NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1#
Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1. This exercise covers forming and solving pairs of linear equations graphically, checking consistency of pairs of equations, and understanding the geometric interpretation of solutions.
Consistency of a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
- Unique solution (consistent): $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ — lines intersect
- Infinitely many solutions (consistent/dependent): $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ — lines coincide
- No solution (inconsistent): $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$ — lines are parallel
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.
Let number of girls = $x$ and number of boys = $y$.
10 students took part in all, so:
$$x + y = 10 \quad \cdots (1)$$The number of girls is 4 more than the number of boys, so:
$$x - y = 4 \quad \cdots (2)$$Table of values:
From (1): $x = 10 - y$
| $y$ | 0 | 5 | 10 |
|---|---|---|---|
| $x$ | 10 | 5 | 0 |
From (2): $x = 4 + y$
| $y$ | 0 | 3 | 6 |
|---|---|---|---|
| $x$ | 4 | 7 | 10 |
Plotting and finding the intersection:
From the graph, the lines intersect at $(7, 3)$.
Verification:
Adding equation (1) and equation (2):
$$\Rightarrow x = 7$$
Substituting $x = 7$ in equation (1):
$$x + y = 10$$$$7 + y = 10$$
$$\Rightarrow y = 3$$
∴ Girls = 7, Boys = 3
(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.
Let cost of one pencil = ₹$x$ and cost of one pen = ₹$y$.
5 pencils and 7 pens cost ₹50, so:
$$5x + 7y = 50 \quad \cdots (1)$$7 pencils and 5 pens cost ₹46, so:
$$7x + 5y = 46 \quad \cdots (2)$$Table of values:
From (1): $y = \dfrac{50 - 5x}{7}$
| $x$ | 3 | -4 | 10 |
|---|---|---|---|
| $y$ | 5 | 10 | 0 |
From (2): $y = \dfrac{46 - 7x}{5}$
| $x$ | 3 | 8 | -2 |
|---|---|---|---|
| $y$ | 5 | -2 | 12 |
Both lines intersect at $(3, 5)$.
∴ Cost of one pencil = ₹3, Cost of one pen = ₹5
2. On comparing the ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$, $\dfrac{c_1}{c_2}$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: #
(i) $5x - 4y + 8 = 0$, $7x + 6y - 9 = 0$
$a_1 = 5,\ b_1 = -4,\ c_1 = 8$
$a_2 = 7,\ b_2 = 6,\ c_2 = -9$
$$\dfrac{b_1}{b_2} = \dfrac{-4}{6} = \dfrac{-2}{3}$$
$$\dfrac{c_1}{c_2} = \dfrac{8}{-9} = \dfrac{-8}{9}$$$$\Rightarrow \dfrac{5}{7} \neq \dfrac{-2}{3}$$
∴ $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the lines intersect at a point.
(ii) $9x + 3y + 12 = 0$, $18x + 6y + 24 = 0$
$a_1 = 9,\ b_1 = 3,\ c_1 = 12$
$a_2 = 18,\ b_2 = 6,\ c_2 = 24$
$$\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{12}{24} = \dfrac{1}{2}$$$$\Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, so the lines are coincident.
(iii) $6x - 3y + 10 = 0$, $2x - y + 9 = 0$
$a_1 = 6,\ b_1 = -3,\ c_1 = 10$
$a_2 = 2,\ b_2 = -1,\ c_2 = 9$
$$\dfrac{b_1}{b_2} = \dfrac{-3}{-1} = 3$$
$$\dfrac{c_1}{c_2} = \dfrac{10}{9}$$$$\Rightarrow 3 = 3 \neq \dfrac{10}{9}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the lines are parallel.
3. On comparing the ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$, find out whether the following pair of linear equations are consistent, or inconsistent: #
(i) $3x + 2y = 5$; $2x - 3y = 7$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = 3,\ b_1 = 2,\ c_1 = -5$
$a_2 = 2,\ b_2 = -3,\ c_2 = -7$
$$\dfrac{b_1}{b_2} = \dfrac{2}{-3} = \dfrac{-2}{3}$$
$$\dfrac{c_1}{c_2} = \dfrac{-5}{-7} = \dfrac{5}{7}$$$$\Rightarrow \dfrac{3}{2} \neq \dfrac{-2}{3}$$
∴ $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the pair is Consistent (unique solution).
(ii) $2x - 3y = 8$; $4x - 6y = 9$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = 2,\ b_1 = -3,\ c_1 = -8$
$a_2 = 4,\ b_2 = -6,\ c_2 = -9$
$$\dfrac{b_1}{b_2} = \dfrac{-3}{-6} = \dfrac{1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{-8}{-9} = \dfrac{8}{9}$$$$\Rightarrow \dfrac{1}{2} = \dfrac{1}{2} \neq \dfrac{8}{9}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the pair is Inconsistent (no solution).
(iii) $\dfrac{3}{2}x + \dfrac{5}{3}y = 7$; $9x - 10y = 14$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = \dfrac{3}{2},\ b_1 = \dfrac{5}{3},\ c_1 = -7$
$a_2 = 9,\ b_2 = -10,\ c_2 = -14$
$$\dfrac{b_1}{b_2} = \dfrac{5/3}{-10} = \dfrac{-5}{30} = \dfrac{-1}{6}$$
$$\dfrac{c_1}{c_2} = \dfrac{-7}{-14} = \dfrac{1}{2}$$$$\Rightarrow \dfrac{1}{6} \neq \dfrac{-1}{6}$$
∴ $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the pair is Consistent (unique solution).
(iv) $5x - 3y = 11$; $-10x + 6y = -22$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = 5,\ b_1 = -3,\ c_1 = -11$
$a_2 = -10,\ b_2 = 6,\ c_2 = 22$
$$\dfrac{b_1}{b_2} = \dfrac{-3}{6} = \dfrac{-1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{-11}{22} = \dfrac{-1}{2}$$$$\Rightarrow \dfrac{-1}{2} = \dfrac{-1}{2} = \dfrac{-1}{2}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, so the pair is Consistent (infinitely many solutions).
(v) $\dfrac{4}{3}x + 2y = 8$; $2x + 3y = 12$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = \dfrac{4}{3},\ b_1 = 2,\ c_1 = -8$
$a_2 = 2,\ b_2 = 3,\ c_2 = -12$
$$\dfrac{b_1}{b_2} = \dfrac{2}{3}$$
$$\dfrac{c_1}{c_2} = \dfrac{-8}{-12} = \dfrac{2}{3}$$$$\Rightarrow \dfrac{2}{3} = \dfrac{2}{3} = \dfrac{2}{3}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, so the pair is Consistent (infinitely many solutions).
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: #
(i) $x + y = 5$, $2x + 2y = 10$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = 1,\ b_1 = 1,\ c_1 = -5$
$a_2 = 2,\ b_2 = 2,\ c_2 = -10$
$$\dfrac{b_1}{b_2} = \dfrac{1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{-5}{-10} = \dfrac{1}{2}$$$$\Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, so the pair is Consistent (infinitely many solutions).
The solution is given by $y = 5 - x$, where $x$ can take any value.
(ii) $x - y = 8$, $3x - 3y = 16$
Writing both equations in the form $ax + by + c = 0$:
$a_1 = 1,\ b_1 = -1,\ c_1 = -8$
$a_2 = 3,\ b_2 = -3,\ c_2 = -16$
$$\dfrac{b_1}{b_2} = \dfrac{-1}{-3} = \dfrac{1}{3}$$
$$\dfrac{c_1}{c_2} = \dfrac{-8}{-16} = \dfrac{1}{2}$$$$\Rightarrow \dfrac{1}{3} = \dfrac{1}{3} \neq \dfrac{1}{2}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the pair is Inconsistent (no solution).
(iii) $2x + y - 6 = 0$, $4x - 2y - 4 = 0$
$a_1 = 2,\ b_1 = 1,\ c_1 = -6$
$a_2 = 4,\ b_2 = -2,\ c_2 = -4$
$$\dfrac{b_1}{b_2} = \dfrac{1}{-2} = \dfrac{-1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{-6}{-4} = \dfrac{3}{2}$$$$\Rightarrow \dfrac{1}{2} \neq \dfrac{-1}{2}$$
∴ $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, so the pair is Consistent (unique solution).
Solving:
$$2x + y - 6 = 0 \quad \cdots (1)$$$$4x - 2y - 4 = 0 \quad \cdots (2)$$
From equation (1):
$$2x + y - 6 = 0$$$$\Rightarrow y = 6 - 2x$$
Substituting this value of $y$ in equation (2):
$$4x - 2y - 4 = 0$$$$4x - 2(6 - 2x) - 4 = 0$$
$$\Rightarrow 4x - 12 + 4x - 4 = 0$$
$$\Rightarrow 8x - 16 = 0$$
$$\Rightarrow 8x = 16$$
$$\Rightarrow x = 2$$
Substituting $x = 2$ in $y = 6 - 2x$:
$$y = 6 - 2x$$$$y = 6 - 2(2) = 2$$
∴ Solution: $x = 2,\ y = 2$
(iv) $2x - 2y - 2 = 0$, $4x - 4y - 5 = 0$
$a_1 = 2,\ b_1 = -2,\ c_1 = -2$
$a_2 = 4,\ b_2 = -4,\ c_2 = -5$
$$\dfrac{b_1}{b_2} = \dfrac{-2}{-4} = \dfrac{1}{2}$$
$$\dfrac{c_1}{c_2} = \dfrac{-2}{-5} = \dfrac{2}{5}$$$$\Rightarrow \dfrac{1}{2} = \dfrac{1}{2} \neq \dfrac{2}{5}$$
∴ $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, so the pair is Inconsistent (no solution).
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. #
Answer
Let length = $x$ m and width = $y$ m.
The length is 4 m more than the width, so:
$$x - y = 4 \quad \cdots (1)$$Half the perimeter is 36 m, so:
$$x + y = 36 \quad \cdots (2)$$Adding equation (1) and equation (2):
$$\begin{array}{rcl} x - y &=& 4 \\ x + y &=& 36 \\ \hline 2x &=& 40 \end{array}$$$$\Rightarrow x = 20$$
Substituting $x = 20$ in equation (2):
$$x + y = 36$$$$20 + y = 36$$
$$\Rightarrow y = 16$$
∴ Length = 20 m and Width = 16 m
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
We need $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.
One possible answer: $\mathbf{3x + 2y - 7 = 0}$
Check: $\dfrac{a_1}{a_2} = \dfrac{2}{3}$
$\dfrac{b_1}{b_2} = \dfrac{3}{2}$
Since $\dfrac{2}{3} \neq \dfrac{3}{2}$ ✓
(ii) parallel lines
We need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
One possible answer: $\mathbf{2x + 3y - 12 = 0}$
Check: $\dfrac{a_1}{a_2} = \dfrac{2}{2}$
$\dfrac{b_1}{b_2} = \dfrac{3}{3}$
$\dfrac{c_1}{c_2} = \dfrac{-8}{-12}$
Here both $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$ equal $1$, but $\dfrac{c_1}{c_2} = \dfrac{2}{3}$, which is not $1$ ✓
(iii) coincident lines
We need $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
One possible answer: $\mathbf{4x + 6y - 16 = 0}$
Check: $\dfrac{a_1}{a_2} = \dfrac{2}{4}$
$\dfrac{b_1}{b_2} = \dfrac{3}{6}$
$\dfrac{c_1}{c_2} = \dfrac{-8}{-16}$
Each ratio equals $\dfrac{1}{2}$ ✓
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. #
Answer
For $x - y + 1 = 0$, i.e., $y = x + 1$:
| $x$ | 0 | 1 | 2 |
|---|---|---|---|
| $y$ | 1 | 2 | 3 |
For $3x + 2y - 12 = 0$, i.e., $y = \dfrac{12 - 3x}{2}$:
| $x$ | 0 | 2 | 4 |
|---|---|---|---|
| $y$ | 6 | 3 | 0 |
Finding vertices:
Vertex 1: intersection of the two lines.
$$y = x + 1 \quad \cdots (1)$$$$3x + 2y = 12 \quad \cdots (2)$$
Substituting equation (1) in equation (2):
$$3x + 2(x + 1) = 12$$$$\Rightarrow 3x + 2x + 2 = 12$$
$$\Rightarrow 5x = 10$$
$$\Rightarrow x = 2$$
Substituting $x = 2$ in equation (1):
$$y = x + 1$$$$y = 2 + 1 = 3$$
Point: $(2, 3)$
Vertex 2: intersection of $x - y + 1 = 0$ with the x-axis ($y = 0$):
$$x + 1 = 0$$$$\Rightarrow x = -1$$
Point: $(-1, 0)$
Vertex 3: intersection of $3x + 2y - 12 = 0$ with the x-axis ($y = 0$):
$$3x = 12$$$$\Rightarrow x = 4$$
Point: $(4, 0)$
∴ Vertices of the triangle: $(-1, 0)$, $(4, 0)$ and $(2, 3)$