Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 13 Exercise Set 13.2 (Questions 1–7). Covers verifying ordered pair solutions, finding multiple solutions, determining unknown constants, quadrant analysis, geometric interpretation of linear graphs, True/False conceptual analysis, and equivalence of linear equations under non-zero scalar multiplication.
An ordered pair (u, v) is a solution of a linear equation ax + by + c = 0 if substituting x = u and y = v satisfies the equation, making the left-hand side equal to the right-hand side.
Every linear equation in two variables has infinitely many solutions. For each real number chosen for x, there is a unique corresponding real value for y determined by solving the equation.
When a point (u, v) is given as a solution to an equation containing an unknown constant (such as m or n), substitute x = u and y = v into the equation and solve the resulting single-variable equation for the unknown.
The Cartesian axes divide the coordinate plane into four quadrants determined by the signs of the coordinates: Quadrant I (+, +), Quadrant II (−, +), Quadrant III (−, −), and Quadrant IV (+, −). Points lying on the axes do not belong to any quadrant.
The graph of every linear equation in two variables is a straight line. Every point on the line is a solution of the equation, and conversely, every point that satisfies the equation lies on the line. A point not on the line does not satisfy the equation.
Multiplying or dividing both sides of a linear equation ax + by = c by any non-zero real number k produces an equivalent equation kax + kby = kc that has the exact same set of solutions and represents the identical straight line.
Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.
The given equation is
For the ordered pair (4, 3), x = 4 and y = 3.
5(4) − 6(3) = 20 − 18 = 2
The left-hand side equals the right-hand side.
Find any two solutions for each of the following equations: (i) 7x − 3y = 21 (ii) 2x + 3y = 5
Let y = 0. Then
One solution is (3, 0).
Let x = 0. Then
Another solution is (0, −7).
Let x = 1. Then
2 + 3y = 5 ⇒ y = 1
One solution is (1, 1).
Let x = 4. Then
8 + 3y = 5 ⇒ y = −1
Another solution is (4, −1).
In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = −10. If (2, −1) is the solution of both equations, find the values of m and n.
Since (2, −1) is a solution of both equations, substitute x = 2 and y = −1.
For the first equation,
For the second equation,
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 Verify your solutions by representing the linear equations on a graph paper.
Recall the signs of coordinates in each quadrant:
−5 + 3y = 7 ⇒ y = 4
So (−1, 4) lies in Quadrant II.
10 + 3y = 7 ⇒ y = −1
So (2, −1) lies in Quadrant IV.
−5 − 3y = 7 ⇒ y = −4
So (−1, −4) lies in Quadrant III.
10 − 3y = 7 ⇒ y = 1
So (2, 1) lies in Quadrant I.
10 + 3y = 7 ⇒ y = −1
So (−2, −1) lies in Quadrant III.
−5 + 3y = 7 ⇒ y = 4
So (1, 4) lies in Quadrant I.
10 − 3y = 7 ⇒ y = 1
So (−2, 1) lies in Quadrant II.
−5 − 3y = 7 ⇒ y = −4
So (1, −4) lies in Quadrant IV.
Consider the graph of the equation 3x − 7y = 21 shown below. 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 the linear equation 3x − 7y = 21 with line passing through A(7, 0) and B(0, −3), and point C(2, 3).
The given equation is
For C(2, 3), substitute x = 2 and y = 3:
3(2) − 7(3) = 6 − 21 = −15
Since −15 ≠ 21, point C(2, 3) does not satisfy the equation. Therefore, it does not lie on the line.
Every point on the graph of a linear equation represents a solution of that equation, and every solution corresponds to a point on the graph.
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.
A linear equation in two variables has infinitely many solutions. For example, both (2, 1) and (5, −1) satisfy 2x + 3y = 7.
The origin is (0, 0). For 2x + 3y = 7, substituting x = 0, y = 0 gives 0 ≠ 7. Therefore, this line does not pass through the origin.
The pair (2, 1) is a rational solution of 2x + 3y = 7, because 2(2) + 3(1) = 7.
The equation x = 3 can be written as x + 0y − 3 = 0, which is a linear equation in two variables.
The equation 2x + 3y = 7 has infinitely many solutions. For example, (2, 1) and (5, −1) both satisfy it. There are infinitely many other solutions.
Substituting (1, 2) gives
2(1) + 3(2) = 2 + 6 = 8 ≠ 7
Therefore, (1, 2) is not a solution.
(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other. (ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
Let (u, v) be any solution of 3x + 4y = 7. Then
Multiplying both sides by 2,
Therefore, (u, v) also satisfies 6x + 8y = 14. Thus, every solution of the first equation is a solution of the second.
Conversely, let (u, v) be any solution of 6x + 8y = 14. Then
Dividing both sides by 2,
Therefore, (u, v) also satisfies 3x + 4y = 7. Thus, every solution of the second equation is a solution of the first.
Since both directions hold, the two equations have the same set of solutions.
Let (u, v) be any solution of ax + by = c. Then
Multiplying both sides by k,
Therefore, (u, v) also satisfies kax + kby = kc. Thus, every solution of the first equation is a solution of the second.
Conversely, let (u, v) be any solution of kax + kby = kc. Then
Since k ≠ 0, divide both sides by k:
Therefore, (u, v) also satisfies ax + by = c. Thus, every solution of the second equation is a solution of the first.
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