Ganita Manjari · Class 9 · Part II · Chapter 13
Seven questions on solutions of linear equations in two variables: verifying a solution, finding solutions, unknown constants, quadrants, graphs and true or false.
Brilliant! You have finished every question in this exercise.
Solution: an ordered pair (x, y) that satisfies the equation.
Ordered pair: the order matters: first x, then y.
Quadrant: one of the four regions of the plane, I (+, +), II (−, +), III (−, −), IV (+, −).
The left side equals the right side, so the pair satisfies the equation. A solution is exactly an ordered pair that satisfies it.
Yes: 5(4) − 6(3) = 2, so (4, 3) is a solution.
Put x = 0: −3y = 21, so y = −7. Put y = 0: 7x = 21, so x = 3.
| x | 0 | 3 |
|---|---|---|
| y | −7 | 0 |
Check (0, −7): 0 + 21 = 21 ✓. Check (3, 0): 21 − 0 = 21 ✓.
Put x = 1: 2 + 3y = 5, so y = 1. Put x = −2: −4 + 3y = 5, so y = 3.
| x | 1 | −2 |
|---|---|---|
| y | 1 | 3 |
Check (1, 1): 2 + 3 = 5 ✓. Check (−2, 3): −4 + 9 = 5 ✓. Other solutions are (0, 53) and (52, 0).
(i) (0, −7) and (3, 0). (ii) (1, 1) and (−2, 3) (many other pairs also work).
With m = 52: 2 × 52 × 2 + 3(−1) = 10 − 3 = 7 ✓. With n = 18: 8 − 18 = −10 ✓.
m = 52 and n = 18.
Choose a value of x, find y, and look at the signs. Quadrant I is (+, +), II is (−, +), III is (−, −) and IV is (+, −).
| Equation | Solution 1 | Solution 2 |
|---|---|---|
| (i) 5x + 3y = 7 | (2, −1) in IV (10 − 3 = 7 ✓) | (−1, 4) in II (−5 + 12 = 7 ✓) |
| (ii) 5x − 3y = 7 | (2, 1) in I (10 − 3 = 7 ✓) | (−1, −4) in III (−5 + 12 = 7 ✓) |
| (iii) −5x + 3y = 7 | (1, 4) in I (−5 + 12 = 7 ✓) | (−2, −1) in III (10 − 3 = 7 ✓) |
| (iv) −5x − 3y = 7 | (−2, 1) in II (10 − 3 = 7 ✓) | (1, −4) in IV (−5 + 12 = 7 ✓) |
A line with a non-zero slope that does not pass through the origin goes through exactly three quadrants, so there are always two quadrants to choose from.
(i) (2, −1) IV and (−1, 4) II. (ii) (2, 1) I and (−1, −4) III. (iii) (1, 4) I and (−2, −1) III. (iv) (−2, 1) II and (1, −4) IV.
C is clearly above the line in the graph, and it does not satisfy the equation. The points A(7, 0) and B(0, −3) on the line do satisfy it: 21 − 0 = 21 and 0 + 21 = 21.
A point off the line cannot satisfy the equation. Every solution of the equation is a point on its graph, and every point on the graph is a solution. So a point is on the line exactly when its coordinates satisfy the equation.
C(2, 3) is not on the line and does not satisfy the equation. No point off the line satisfies it.
| Statement | True or False | Reason |
|---|---|---|
| (i) only one solution | False | Any x gives a y, so there are infinitely many solutions |
| (ii) always through the origin | False | Only when c = 0. For x + y = 1 the point (0, 0) gives 0 ≠ 1 |
| (iii) never rational solutions | False | 2x + y = 1 has the rational solution (12, 0) |
| (iv) x = 3 is valid | True | It is x + 0y − 3 = 0 with a = 1, b = 0; its graph is a vertical line |
| (v) infinitely many solutions | True | Choose any x, then y = 7 − 2x3 |
| (vi) (1, 2) is a solution | False | 2(1) + 3(2) = 8 ≠ 7 |
(i) False, (ii) False, (iii) False, (iv) True, (v) True, (vi) False.
The second equation is 2 times the first: 6x + 8y = 2(3x + 4y) and 14 = 2 × 7.
Example: (1, 1) satisfies both, since 3 + 4 = 7 and 6 + 8 = 14.
Each equation follows from the other, so their solutions are the same. We need k ≠ 0: if k = 0, the second equation is 0 = 0, which every pair satisfies.
Multiplying or dividing both sides by a non-zero number does not change the solutions, so the two equations have the same set of solutions.
A pair is a solution only if it satisfies the equation; the graph shows all of them.
| Question | What is asked | Key idea | Answer |
|---|---|---|---|
| Q1 | (4, 3) in 5x − 6y = 2 | Substitute | Yes |
| Q2 | Two solutions | Choose x, find y | (0, −7), (3, 0); (1, 1), (−2, 3) |
| Q3 | Unknown m and n | Substitute (2, −1) | m = 5/2, n = 18 |
| Q4 | Different quadrants | Sign of x and y | See the table |
| Q5 | Point off the line | Substitute | C(2, 3) is not a solution |
| Q6 | True or false | Examples | F, F, F, T, T, F |
| Q7 | Same solutions | Multiply by k ≠ 0 | Equivalent equations |
Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 13, Exercise Set 13.2. The solutions, explanations and diagrams on this page are our own working.