Ex 13.2 Class 9 Maths Ganita Manjari Solutions | Sid Classes

Ganita Manjari Class 9 Exercise 13.2 solutions: solutions of linear equations in two variables, with stepwise CBSE answers. Sid Classes.

Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.2 Solutions, Solutions of Linear Equations, Graphing Coordinates, and True/False Statements. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers verifying solutions, finding coordinate pairs across different quadrants, analyzing linear equation graphs, and testing algebraic equivalence entirely clear.

🔑 Key Concepts & Principles:

  • Solutions of Linear Equations: An ordered pair \((x, y)\) is a solution if substituting its values into the equation satisfies equality (\(\text{LHS} = \text{RHS}\)).
  • Graphical Representation: Every linear equation in two variables represents a straight line where every point on the line is a solution, and every solution lies on the line.

NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.2 Solutions

Question 1

Verify if the ordered pair \((4, 3)\) is a solution of \(5x - 6y = 2\). Explain your reasoning.

Solution:

To verify if \((4, 3)\) is a solution, we substitute \(x = 4\) and \(y = 3\) into the left-hand side (LHS) of the equation:

\(\text{LHS} = 5(4) - 6(3)\)
\(\text{LHS} = 20 - 18 = 2\)

Since \(\text{LHS} = \text{RHS} (2 = 2)\), the ordered pair \(\mathbf{(4, 3)\text{ is a solution}}\) of the given linear equation.

Question 2

Find any two solutions for each of the following equations:
(i) \(7x - 3y = 21\)
(ii) \(2x + 3y = 5\)

Solution:

(i) For \(7x - 3y = 21\):
* Put \(x = 0\): \(7(0) - 3y = 21 \implies -3y = 21 \implies y = -7\). Solution: \(\mathbf{(0, -7)}\).
* Put \(y = 0\): \(7x - 3(0) = 21 \implies 7x = 21 \implies x = 3\). Solution: \(\mathbf{(3, 0)}\).

(ii) For \(2x + 3y = 5\):
* Put \(x = 1\): \(2(1) + 3y = 5 \implies 2 + 3y = 5 \implies 3y = 3 \implies y = 1\). Solution: \(\mathbf{(1, 1)}\).
* Put \(x = -2\): \(2(-2) + 3y = 5 \implies -4 + 3y = 5 \implies 3y = 9 \implies y = 3\). Solution: \(\mathbf{(-2, 3)}\).

Question 3

In the equations given below, \(m\) and \(n\) are unknown constants: \(2mx + 3y = 7\) and \(4x + ny = -10\). If \((2, -1)\) is the solution of both equations, find the values of \(m\) and \(n\).

Solution:

Since \((2, -1)\) is a solution to both equations, we substitute \(x = 2\) and \(y = -1\) into each equation:

Step 1: Find \(m\) from \(2mx + 3y = 7\):
\(2m(2) + 3(-1) = 7\)
\(4m - 3 = 7\)
\(4m = 10 \implies m = \frac{10}{4} = \mathbf{\frac{5}{2}}\)

Step 2: Find \(n\) from \(4x + ny = -10\):
\(4(2) + n(-1) = -10\)
\(8 - n = -10\)
\(-n = -10 - 8 \implies -n = -18 \implies \mathbf{n = 18}\).

Question 4

Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie:
(i) \(5x + 3y = 7\)
(ii) \(5x - 3y = 7\)
(iii) \(-5x + 3y = 7\)
(iv) \(-5x - 3y = 7\)

Solution:

* (i) \(5x + 3y = 7\):
- If \(x = 2\), \(5(2) + 3y = 7 \implies 10 + 3y = 7 \implies 3y = -3 \implies y = -1\). Point \((2, -1)\) lies in Quadrant IV.
- If \(y = 4\), \(5x + 3(4) = 7 \implies 5x + 12 = 7 \implies 5x = -5 \implies x = -1\). Point \((-1, 4)\) lies in Quadrant II.

* (ii) \(5x - 3y = 7\):
- If \(x = 2\), \(5(2) - 3y = 7 \implies 10 - 3y = 7 \implies -3y = -3 \implies y = 1\). Point \((2, 1)\) lies in Quadrant I.
- If \(x = -1\), \(5(-1) - 3y = 7 \implies -5 - 3y = 7 \implies -3y = 12 \implies y = -4\). Point \((-1, -4)\) lies in Quadrant III.

* (iii) \(-5x + 3y = 7\):
- If \(x = 1\), \(-5(1) + 3y = 7 \implies -5 + 3y = 7 \implies 3y = 12 \implies y = 4\). Point \((1, 4)\) lies in Quadrant I.
- If \(x = -2\), \(-5(-2) + 3y = 7 \implies 10 + 3y = 7 \implies 3y = -3 \implies y = -1\). Point \((-2, -1)\) lies in Quadrant III.

* (iv) \(-5x - 3y = 7\):
- If \(x = -2\), \(-5(-2) - 3y = 7 \implies 10 - 3y = 7 \implies -3y = -3 \implies y = 1\). Point \((-2, 1)\) lies in Quadrant II.
- If \(x = 1\), \(-5(1) - 3y = 7 \implies -5 - 3y = 7 \implies -3y = 12 \implies y = -4\). Point \((1, -4)\) lies in Quadrant IV.

Question 5

Consider the graph of the equation \(3x - 7y = 21\). Does the point \(C(2, 3)\) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?

Graph of \(3x - 7y = 21\) (with Point C)
C(2, 3) A(7, 0) B(0, -3)

Solution:

* Does point C(2, 3) lie on the line?
Looking at the graph, the point \(C(2, 3)\) lies above the straight line representing \(3x - 7y = 21\), so it does not lie on the line.

* Does it satisfy the equation?
Substituting \(x = 2\) and \(y = 3\) into \(\text{LHS} = 3(2) - 7(3) = 6 - 21 = -15\). Since \(-15 \neq 21\), it does not satisfy the equation.

* Can points that do not lie on the line satisfy the equation?
No. Only points that lie directly on the graphical line represent valid solutions to the linear equation.

Question 6

State whether the following sentences are True or False. Justify your answer:
(i) A linear equation in two variables has only one solution.
(ii) The graph of a linear equation in two variables always passes through the origin.
(iii) A linear equation in two variables can never have rational solutions.
(iv) \(x = 3\) is a valid linear equation in two variables.
(v) The equation \(2x + 3y = 7\) has infinitely many solutions.
(vi) The point \((1, 2)\) is a solution of the equation \(2x + 3y = 7\).

Solution:

* (i) False. A linear equation in two variables has infinitely many solutions corresponding to every point on its line.
* (ii) False. The graph passes through the origin only if the constant term \(c = 0\).
* (iii) False. Linear equations can have infinitely many rational and irrational solutions.
* (iv) True. It can be expressed in two variables as \(1x + 0y = 3\).
* (v) True. Every linear equation in two variables has infinitely many solutions.
* (vi) False. Substituting \((1, 2)\) gives \(2(1) + 3(2) = 8 \neq 7\).

Question 7

(i) Compare the solutions of the equations \(3x + 4y = 7\) and \(6x + 8y = 14\). Argue that they have the same set of solutions.
(ii) Show that the equations \(ax + by = c\) and \(kax + kby = kc\), with \(k \neq 0\), have the same set of solutions.

Solution:

* (i) Multiplying both sides of \(3x + 4y = 7\) by \(2\) gives \(6x + 8y = 14\). Since multiplying an equation by a non-zero constant preserves equality, any pair \((x, y)\) satisfying the first equation will automatically satisfy the second, and vice versa.
* (ii) Similarly, multiplying \(ax + by = c\) by any non-zero constant \(k\) yields \(kax + kby = kc\). The solution set remains identical because the ratio and balance of variables are unchanged.

Frequently Asked Questions (FAQs)

Q1: How many solutions does a linear equation in two variables have?
A: Every linear equation in two variables has infinitely many solutions.

Q2: Does every point on the graph of a linear equation represent a solution?
A: Yes, every coordinate point lying on the straight line is a valid solution to that equation.

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