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 = $y$.
Equations:
$$x + y = 10 \quad \cdots (1)$$$$x - y = 4 \quad \cdots (2)$$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:
Solving (1) and (2): Adding both equations:
$$2x = 14 \implies x = 7, \quad y = 3$$∴ 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 = $₹y$.
Equations:
$$5x + 7y = 50 \quad \cdots (1)$$$$7x + 5y = 46 \quad \cdots (2)$$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) $5x - 4y + 8 = 0$, $7x + 6y - 9 = 0$
$a_1 = 5,\ b_1 = -4,\ c_1 = 8$ $a_2 = 7,\ b_2 = 6,\ c_2 = -9$
$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$$Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the lines intersect at a point.
(ii) $9x + 3y + 12 = 0$, $18x + 6y + 24 = 0$
$a_1 = 9,\ b_1 = 3,\ c_1 = 12$ $a_2 = 18,\ b_2 = 6,\ c_2 = 24$
$$\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$$Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines are coincident.
(iii) $6x - 3y + 10 = 0$, $2x - y + 9 = 0$
$a_1 = 6,\ b_1 = -3,\ c_1 = 10$ $a_2 = 2,\ b_2 = -1,\ c_2 = 9$
$$\frac{a_1}{a_2} = \frac{6}{2} = 3, \quad \frac{b_1}{b_2} = \frac{-3}{-1} = 3, \quad \frac{c_1}{c_2} = \frac{10}{9}$$Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, 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) $3x + 2y = 5$; $2x - 3y = 7$
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is Consistent (unique solution).
(ii) $2x - 3y = 8$; $4x - 6y = 9$
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the pair is Inconsistent (no solution).
(iii) $\dfrac{3}{2}x + \dfrac{5}{3}y = 7$; $9x - 10y = 14$
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is Consistent (unique solution).
(iv) $5x - 3y = 11$; $-10x + 6y = -22$
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the pair is Consistent (infinitely many solutions).
(v) $\dfrac{4}{3}x + 2y = 8$; $2x + 3y = 12$
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, 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) $x + y = 5$, $2x + 2y = 10$
Consistent (infinitely many solutions). The solution is given by $y = 5 - x$, where $x$ can take any value.
(ii) $x - y = 8$, $3x - 3y = 16$
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the pair is Inconsistent (no solution).
(iii) $2x + y - 6 = 0$, $4x - 2y - 4 = 0$
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the pair is Consistent (unique solution).
Solving: From equation 1: $y = 6 - 2x$. Substituting in equation 2:
$$4x - 2(6 - 2x) - 4 = 0 \implies 4x - 12 + 4x - 4 = 0 \implies 8x = 16 \implies x = 2,\ y = 2$$∴ Solution: $x = 2,\ y = 2$
(iv) $2x - 2y - 2 = 0$, $4x - 4y - 5 = 0$
Since $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, 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.
Condition 1: Length is 4 m more than width:
$$x - y = 4 \quad \cdots (1)$$Condition 2: Half the perimeter = 36 m:
$$x + y = 36 \quad \cdots (2)$$Solving graphically / algebraically:
Adding (1) and (2):
$$2x = 40 \implies x = 20$$Substituting in (2):
$$y = 36 - 20 = 16$$∴ Length = 20 m and Breadth = 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: $\dfrac{2}{3} \neq \dfrac{3}{2}$ ✓
(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: $\dfrac{2}{2} = \dfrac{3}{3} = 1$, but $\dfrac{8}{12} = \dfrac{2}{3} \neq 1$ ✓
(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: $\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{8}{16} = \dfrac{1}{2}$ ✓
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:
Intersection of the two lines: From $y = x + 1$ and $3x + 2y = 12$:
$$3x + 2(x + 1) = 12 \implies 5x = 10 \implies x = 2,\ y = 3$$Point: $(2, 3)$
Intersection of $x - y + 1 = 0$ with x-axis ($y = 0$):
$$x + 1 = 0 \implies x = -1$$Point: $(-1, 0)$
Intersection of $3x + 2y - 12 = 0$ with x-axis ($y = 0$):
$$3x = 12 \implies x = 4$$Point: $(4, 0)$
∴ Vertices of the triangle: $(-1, 0)$, $(4, 0)$ and $(2, 3)$