SEBA Class 10 Maths Chapter 3 Exercise 3.2 Solutions – Linear Equations in Two Variables

Q1. Solve the following pair of linear equations by the substitution method.

(i) \(x + y = 14\) and \(x - y = 4\)

Answer:
Given:
\[ x + y = 14 \quad ...(1) \] \[ x - y = 4 \quad ...(2) \]
From (2),
\[ x = y + 4 \quad ...(3) \]
Substituting the value of x from (3) in (1), we get:
\[ (y + 4) + y = 14 \] \[ 2y + 4 = 14 \] \[ 2y = 10 \Rightarrow y = 5 \]
Now substituting the value of y in (3), we get:
\[ x = 5 + 4 = 9 \]

Therefore, \(x = 9,\; y = 5\)

(ii) \(5 - x = 3\) and \( \frac{5}{3} + \frac{x}{2} = 6\)

Answer:
Given:
\[ 5 - x = 3 \quad ...(1) \] \[ \frac{5}{3} + \frac{x}{2} = 6 \quad ...(2) \]
From (1),
\[ x = 2 \quad ...(3) \]
Substituting the value of x in (2), we get:
\[ \frac{5}{3} + \frac{2}{2} = 6 \] \[ \frac{5}{3} + 1 = \frac{8}{3} \ne 6 \]
Since both equations are not satisfied together,
Therefore, the system has no solution.

(iii) \(3x - y = 3\) and \(9x - 3y = 9\)

Answer:
Given:
\[ 3x - y = 3 \quad ...(1) \] \[ 9x - 3y = 9 \quad ...(2) \]
From (1),
\[ y = 3x - 3 \quad ...(3) \]
Substituting the value of y from (3) in (2), we get:
\[ 9x - 3(3x - 3) = 9 \] \[ 9x - 9x + 9 = 9 \] \[ 9 = 9 \]
This is true for all values of x.

Therefore, the system has infinitely many solutions.

(iv) \(0.2x + 0.3y = 1.3\) and \(0.4x + 0.5y = 2.3\)

Answer:
Given:
\[ 0.2x + 0.3y = 1.3 \quad ...(1) \] \[ 0.4x + 0.5y = 2.3 \quad ...(2) \]
Multiplying both equations by 10, we get:
\[ 2x + 3y = 13 \quad ...(1) \] \[ 4x + 5y = 23 \quad ...(2) \]
From (1),
\[ 2x = 13 - 3y \Rightarrow x = \frac{13 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ 4\left(\frac{13 - 3y}{2}\right) + 5y = 23 \] \[ 2(13 - 3y) + 5y = 23 \] \[ 26 - 6y + 5y = 23 \] \[ 26 - y = 23 \Rightarrow y = 3 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{13 - 9}{2} = 2 \]

Therefore, \(x = 2,\; y = 3\)

(v) \(\sqrt{2}x + \sqrt{3}y = 0\) and \(\sqrt{3}x - \sqrt{8}y = 0\)

Answer:
Given:
\[ \sqrt{2}x + \sqrt{3}y = 0 \quad ...(1) \] \[ \sqrt{3}x - \sqrt{8}y = 0 \quad ...(2) \]
From (1),
\[ \sqrt{2}x = -\sqrt{3}y \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ \sqrt{3}\left(-\frac{\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0 \] \[ -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0 \]
Taking LCM and simplifying, we get:
\[ y = 0 \]
Now substituting the value of y in (3), we get:
\[ x = 0 \]

Therefore, \(x = 0,\; y = 0\)

(vi) \( \frac{3x}{2} - \frac{5y}{3} = -2\) and \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)

Answer:
Given:
\[ \frac{3x}{2} - \frac{5y}{3} = -2 \quad ...(1) \] \[ \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad ...(2) \]
Multiplying both equations by 6, we get:
\[ 9x - 10y = -12 \quad ...(1) \] \[ 2x + 3y = 13 \quad ...(2) \]
From (2),
\[ 2x = 13 - 3y \Rightarrow x = \frac{13 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (1), we get:
\[ 9\left(\frac{13 - 3y}{2}\right) - 10y = -12 \] \[ \frac{117 - 27y}{2} - 10y = -12 \] \[ 117 - 27y - 20y = -24 \] \[ 117 - 47y = -24 \Rightarrow y = 3 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{13 - 9}{2} = 2 \]
Therefore, \(x = 2,\; y = 3\)

Q2. Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of ‘m’ for which \(y = mx + 3\).

Answer:
Given:
\[ 2x + 3y = 11 \quad ...(1) \] \[ 2x - 4y = -24 \quad ...(2) \]
From (1),
\[ 2x = 11 - 3y \Rightarrow x = \frac{11 - 3y}{2} \quad ...(3) \]
Substituting the value of x from (3) in (2), we get:
\[ 2\left(\frac{11 - 3y}{2}\right) - 4y = -24 \] \[ 11 - 3y - 4y = -24 \] \[ 11 - 7y = -24 \Rightarrow -7y = -35 \Rightarrow y = 5 \]
Now substituting the value of y in (3), we get:
\[ x = \frac{11 - 15}{2} = -2 \]

Now given: \[ y = mx + 3 \] Substitute values:
\[ 5 = m(-2) + 3 \] \[ 5 = -2m + 3 \Rightarrow -2m = 2 \Rightarrow m = -1 \]

Therefore, \(x = -2,\; y = 5\) and \(m = -1\)

Q3. 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.

Answer:
Let the two numbers be x and y.

Given:
\[ x - y = 26 \quad ...(1) \] One number is three times the other:
\[ x = 3y \quad ...(2) \]
Substituting (2) in (1), we get:
\[ 3y - y = 26 \] \[ 2y = 26 \Rightarrow y = 13 \]
Now from (2):
\[ x = 3 \times 13 = 39 \]

Therefore, the numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18°. Find them.

Answer:
Let thee smaller angle be x and larger angle be y.

Since angles are supplementary:
\[ x + y = 180 \quad ...(1) \] Given:
\[ y = x + 18 \quad ...(2) \]
Substituting (2) in (1), we get:
\[ x + (x + 18) = 180 \] \[ 2x + 18 = 180 \] \[ 2x = 162 \Rightarrow x = 81 \]
Now from (2):
\[ y = 81 + 18 = 99 \]

Therefore, the angles are 81° and 99°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, he buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and ball.

Answer:
Let cost of one bat be x and cost of one ball be y.

\[ 7x + 6y = 3800 \quad ...(1) \] \[ 3x + 5y = 1750 \quad ...(2) \]
From (2),
\[ 3x = 1750 - 5y \Rightarrow x = \frac{1750 - 5y}{3} \quad ...(3) \]
Substituting (3) in (1), we get:
\[ 7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800 \] \[ \frac{12250 - 35y}{3} + 6y = 3800 \] \[ 12250 - 35y + 18y = 11400 \] \[ 12250 - 17y = 11400 \Rightarrow 17y = 850 \Rightarrow y = 50 \]
Now from (3):
\[ x = \frac{1750 - 250}{3} = 500 \]

Therefore, cost of one bat = Rs 500 and one ball = Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for distance covered. For a journey of 10 km, the charge paid is Rs 105 and for 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km?

Answer:
Let fixed charge of the taxi be x and charge per km be y.

\[ x + 10y = 105 \quad ...(1) \] \[ x + 15y = 155 \quad ...(2) \]
From (1),
\[ x = 105 - 10y \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 105 - 10y + 15y = 155 \] \[ 105 + 5y = 155 \] \[ 5y = 50 \Rightarrow y = 10 \]
Now from (3):
\[ x = 105 - 100 = 5 \]

Therefore, fixed charge = Rs 5 and charge per km = Rs 10.

(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.

Answer:
Let the numerator be x and denominator be y.
Given:
\[ \frac{x+2}{y+2} = \frac{9}{11} \quad ...(1) \] \[ \frac{x+3}{y+3} = \frac{5}{6} \quad ...(2) \]
From (1):
\[ 11(x+2) = 9(y+2) \] \[ 11x + 22 = 9y + 18 \] \[ 11x - 9y = -4 \quad ...(3) \]
From (2):
\[ 6(x+3) = 5(y+3) \] \[ 6x + 18 = 5y + 15 \] \[ 6x - 5y = -3 \quad ...(4) \]
From (4):
\[ 6x = 5y - 3 \Rightarrow x = \frac{5y - 3}{6} \quad ...(5) \]
Substituting (5) in (3), we get:
\[ 11\left(\frac{5y - 3}{6}\right) - 9y = -4 \] \[ \frac{55y - 33}{6} - 9y = -4 \] \[ 55y - 33 - 54y = -24 \] \[ y - 33 = -24 \Rightarrow y = 9 \]
Now from (5):
\[ x = \frac{45 - 3}{6} = 7 \]
Therefore, the fraction is \( \frac{7}{9} \).

(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. Find their present ages.

Answer:
Let the present age of Jacob be x years and his son be y years.
Given:
\[ x + 5 = 3(y + 5) \quad ...(1) \] \[ x - 5 = 7(y - 5) \quad ...(2) \]
From (1):
\[ x + 5 = 3y + 15 \] \[ x = 3y + 10 \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 3y + 10 - 5 = 7(y - 5) \] \[ 3y + 5 = 7y - 35 \] \[ 40 = 4y \Rightarrow y = 10 \]
Now from (3):
\[ x = 3(10) + 10 = 40 \]
Therefore, Jacob’s present age is 40 years and his son’s age is 10 years.

Q4. If \(x + 3\) is a factor of \(x^3 + ax^2 - bx + 6\) and \(a + b = 7\), find the values of a and b.

Answer:
Since \(x + 3\) is a factor of the given polynomial,
by Factor Theorem, we have:
\[ (-3)^3 + a(-3)^2 - b(-3) + 6 = 0 \] \[ -27 + 9a + 3b + 6 = 0 \] \[ 9a + 3b - 21 = 0 \Rightarrow 3a + b = 7 \quad ...(1) \]
Given:
\[ a + b = 7 \quad ...(2) \]
From (2),
\[ b = 7 - a \quad ...(3) \]
Substituting the value of b from (3) in (1), we get:
\[ 3a + (7 - a) = 7 \] \[ 2a + 7 = 7 \Rightarrow a = 0 \]
Now substituting the value of a in (3), we get:
\[ b = 7 - 0 = 7 \]
Therefore, \(a = 0\) and \(b = 7\).

Q5. Consider the following pair of linear equations of two variables, then choose the correct option from the given alternatives.

\[ 2x = y \quad \text{and} \quad -5x + 2y - 3 = 0 \]

(i) The graphs of the equations intersect at a point.
(ii) The graphs of the equations are parallel to each other.
(iii) The graphs of the equations coincide.
(iv) The equations have a unique solution.

(a) Both (i) and (ii) are true
(b) Both (i) and (iii) are true
(c) Both (ii) and (iii) are true
(d) Both (i) and (iv) are true

Answer: (d) Both (i) and (iv) are true

Q6. Match the items of column I with column II and choose the correct option.

Column I
A. \(x - y = 0\)
B. \(2x - 3y = 5\) and \(x - y = 1\)
C. \(x + 2y = 6\) and \(4x + 8y = 24\)
D. \(2x + 3y = 6\) and \(4x + 6y = 10\)

Column II
(i) unique solution
(ii) linear equation in two variables
(iii) no solution
(iv) infinitely many solutions and lines coincide

(a) A → (ii), B → (i), C → (iv), D → (iii)
(b) A → (ii), B → (iv), C → (i), D → (iii)
(c) A → (i), B → (i), C → (ii), D → (iii)
(d) A → (iv), B → (ii), C → (i), D → (iii)

Answer: (a)

Q7. The value of x for which the pair (x, 4) satisfies the equation \(3x + y = 19\) is:

(a) 6
(b) 5
(c) 3
(d) 4

Answer: (b) 5

Q8. Two equations are given as \(2x + 4y = 10\) and \(kx + 8y = 20\). For which value of k will the system have infinitely many solutions?

(a) 4
(b) 3
(c) 2
(d) 1

Answer: (a) 4