$, $\angle C = -4x$, $\angle D = -7x + 5$. [CBSE 2020]
Answer:
In a cyclic quadrilateral, opposite angles sum to $180^\circ$:
Condition 1: $\angle A + \angle C = 180^\circ$
$$\begin{aligned} (4y + 20) + (-4x) &= 180 \\ -4x + 4y &= 160 \\ -x + y &= 40 \implies y = x + 40 \quad \text{— (Equation 1)} \end{aligned}$$
Condition 2: $\angle B + \angle D = 180^\circ$
$$\begin{aligned} (3y – 5) + (-7x + 5) &= 180 \\ -7x + 3y &= 180 \quad \text{— (Equation 2)} \end{aligned}$$
Substitute Equation 1 into Equation 2:
$$\begin{aligned} -7x + 3(x + 40) &= 180 \\ -7x + 3x + 120 &= 180 \\ -4x &= 60 \implies x = -15 \end{aligned}$$
Substitute $x = -15$ into Equation 1:
$$y = -15 + 40 = 25$$
Calculating the angles:
- $\angle A = 4(25) + 20 = 100 + 20 = \underline{120^\circ}$
- $\angle B = 3(25) – 5 = 75 – 5 = \underline{70^\circ}$
- $\angle C = -4(-15) = \underline{60^\circ}$
- $\angle D = -7(-15) + 5 = 105 + 5 = \underline{110^\circ}$
FAQ 6 (Assertion-Reasoning).
- Assertion (A): The pair of equations $x + 2y – 5 = 0$ and $-3x – 6y + 15 = 0$ has infinitely many solutions.
- Reason (R): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the pair of linear equations is consistent and dependent.
Answer:
Evaluating the ratios:
$$\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-5}{15} = -\frac{1}{3}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{3}$, the lines are coincident and have infinitely many solutions.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
FAQ 7 (Speed-Distance Relative Motion). Points $A$ and $B$ are $100\text{ km}$ apart on a highway. One car starts from $A$ and another from $B$ at the same time. If the cars travel in the same direction at different speeds, they meet in $5\text{ hours}$. If they travel towards each other, they meet in $1\text{ hour}$. What are the speeds of the two cars? [CBSE 2018, 2022]
Answer:
Let the speed of the car from $A$ be $u\text{ km/h}$ and the car from $B$ be $v\text{ km/h}$ ($u > v$).
- Case 1: Same direction ($5\text{ hours}$)Relative speed $= u – v$. Distance $= 100\text{ km}$.
$$5(u – v) = 100 \implies u – v = 20 \quad \text{— (Equation 1)}$$
- Case 2: Opposite directions ($1\text{ hour}$)Relative speed $= u + v$. Distance $= 100\text{ km}$.
$$1(u + v) = 100 \implies u + v = 100 \quad \text{— (Equation 2)}$$
Add Equation 1 and Equation 2:
$$2u = 120 \implies u = 60\text{ km/h}$$
Substitute $u = 60$ into Equation 2:
$$60 + v = 100 \implies v = 40\text{ km/h}$$
The speed of the car from $A$ is $60\text{ km/h}$ and the speed of the car from $B$ is $40\text{ km/h}$.
FAQ 8 (Rectangle Dimension HOTS). The area of a rectangle gets reduced by $9\text{ sq units}$, if its length is reduced by $5\text{ units}$ and breadth is increased by $3\text{ units}$. If we increase the length by $3\text{ units}$ and the breadth by $2\text{ units}$, the area increases by $67\text{ sq units}$. Find the dimensions of the rectangle. [CBSE 2019, 2023]
Answer:
Let length $= x$ and breadth $= y$. Initial Area $= xy$.
Condition 1:
$$\begin{aligned} (x – 5)(y + 3) &= xy – 9 \\ xy + 3x – 5y – 15 &= xy – 9 \\ 3x – 5y &= 6 \quad \text{— (Equation 1)} \end{aligned}$$
Condition 2:
$$\begin{aligned} (x + 3)(y + 2) &= xy + 67 \\ xy + 2x + 3y + 6 &= xy + 67 \\ 2x + 3y &= 61 \quad \text{— (Equation 2)} \end{aligned}$$
Multiply Equation 1 by $3$ and Equation 2 by $5$:
$$\begin{aligned} 9x – 15y &= 18 \\ 10x + 15y &= 305 \end{aligned}$$
Adding the two equations:
$$19x = 323 \implies x = 17\text{ units}$$
Substitute $x = 17$ into Equation 1:
$$\begin{aligned} 3(17) – 5y &= 6 \\ 51 – 5y &= 6 \\ 5y &= 45 \implies y = 9\text{ units} \end{aligned}$$
The length is $17\text{ units}$ and the breadth is $9\text{ units}$.
FAQ 9 (Determinant Condition for Trivial Zero Solution). If a homogeneous system $a_1x + b_1y = 0$ and $a_2x + b_2y = 0$ has only the trivial solution $(0, 0)$, what condition must the coefficients satisfy?
Answer:
A homogeneous linear system passes through the origin $(0, 0)$. It has only the trivial unique solution $(0, 0)$ if and only if the lines are non-parallel and non-coincident:
$$\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \iff a_1 b_2 – a_2 b_1 \neq 0$$
FAQ 10 (Symmetric Cross-Summation Technique). Solve the following system of linear equations:
$$152x – 378y = -74; \quad -378x + 152y = -604$$
[CBSE 2020, 2023]
Answer:
Let the equations be:
$$\begin{aligned} 152x – 378y &= -74 \quad \text{— (Equation 1)} \\ -378x + 152y &= -604 \quad \text{— (Equation 2)} \end{aligned}$$
Step 1: Add Equation 1 and Equation 2
$$\begin{aligned} -226x – 226y &= -678 \\ -226(x + y) &= -678 \\ x + y &= 3 \quad \text{— (Equation 3)} \end{aligned}$$
Step 2: Subtract Equation 2 from Equation 1
$$\begin{aligned} 530x – 530y &= 530 \\ 530(x – y) &= 530 \\ x – y &= 1 \quad \text{— (Equation 4)} \end{aligned}$$
Step 3: Solve Equations 3 and 4
Adding Equations 3 and 4:
$$2x = 4 \implies x = 2$$
Subtracting Equation 4 from Equation 3:
$$2y = 2 \implies y = 1$$
Therefore, $x = 2, y = 1$.
FAQ 11 (Age Ratio Problem). Six years hence, a man’s age will be three times the age of his son, and three years ago, he was nine times as old as his son. Find their present ages.
Answer:
Let the father’s present age be $x\text{ years}$ and the son’s present age be $y\text{ years}$.
- Condition 1 ($6\text{ years}$ hence): $x + 6 = 3(y + 6) \implies x – 3y = 12$ — (Equation 1)
- Condition 2 ($3\text{ years}$ ago): $x – 3 = 9(y – 3) \implies x – 9y = -24$ — (Equation 2)
Subtract Equation 2 from Equation 1:
$$6y = 36 \implies y = 6\text{ years}$$
Substitute $y = 6$ into Equation 1:
$$x – 3(6) = 12 \implies x – 18 = 12 \implies x = 30\text{ years}$$
The father’s present age is $30\text{ years}$ and the son’s present age is $6\text{ years}$.
FAQ 12 (Case Study: Manufacturing Production). A manufacturer of TV sets produced $600$ sets in the third year and $700$ sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production in the $1\text{st}$ year and the fixed annual increase.
Answer:
Let the initial production in the $1\text{st}$ year be $a$, and the constant annual increase be $d$.
- Production in $3\text{rd}$ year: $a + 2d = 600$ — (Equation 1)
- Production in $7\text{th}$ year: $a + 6d = 700$ — (Equation 2)
Subtract Equation 1 from Equation 2:
$$4d = 100 \implies d = 25\text{ sets}$$
Substitute $d = 25$ into Equation 1:
$$a + 2(25) = 600 \implies a + 50 = 600 \implies a = 550\text{ sets}$$
The production in the $1\text{st}$ year is $550\text{ sets}$ and the fixed annual increase is $25\text{ sets}$.
FAQ 13 (Intersection on $y$-axis). Find the condition under which the pair of equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ intersect on the $y$-axis.
Answer:
Any point on the $y$-axis has an $x$-coordinate of $0$ ($x = 0$).
Substituting $x = 0$ into both equations gives:
$$b_1y + c_1 = 0 \implies y = -\frac{c_1}{b_1}$$
$$b_2y + c_2 = 0 \implies y = -\frac{c_2}{b_2}$$
For both lines to cross the $y$-axis at the same point:
$$-\frac{c_1}{b_1} = -\frac{c_2}{b_2} \iff \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
FAQ 14 (HOTS / Linear Combinations). If $2x + y = 23$ and $4x – y = 19$, evaluate the expression $5y – 2x$.
Answer:
Add the two equations:
$$(2x + y) + (4x – y) = 23 + 19 \implies 6x = 42 \implies x = 7$$
Substitute $x = 7$ into the first equation:
$$2(7) + y = 23 \implies 14 + y = 23 \implies y = 9$$
Evaluating the expression:
$$5y – 2x = 5(9) – 2(7) = 45 – 14 = 31$$
The value of $5y – 2x$ is $31$.
FAQ 15 (No Solution Region on a Graph). How does a system with no solution appear geometrically when plotted on Cartesian coordinates?
Answer:
A system with no solution ($\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$) appears as two distinct parallel straight lines on the Cartesian plane. Because parallel lines have identical slopes ($m_1 = m_2$) but different $y$-intercepts ($c_1 \neq c_2$), they remain equidistant and never intersect at any point in $\mathbb{R}^2$.
To score full marks on Pair of Linear Equations questions in the CBSE Class 10 Board Examinations, present every step clearly and systematically. In word problems, always begin with explicit variable definitions that include appropriate units ($₹$, $\text{km/h}$, $\text{years}$, $\text{m}$). Show every intermediate algebraic operation when using substitution or elimination. In ratio comparison questions, write out the individual fractions $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$ separately before writing the compound equality or inequality statement. For graphical questions, provide a complete table with at least three coordinate points per line, clearly label the axes, and identify the point of intersection on the graph.