NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 13 Two Variables One Line Exercise 13.2

Exercise Set 13.2

Question 1. Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.
Answer:
A pair is a solution when substituting it makes the left side equal to the right side.
LHS = 5(4) − 6(3) = 20 − 18 = 2 = RHS.
So (4, 3) is a solution of 5x − 6y = 2.

Question 2. Find any two solutions for each of the following equations:
(i) 7x − 3y = 21   (ii) 2x + 3y = 5
Answer:
(i) x = 0: −3y = 21, y = −7 → (0, −7). y = 0: 7x = 21, x = 3 → (3, 0).
Check: 7(0) − 3(−7) = 21 ✓; 7(3) − 0 = 21 ✓. Two solutions: (0, −7) and (3, 0) (others, like (6, 7), also work).
(ii) x = 1: 2 + 3y = 5, 3y = 3, y = 1 → (1, 1). x = 4: 8 + 3y = 5, 3y = −3, y = −1 → (4, −1).
Check: 2 + 3 = 5 ✓; 8 − 3 = 5 ✓. Two solutions: (1, 1) and (4, −1) (others, like (0, \( \frac{5}{3} \)) or (−2, 3), also work).

Question 3. 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.
Answer:
Put x = 2, y = −1 in both equations.
First: 2m(2) + 3(−1) = 7 → 4m − 3 = 7 → 4m = 10 → m = \( \frac{10}{4} \) = \( \frac{5}{2} \).
Second: 4(2) + n(−1) = −10 → 8 − n = −10 → n = 18.
Check: 2 × \( \frac{5}{2} \) × 2 − 3 = 10 − 3 = 7 ✓; 8 − 18 = −10 ✓.
m = \( \frac{5}{2} \), 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
Verify your solutions by representing the linear equations on a graph paper.
Answer:
Signs of coordinates: Quadrant I (+, +), II (−, +), III (−, −), IV (+, −). We pick x-values that give whole-number y.
(i) y = \( \frac{7 - 5x}{3} \). x = −1: y = 4 → (−1, 4), Quadrant II. x = 2: y = −1 → (2, −1), Quadrant IV.
Check: −5 + 12 = 7 ✓; 10 − 3 = 7 ✓.
(ii) y = \( \frac{5x - 7}{3} \). x = 2: y = 1 → (2, 1), Quadrant I. x = −1: y = −4 → (−1, −4), Quadrant III.
Check: 10 − 3 = 7 ✓; −5 + 12 = 7 ✓.
(iii) y = \( \frac{7 + 5x}{3} \). x = 1: y = 4 → (1, 4), Quadrant I. x = −2: y = −1 → (−2, −1), Quadrant III.
Check: −5 + 12 = 7 ✓; 10 − 3 = 7 ✓.
(iv) y = \( \frac{-7 - 5x}{3} \). x = −2: y = 1 → (−2, 1), Quadrant II. x = 1: y = −4 → (1, −4), Quadrant IV.
Check: 10 − 3 = 7 ✓; −5 + 12 = 7 ✓.
Graph check: in each graph below, both chosen points lie on the line of the equation, confirming they are solutions.

Teacher's Note:
1. Writing y in terms of x first makes it easy to spot x-values that give whole numbers.
2. Many other correct pairs exist; any two points on the line in different quadrants are acceptable.

Question 5. 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?

Answer:
On the line? The line goes through A(7, 0) and B(0, −3). At x = 2 the line is below the x-axis, whereas C(2, 3) is above it in Quadrant I. So C does not lie on the line.
Satisfies the equation? 3(2) − 7(3) = 6 − 21 = −15 ≠ 21. So C does not satisfy the equation.
(The point of the line directly below C is (2, \( -\frac{15}{7} \)), since −7y = 21 − 6 = 15.)
Can off-line points satisfy it? No. The graph of a linear equation is exactly the set of all its solutions: every solution is on the line and every point of the line is a solution. So a point not on the line can never satisfy the 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.
Answer:
(i) False. Each value of x gives a matching y, so there are infinitely many solutions; e.g. x + y = 5 has (0, 5), (1, 4), (2, 3), ….
(ii) False. The line passes through the origin only when c = 0. For x + y = 5, (0, 0) gives 0 ≠ 5.
(iii) False. 2x + 3y = 7 has the rational solution (2, 1), and x + y = 1 has (\( \frac{1}{2} \), \( \frac{1}{2} \)).
(iv) True. x = 3 is 1·x + 0·y − 3 = 0, with a = 1, b = 0, c = −3 (a and b not both zero). Its graph is a vertical line.
(v) True. For every x there is a y = \( \frac{7 - 2x}{3} \), so there are infinitely many solutions.
(vi) False. 2(1) + 3(2) = 8 ≠ 7, so (1, 2) is not a solution.

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, 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.
Answer:
(i) The second equation is the first with every term doubled.
If (p, q) satisfies 3p + 4q = 7, doubling both sides gives 6p + 8q = 14, so (p, q) satisfies the second equation.
If (p, q) satisfies 6p + 8q = 14, halving both sides gives 3p + 4q = 7, so (p, q) satisfies the first equation.
Examples: (1, 1) gives 7 and 14 ✓; (5, −2) gives 15 − 8 = 7 and 30 − 16 = 14 ✓.
So the two equations have exactly the same solutions, and their graphs are the same line.
(ii) Let k ≠ 0. If ap + bq = c, multiplying by k gives kap + kbq = kc, so (p, q) solves the second equation.
Conversely, if kap + kbq = kc, then k(ap + bq) = kc; since k ≠ 0 we may divide by k to get ap + bq = c.
Hence both equations have the same set of solutions.
Why k ≠ 0? With k = 0 the second equation becomes 0 = 0, which every (x, y) satisfies, and division by k would be impossible.

Teacher's Note:
1. A 'same set' argument needs both directions: first ⇒ second and second ⇒ first.
2. The condition k ≠ 0 is used only in the reverse direction, when dividing.