📘 Class 10 Maths Chapter 3 Exercise 3.2 Solutions | SEBA 2026–2027
📑 Table of Contents
📌 What is Substitution Method?
The substitution method is used to solve a pair of linear equations by expressing one variable in terms of another and substituting it into the second equation.
📚 Chapter 3 Navigation
📊 Steps of Substitution Method
- Make one variable subject
- Substitute into second equation
- Solve the equation
- Find second variable
📝 Exercise 3.2 solutions
1. Solve the following pair of linear equations by the substitution method:
(i)
$x + y = 14$
$x - y = 4$
$x - y = 4$
(ii)
$5 - t = 3$
$\frac{s}{3} + \frac{t}{2} = 6$
$\frac{s}{3} + \frac{t}{2} = 6$
(iii)
$3x - y = 3$
$9x - 3y = 9$
$9x - 3y = 9$
(iv)
$0.2x + 0.3y = 1.3$
$0.4x + 0.5y = 2.3$
$0.4x + 0.5y = 2.3$
(v)
$\sqrt{2}x + \sqrt{3}y = 0$
$\sqrt{3}x - \sqrt{8}y = 0$
$\sqrt{3}x - \sqrt{8}y = 0$
(vi)
$\frac{3x}{2} - \frac{5y}{3} = -2$
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Solution (i)
Given:
\( x + y = 14 \quad \dots (1) \)
\( x - y = 4 \quad \dots (2) \)
From equation (1)
\( x - y = 4 \quad \dots (2) \)
\( \Rightarrow y = 14 - x \quad \dots (3) \)
Substitute the value of \( y \) in equation (2)
\( \Rightarrow x - (14 - x) = 4 \)
\( \Rightarrow x - 14 + x = 4 \)
\( \Rightarrow 2x - 14 = 4 \)
\( \Rightarrow 2x = 4 + 14 \)
\( \Rightarrow 2x = 18 \)
\( \Rightarrow x = \frac{18}{2} \)
\( \therefore x = 9 \)
Substitute the value of \( x \) in equation (3)
\( \Rightarrow x - 14 + x = 4 \)
\( \Rightarrow 2x - 14 = 4 \)
\( \Rightarrow 2x = 4 + 14 \)
\( \Rightarrow 2x = 18 \)
\( \Rightarrow x = \frac{18}{2} \)
\( \therefore x = 9 \)
\( \Rightarrow y = 14 - 9 \)
\( \therefore y = 5 \)
\( \therefore y = 5 \)
Final Answer: \( x = 9 \) and \( y = 5 \).
Solution (ii)
Given:
\( s - t = 3 \quad \dots (1) \)
\( \frac{s}{3} + \frac{t}{2} = 6 \quad \dots (2) \)
Step 1: From equation (1)
\( \frac{s}{3} + \frac{t}{2} = 6 \quad \dots (2) \)
\( \Rightarrow s = 3 + t \quad \dots (3) \)
Step 2: Substitute the value of \( s \) in equation (2)
\( \Rightarrow \frac{3 + t}{3} + \frac{t}{2} = 6 \)
\( \Rightarrow \frac{2(3 + t) + 3t}{6} = 6 \)
\( \Rightarrow \frac{6 + 2t + 3t}{6} = 6 \)
\( \Rightarrow 6 + 5t = 36 \)
\( \Rightarrow 5t = 36 - 6 \)
\( \Rightarrow 5t = 30 \)
\( \Rightarrow t = \frac{30}{5} \)
\( \therefore t = 6 \)
Step 3: Substitute the value of \( t \) in equation (3)
\( \Rightarrow \frac{2(3 + t) + 3t}{6} = 6 \)
\( \Rightarrow \frac{6 + 2t + 3t}{6} = 6 \)
\( \Rightarrow 6 + 5t = 36 \)
\( \Rightarrow 5t = 36 - 6 \)
\( \Rightarrow 5t = 30 \)
\( \Rightarrow t = \frac{30}{5} \)
\( \therefore t = 6 \)
\( \Rightarrow s = 3 + 6 \)
\( \therefore s = 9 \)
\( \therefore s = 9 \)
Final Answer: \( s = 9 \) and \( t = 6 \).
Solution (iii)
Given:
\( 3x - y = 3 \quad \dots (1) \)
\( 9x - 3y = 9 \quad \dots (2) \)
From equation (1)
\( 9x - 3y = 9 \quad \dots (2) \)
\( \Rightarrow 3x = 3 + y \)
\( \Rightarrow x = \frac{3 + y}{3} \quad \dots (3) \)
Substitute the value of \( x \) in equation (2)
\( \Rightarrow x = \frac{3 + y}{3} \quad \dots (3) \)
\( \Rightarrow 9 \left( \frac{3 + y}{3} \right) - 3y = 9 \)
\( \Rightarrow 3(3 + y) - 3y = 9 \)
\( \Rightarrow 9 + 3y - 3y = 9 \)
\( \Rightarrow 9 = 9 \)
\( \Rightarrow 3(3 + y) - 3y = 9 \)
\( \Rightarrow 9 + 3y - 3y = 9 \)
\( \Rightarrow 9 = 9 \)
Since no specific values of the variables are obtained and the statement is always true (\(9 = 9\)), the pair of equations has infinitely many solutions.
Solution (iv)
Given:
\( 0.2x + 0.3y = 1.3 \quad \dots (1) \)
\( 0.4x + 0.5y = 2.3 \quad \dots (2) \)
From equation (1)
\( 0.4x + 0.5y = 2.3 \quad \dots (2) \)
\( \Rightarrow 0.2x = 1.3 - 0.3y \)
\( \Rightarrow x = \frac{1.3 - 0.3y}{0.2} \quad \dots (3) \)
Substitute the value of \( x \) in equation (2)
\( \Rightarrow x = \frac{1.3 - 0.3y}{0.2} \quad \dots (3) \)
\( \Rightarrow 0.4 \left( \frac{1.3 - 0.3y}{0.2} \right) + 0.5y = 2.3 \)
\( \Rightarrow 2(1.3 - 0.3y) + 0.5y = 2.3 \)
\( \Rightarrow 2.6 - 0.6y + 0.5y = 2.3 \)
\( \Rightarrow 2.6 - 0.1y = 2.3 \)
\( \Rightarrow -0.1y = 2.3 - 2.6 \)
\( \Rightarrow -0.1y = -0.3 \)
\( \Rightarrow y = \frac{0.3}{0.1} \)
\( \therefore y = 3 \)
Substitute the value of \( y \) in equation (3)
\( \Rightarrow 2(1.3 - 0.3y) + 0.5y = 2.3 \)
\( \Rightarrow 2.6 - 0.6y + 0.5y = 2.3 \)
\( \Rightarrow 2.6 - 0.1y = 2.3 \)
\( \Rightarrow -0.1y = 2.3 - 2.6 \)
\( \Rightarrow -0.1y = -0.3 \)
\( \Rightarrow y = \frac{0.3}{0.1} \)
\( \therefore y = 3 \)
\( \Rightarrow x = \frac{1.3 - 0.3(3)}{0.2} \)
\( \Rightarrow x = \frac{1.3 - 0.9}{0.2} \)
\( \Rightarrow x = \frac{0.4}{0.2} \)
\( \therefore x = 2 \)
\( \Rightarrow x = \frac{1.3 - 0.9}{0.2} \)
\( \Rightarrow x = \frac{0.4}{0.2} \)
\( \therefore x = 2 \)
Final Answer: \( x = 2 \) and \( y = 3 \).
Solution (v)
Given:
\( \sqrt{2}x + \sqrt{3}y = 0 \quad \dots (1) \)
\( \sqrt{3}x - \sqrt{8}y = 0 \quad \dots (2) \)
From equation (1)
\( \sqrt{3}x - \sqrt{8}y = 0 \quad \dots (2) \)
\( \Rightarrow \sqrt{2}x = -\sqrt{3}y \)
\( \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad \dots (3) \)
Substitute the value of \( x \) in equation (2)
\( \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}y \quad \dots (3) \)
\( \Rightarrow \sqrt{3} \left( -\frac{\sqrt{3}}{\sqrt{2}}y \right) - \sqrt{8}y = 0 \)
\( \Rightarrow -\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0 \)
\( \Rightarrow y \left( -\frac{3}{\sqrt{2}} - \sqrt{8} \right) = 0 \)
\( \therefore y = 0 \)
Substitute the value of \( y \) in equation (3)
\( \Rightarrow -\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0 \)
\( \Rightarrow y \left( -\frac{3}{\sqrt{2}} - \sqrt{8} \right) = 0 \)
\( \therefore y = 0 \)
\( \Rightarrow x = -\frac{\sqrt{3}}{\sqrt{2}}(0) \)
\( \therefore x = 0 \)
\( \therefore x = 0 \)
Final Answer: \( x = 0 \) and \( y = 0 \).
Solution (vi)
Given:
\( \frac{3x}{2} - \frac{5y}{3} = -2 \quad \dots (1) \)
\( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad \dots (2) \)
From equation (1)
\( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \quad \dots (2) \)
\( \Rightarrow \frac{3x}{2} = -2 + \frac{5y}{3} \)
\( \Rightarrow \frac{3x}{2} = \frac{-6 + 5y}{3} \)
\( \Rightarrow x = \frac{2(-6 + 5y)}{9} \)
\( \Rightarrow x = \frac{-12 + 10y}{9} \quad \dots (3) \)
Substitute the value of \( x \) in equation (2)
\( \Rightarrow \frac{3x}{2} = \frac{-6 + 5y}{3} \)
\( \Rightarrow x = \frac{2(-6 + 5y)}{9} \)
\( \Rightarrow x = \frac{-12 + 10y}{9} \quad \dots (3) \)
\( \Rightarrow \frac{1}{3} \left( \frac{-12 + 10y}{9} \right) + \frac{y}{2} = \frac{13}{6} \)
\( \Rightarrow \frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6} \)
\( \Rightarrow \frac{2(-12 + 10y) + 27y}{54} = \frac{13}{6} \)
\( \Rightarrow \frac{-24 + 20y + 27y}{54} = \frac{13}{6} \)
\( \Rightarrow \frac{-24 + 47y}{54} = \frac{13}{6} \)
\( \Rightarrow -24 + 47y = \frac{13 \times 54}{6} \)
\( \Rightarrow -24 + 47y = 13 \times 9 \)
\( \Rightarrow 47y = 117 + 24 \)
\( \Rightarrow 47y = 141 \)
\( \Rightarrow y = \frac{141}{47} \)
\( \therefore y = 3 \)
Substitute the value of \( y \) in equation (3)
\( \Rightarrow \frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6} \)
\( \Rightarrow \frac{2(-12 + 10y) + 27y}{54} = \frac{13}{6} \)
\( \Rightarrow \frac{-24 + 20y + 27y}{54} = \frac{13}{6} \)
\( \Rightarrow \frac{-24 + 47y}{54} = \frac{13}{6} \)
\( \Rightarrow -24 + 47y = \frac{13 \times 54}{6} \)
\( \Rightarrow -24 + 47y = 13 \times 9 \)
\( \Rightarrow 47y = 117 + 24 \)
\( \Rightarrow 47y = 141 \)
\( \Rightarrow y = \frac{141}{47} \)
\( \therefore y = 3 \)
\( \Rightarrow x = \frac{-12 + 10(3)}{9} \)
\( \Rightarrow x = \frac{-12 + 30}{9} \)
\( \Rightarrow x = \frac{18}{9} \)
\( \therefore x = 2 \)
\( \Rightarrow x = \frac{-12 + 30}{9} \)
\( \Rightarrow x = \frac{18}{9} \)
\( \therefore x = 2 \)
Final Answer: \( x = 2 \) and \( y = 3 \).
2. Solve \(2x + 3y = 11\) and \(2x - 4y = -24\) and hence find the value of \(m\) for which \(y = mx + 3\).
Solution
The given pair of equations is:
\( 2x + 3y = 11 \quad \dots (I) \)
\( 2x - 4y = -24 \quad \dots (II) \)
Find the value of \( x \) from equation (I)
\( 2x - 4y = -24 \quad \dots (II) \)
\( \Rightarrow 2x = 11 - 3y \)
\( \Rightarrow x = \frac{11 - 3y}{2} \quad \dots (III) \)
Substitute the value of \( x \) in equation (II) to find \( y \)
\( \Rightarrow x = \frac{11 - 3y}{2} \quad \dots (III) \)
\( \Rightarrow 2 \left( \frac{11 - 3y}{2} \right) - 4y = -24 \)
\( \Rightarrow 11 - 3y - 4y = -24 \)
\( \Rightarrow 11 - 7y = -24 \)
\( \Rightarrow -7y = -24 - 11 \)
\( \Rightarrow -7y = -35 \)
\( \Rightarrow y = \frac{-35}{-7} \)
\( \therefore y = 5 \)
Substitute the value of \( y \) in equation (III) to find \( x \)
\( \Rightarrow 11 - 3y - 4y = -24 \)
\( \Rightarrow 11 - 7y = -24 \)
\( \Rightarrow -7y = -24 - 11 \)
\( \Rightarrow -7y = -35 \)
\( \Rightarrow y = \frac{-35}{-7} \)
\( \therefore y = 5 \)
\( \Rightarrow x = \frac{11 - 3(5)}{2} \)
\( \Rightarrow x = \frac{11 - 15}{2} \)
\( \Rightarrow x = \frac{-4}{2} \)
\( \therefore x = -2 \)
Find the value of \( m \)
\( \Rightarrow x = \frac{11 - 15}{2} \)
\( \Rightarrow x = \frac{-4}{2} \)
\( \therefore x = -2 \)
Now, substitute \( x = -2 \) and \( y = 5 \) into the equation \( y = mx + 3 \):
\( \Rightarrow 5 = m(-2) + 3 \)
\( \Rightarrow 5 = -2m + 3 \)
\( \Rightarrow 2m = 3 - 5 \)
\( \Rightarrow 2m = -2 \)
\( \Rightarrow m = \frac{-2}{2} \)
\( \therefore m = -1 \)
\( \Rightarrow 5 = -2m + 3 \)
\( \Rightarrow 2m = 3 - 5 \)
\( \Rightarrow 2m = -2 \)
\( \Rightarrow m = \frac{-2}{2} \)
\( \therefore m = -1 \)
Final values: \( x = -2 \), \( y = 5 \) and \( m = -1 \).
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.
Step 1: Formulate Equations
Let the two numbers be \(x\) and \(y\), such that \(y > x\).
According to the question:
\(y = 3x \quad \dots (1)\)
\(y - x = 26 \quad \dots (2)\)
Step 2: Solve using Substitution
According to the question:
\(y = 3x \quad \dots (1)\)
\(y - x = 26 \quad \dots (2)\)
Substituting the value of (1) into (2):
\( \Rightarrow 3x - x = 26 \)
\( \Rightarrow 2x = 26 \)
\( \Rightarrow x = 13 \quad \dots (3) \)
Substituting (3) into (1):
\( \Rightarrow y = 3(13) \)
\( \therefore y = 39 \)
\( \Rightarrow 3x - x = 26 \)
\( \Rightarrow 2x = 26 \)
\( \Rightarrow x = 13 \quad \dots (3) \)
Substituting (3) into (1):
\( \Rightarrow y = 3(13) \)
\( \therefore y = 39 \)
Hence, the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Step 1: Formulate Equations
Let the larger angle be \(x\) and the smaller angle be \(y\).
We know the sum of supplementary angles is \(180^\circ\):
\(x + y = 180 \quad \dots (1)\)
According to the question:
\(x - y = 18 \quad \dots (2)\)
Step 2: Solve using Substitution
We know the sum of supplementary angles is \(180^\circ\):
\(x + y = 180 \quad \dots (1)\)
According to the question:
\(x - y = 18 \quad \dots (2)\)
From (1), we get: \(x = 180 - y \quad \dots (3)\)
Substituting (3) into (2):
\( \Rightarrow (180 - y) - y = 18 \)
\( \Rightarrow 180 - 2y = 18 \)
\( \Rightarrow -2y = 18 - 180 \)
\( \Rightarrow -2y = -162 \)
\( \Rightarrow y = 81 \quad \dots (4) \)
Using the value of \(y\) in (3):
\( \Rightarrow x = 180 - 81 = 99 \)
Substituting (3) into (2):
\( \Rightarrow (180 - y) - y = 18 \)
\( \Rightarrow 180 - 2y = 18 \)
\( \Rightarrow -2y = 18 - 180 \)
\( \Rightarrow -2y = -162 \)
\( \Rightarrow y = 81 \quad \dots (4) \)
Using the value of \(y\) in (3):
\( \Rightarrow x = 180 - 81 = 99 \)
Hence, the angles are 99° and 81°.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Step 1: Formulate Equations
Let the cost of a bat be \(x\) and the cost of a ball be \(y\).
\(7x + 6y = 3800 \quad \dots (I)\)
\(3x + 5y = 1750 \quad \dots (II)\)
Step 2: Solve using Substitution
\(7x + 6y = 3800 \quad \dots (I)\)
\(3x + 5y = 1750 \quad \dots (II)\)
From (I), we get: \(y = \frac{3800 - 7x}{6} \quad \dots (III)\)
Substituting (III) into (II):
\( \Rightarrow 3x + 5\left(\frac{3800 - 7x}{6}\right) = 1750 \)
Multiply the entire equation by 6:
\( \Rightarrow 18x + 5(3800 - 7x) = 10500 \)
\( \Rightarrow 18x + 19000 - 35x = 10500 \)
\( \Rightarrow -17x = 10500 - 19000 \)
\( \Rightarrow -17x = -8500 \)
\( \therefore x = 500 \)
Substituting \(x = 500\) into (III):
\( \Rightarrow y = \frac{3800 - 7(500)}{6} = \frac{300}{6} = 50 \)
Substituting (III) into (II):
\( \Rightarrow 3x + 5\left(\frac{3800 - 7x}{6}\right) = 1750 \)
Multiply the entire equation by 6:
\( \Rightarrow 18x + 5(3800 - 7x) = 10500 \)
\( \Rightarrow 18x + 19000 - 35x = 10500 \)
\( \Rightarrow -17x = 10500 - 19000 \)
\( \Rightarrow -17x = -8500 \)
\( \therefore x = 500 \)
Substituting \(x = 500\) into (III):
\( \Rightarrow y = \frac{3800 - 7(500)}{6} = \frac{300}{6} = 50 \)
Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.
(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 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is 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 km?
Step 1: Formulate Equations
Let the fixed charge be \(Rs \: x\) and charge per km be \(Rs \: y\).
\(x + 10y = 105 \quad \dots (1)\)
\(x + 15y = 155 \quad \dots (2)\)
Step 2: Solve for x and y
\(x + 10y = 105 \quad \dots (1)\)
\(x + 15y = 155 \quad \dots (2)\)
From (1): \(x = 105 - 10y \quad \dots (3)\)
Substituting (3) into (2):
\( \Rightarrow (105 - 10y) + 15y = 155 \)
\( \Rightarrow 5y = 50 \)
\( \therefore y = 10 \)
Putting \(y = 10\) in (3):
\( \Rightarrow x = 105 - 10(10) = 5 \)
Step 3: Calculate Charge for 25 km
Substituting (3) into (2):
\( \Rightarrow (105 - 10y) + 15y = 155 \)
\( \Rightarrow 5y = 50 \)
\( \therefore y = 10 \)
Putting \(y = 10\) in (3):
\( \Rightarrow x = 105 - 10(10) = 5 \)
Charge = \(x + 25y = 5 + 25(10) = 5 + 250 = 255\)
Hence, fixed charge is Rs 5, charge per km is Rs 10, and total charge for 25 km is Rs 255.
(v) A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Step 1: Formulate Equations
Let the fraction be \(\frac{x}{y}\).
Case 1: \(\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x - 9y = -4 \quad \dots (1)\)
Case 2: \(\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x - 5y = -3 \quad \dots (2)\)
Step 2: Solve using Substitution
Case 1: \(\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x - 9y = -4 \quad \dots (1)\)
Case 2: \(\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x - 5y = -3 \quad \dots (2)\)
From (1): \(x = \frac{-4 + 9y}{11} \quad \dots (3)\)
Substituting (3) into (2):
\( \Rightarrow 6\left(\frac{-4 + 9y}{11}\right) - 5y = -3 \)
\( \Rightarrow -24 + 54y - 55y = -33 \)
\( \Rightarrow -y = -9 \)
\( \therefore y = 9 \)
Substituting \(y = 9\) into (3):
\( \Rightarrow x = \frac{-4 + 9(9)}{11} = 7 \)
Substituting (3) into (2):
\( \Rightarrow 6\left(\frac{-4 + 9y}{11}\right) - 5y = -3 \)
\( \Rightarrow -24 + 54y - 55y = -33 \)
\( \Rightarrow -y = -9 \)
\( \therefore y = 9 \)
Substituting \(y = 9\) into (3):
\( \Rightarrow x = \frac{-4 + 9(9)}{11} = 7 \)
Hence, the fraction is 7/9.
(vi) Five years hence, Jacob's age will be three times that of his son. Five years ago, it was seven times. Find their present ages.
Step 1: Formulate Equations
Let Jacob's age be \(x\) and his son's age be \(y\).
Future (5 years hence): \(x + 5 = 3(y + 5) \Rightarrow x - 3y = 10 \quad \dots (1)\)
Past (5 years ago): \(x - 5 = 7(y - 5) \Rightarrow x - 7y = -30 \quad \dots (2)\)
Step 2: Solve using Substitution
Future (5 years hence): \(x + 5 = 3(y + 5) \Rightarrow x - 3y = 10 \quad \dots (1)\)
Past (5 years ago): \(x - 5 = 7(y - 5) \Rightarrow x - 7y = -30 \quad \dots (2)\)
From (1): \(x = 3y + 10 \quad \dots (3)\)
Substituting (3) into (2):
\( \Rightarrow (3y + 10) - 7y = -30 \)
\( \Rightarrow -4y = -40 \)
\( \therefore y = 10 \)
Substituting \(y = 10\) into (3):
\( \Rightarrow x = 3(10) + 10 = 40 \)
Substituting (3) into (2):
\( \Rightarrow (3y + 10) - 7y = -30 \)
\( \Rightarrow -4y = -40 \)
\( \therefore y = 10 \)
Substituting \(y = 10\) into (3):
\( \Rightarrow x = 3(10) + 10 = 40 \)
Hence, the present age of Jacob is 40 years and his son is 10 years.
4. If \(x + 3\) is a factor of \(x^3 + ax^2 - bx + 6\) and \(a + b = 7\), then find the values of \(a\) and \(b\).
Solution
Given polynomial:
\( P(x) = x^3 + ax^2 - bx + 6 \)
Applying the Factor Theorem:
Since \( (x + 3) \) is a factor of \( P(x) \), then \( P(-3) = 0 \).
\( \Rightarrow (-3)^3 + a(-3)^2 - b(-3) + 6 = 0 \)
\( \Rightarrow -27 + 9a + 3b + 6 = 0 \)
\( \Rightarrow 9a + 3b - 21 = 0 \)
Divide by 3:
\( \Rightarrow 3a + b = 7 \quad \dots (1) \)
From the second given condition:
\( \Rightarrow (-3)^3 + a(-3)^2 - b(-3) + 6 = 0 \)
\( \Rightarrow -27 + 9a + 3b + 6 = 0 \)
\( \Rightarrow 9a + 3b - 21 = 0 \)
Divide by 3:
\( \Rightarrow 3a + b = 7 \quad \dots (1) \)
\( a + b = 7 \quad \dots (2) \)
Solving the equations:
From equation (2):
\( \Rightarrow b = 7 - a \quad \dots (3) \)
Substitute \( b \) into equation (1):
\( \Rightarrow 3a + (7 - a) = 7 \)
\( \Rightarrow 2a + 7 = 7 \)
\( \Rightarrow 2a = 0 \)
\( \therefore a = 0 \)
Substitute \( a = 0 \) into equation (3):
\( \Rightarrow b = 7 - 0 \)
\( \therefore b = 7 \)
\( \Rightarrow b = 7 - a \quad \dots (3) \)
Substitute \( b \) into equation (1):
\( \Rightarrow 3a + (7 - a) = 7 \)
\( \Rightarrow 2a + 7 = 7 \)
\( \Rightarrow 2a = 0 \)
\( \therefore a = 0 \)
Substitute \( a = 0 \) into equation (3):
\( \Rightarrow b = 7 - 0 \)
\( \therefore b = 7 \)
The values are: \( a = 0 \) and \( b = 7 \).
5. Consider the following pair of linear equations in two variables. Now choose the correct option from the alternatives provided.
\(2x = y\) and \( -5x + 2y - 3 = 0\)
(i) The graphs of the two equations intersect at a single point.
(ii) The graphs of the two equations are parallel to each other.
(iii) The graphs of the two equations coincide.
(iv) The two equations have a unique solution.
(a) Both (i) and (ii) are correct
(b) Both (i) and (iii) are correct
(c) Both (ii) and (iii) are correct
(d) Both (i) and (iv) are correct
Solution
First, write the given equations in standard form \(ax + by + c = 0\):
\( 2x - y + 0 = 0 \quad \dots (1) \)
\( -5x + 2y - 3 = 0 \quad \dots (2) \)
Comparing the coefficients, we get:
\( -5x + 2y - 3 = 0 \quad \dots (2) \)
\( a_1 = 2, \quad b_1 = -1, \quad c_1 = 0 \)
\( a_2 = -5, \quad b_2 = 2, \quad c_2 = -3 \)
Now, let's check the ratios:
\( \frac{a_1}{a_2} = \frac{2}{-5} = -0.4 \)
\( \frac{b_1}{b_2} = \frac{-1}{2} = -0.5 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the pair of linear equations has a unique solution.
Analyzing the properties:
\( a_2 = -5, \quad b_2 = 2, \quad c_2 = -3 \)
Now, let's check the ratios:
\( \frac{a_1}{a_2} = \frac{2}{-5} = -0.4 \)
\( \frac{b_1}{b_2} = \frac{-1}{2} = -0.5 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the pair of linear equations has a unique solution.
1. Because there is a unique solution, the graphs of the two equations will intersect at a single point.
2. This means that both statement (i) and statement (iv) are correct.
2. This means that both statement (i) and statement (iv) are correct.
Correct Option: (d) Both (i) and (iv) are correct.
6. Match the items in Column A and Column B below and then choose the correct option.
| Column A | Column B |
|---|---|
| A. \(x - y = 0\) | (i) Has a unique solution |
| B. \(2x - 3y = 5\) and \(x - y = 1\) | (ii) Linear equation in two variables |
| C. \(x + 2y = 6\) and \(4x + 8y = 24\) | (iii) No solution |
| D. \(2x + 3y = 6\) and \(4x + 6y = 10\) | (iv) Has infinitely many solutions and the graphs are coincident lines. |
(a) A → (ii), B → (i), C → (iv), D → (iii)
(b) A → (ii), B → (iv), C → (i), D → (iii)
(c) A → (iv), B → (i), C → (ii), D → (iii)
(d) A → (iv), B → (ii), C → (i), D → (iii)
Solution
Analysis of Column A items:
A. \(x - y = 0\): This is a single equation with two variables (\(x\) and \(y\)) where the highest power is 1. Therefore, it is a Linear equation in two variables. Matches with (ii).
B. \(2x - 3y = 5\) and \(x - y = 1\):
Comparing ratios: \(\frac{a_1}{a_2} = \frac{2}{1} = 2\) and \(\frac{b_1}{b_2} = \frac{-3}{-1} = 3\).
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), it Has a unique solution. Matches with (i).
Comparing ratios: \(\frac{a_1}{a_2} = \frac{2}{1} = 2\) and \(\frac{b_1}{b_2} = \frac{-3}{-1} = 3\).
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), it Has a unique solution. Matches with (i).
C. \(x + 2y = 6\) and \(4x + 8y = 24\):
Comparing ratios: \(\frac{a_1}{a_2} = \frac{1}{4}\), \(\frac{b_1}{b_2} = \frac{2}{8} = \frac{1}{4}\), \(\frac{c_1}{c_2} = \frac{6}{24} = \frac{1}{4}\).
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), it Has infinitely many solutions and the graphs are coincident lines. Matches with (iv).
Comparing ratios: \(\frac{a_1}{a_2} = \frac{1}{4}\), \(\frac{b_1}{b_2} = \frac{2}{8} = \frac{1}{4}\), \(\frac{c_1}{c_2} = \frac{6}{24} = \frac{1}{4}\).
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), it Has infinitely many solutions and the graphs are coincident lines. Matches with (iv).
D. \(2x + 3y = 6\) and \(4x + 6y = 10\):
Comparing ratios: \(\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{6}{10} = \frac{3}{5}\).
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and there is No solution. Matches with (iii).
Comparing ratios: \(\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{6}{10} = \frac{3}{5}\).
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel and there is No solution. Matches with (iii).
Correct Option: (a) A → (ii), B → (i), C → (iv), D → (iii)
7. If the point \((x, 4)\) satisfies the equation \(3x + y = 19\), then the value of \(x\) will be—
(a) 6
(b) 5
(c) 3
(d) 4
Solution
Given equation and point:
Equation: \( 3x + y = 19 \)
Point: \( (x, 4) \)
Substituting the values:
Point: \( (x, 4) \)
Since the point \( (x, 4) \) satisfies the equation, we substitute \( y = 4 \) into the equation:
\( \Rightarrow 3x + 4 = 19 \)
\( \Rightarrow 3x = 19 - 4 \)
\( \Rightarrow 3x = 15 \)
\( \Rightarrow x = \frac{15}{3} \)
\( \therefore x = 5 \)
\( \Rightarrow 3x + 4 = 19 \)
\( \Rightarrow 3x = 19 - 4 \)
\( \Rightarrow 3x = 15 \)
\( \Rightarrow x = \frac{15}{3} \)
\( \therefore x = 5 \)
Correct Option: (b) 5
8. Two equations \(2x + 4y = 10\) and \(Kx + 8y = 20\) are given. For what value of \(K\) will the pair have infinitely many solutions?
(a) 4
(b) 3
(c) 2
(d) 1
Solution
Given equations:
\( 2x + 4y = 10 \)
\( Kx + 8y = 20 \)
Condition for infinitely many solutions:
\( Kx + 8y = 20 \)
For a pair of linear equations to have infinitely many solutions, the ratio of their coefficients must be equal:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Calculating the value of \(K\):
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Comparing the coefficients:
\( \frac{2}{K} = \frac{4}{8} = \frac{10}{20} \)
Taking the first two ratios:
\( \Rightarrow \frac{2}{K} = \frac{4}{8} \)
\( \Rightarrow \frac{2}{K} = \frac{1}{2} \)
\( \Rightarrow K = 2 \times 2 \)
\( \therefore K = 4 \)
\( \frac{2}{K} = \frac{4}{8} = \frac{10}{20} \)
Taking the first two ratios:
\( \Rightarrow \frac{2}{K} = \frac{4}{8} \)
\( \Rightarrow \frac{2}{K} = \frac{1}{2} \)
\( \Rightarrow K = 2 \times 2 \)
\( \therefore K = 4 \)
Correct Option: (a) 4
❓ FAQs
What is substitution method?
It solves equations by replacing one variable using another equation.
Is graphical method in Exercise 3.2?
No, graphical method is part of Exercise 3.1.
Is this updated syllabus?
Yes, it is updated for SEBA 2026–2027.
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