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Class 9 Maths: Two Variables, One Line, 8 practice questions with answers

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These are 8 practice questions on Chapter 13, Two Variables, One Line, of the new NCERT Class 9 Maths textbook, Ganita Manjari, Part II (First Edition, August 2026). Each has a worked answer and the textbook page it comes from. They cover writing a linear equation in standard form, checking a solution, finding where a line meets the axes, the slope between two points, the slope-intercept form, solving a pair of equations by substitution and by elimination, telling from the coefficients whether two lines meet, are parallel or coincide, and reading the solution of a pair from where their lines cross. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the larger numbers is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Manjari, Grade 9, Part II, Chapter 13, printed pages 85 to 122):

One equation, one line

Question 1. A school canteen buys x kg of rice at ₹47 per kg and y kg of dal at ₹131 per kg, and spends ₹4013 in all. Write this as a linear equation, put it in standard form and give a, b and c. Is (38, 17) a solution? Is (39, 17)?

Answer: 47x + 131y − 4013 = 0, with a = 47, b = 131, c = −4013; (38, 17) is a solution, (39, 17) is not. The cost is 47x + 131y = 4013, which in standard form is 47x + 131y − 4013 = 0. For (38, 17): 47 × 38 + 131 × 17 = 1786 + 2227 = 4013, so it satisfies the equation. For (39, 17): 47 × 39 + 131 × 17 = 1833 + 2227 = 4060 ≠ 4013, so it is not a solution. (Standard form, page 86; solutions as ordered pairs, page 88.)

Question 2. Where does the line 57x + 38y = 2166 meet the x-axis and the y-axis? Find one more solution with x = 76.

Answer: at (38, 0) and (0, 57); another solution is (76, −57). Put y = 0: 57x = 2166, so x = 2166 ÷ 57 = 38. Put x = 0: 38y = 2166, so y = 2166 ÷ 38 = 57. For x = 76: 57 × 76 = 4332, so 38y = 2166 − 4332 = −2166 and y = −57. Check: 57 × 76 + 38 × (−57) = 4332 − 2166 = 2166. (Finding points by putting x = 0 and y = 0, page 90; infinitely many solutions, page 88.)

Question 3. Find the slope of the line through A(−17, 171) and B(47, 44). Does the line go up or down from left to right? Do you get the same slope if you take the points the other way round?

Answer: −127/64; down; yes. Taking A as (x₁, y₁) and B as (x₂, y₂): rise = 44 − 171 = −127 and run = 47 − (−17) = 64, so the slope is −127/64. The slope is negative, so the line goes down from left to right. The other way round: rise = 171 − 44 = 127 and run = −17 − 47 = −64, and 127 ÷ (−64) = −127/64 again. (Slope as rise over run, pages 95 and 96; order of the points, page 98; negative slope, page 94.)

Question 4. Write 68x + 17y − 136 = 0 in slope-intercept form. What are the slope and the y-intercept, and what do they tell you about the line?

Answer: y = −4x + 8; slope −4, y-intercept 8. 17y = −68x + 136, so y = −4x + 8. The slope −4 means that for every 1 unit you move to the right, y goes down by 4 units. The y-intercept 8 means the line crosses the y-axis at (0, 8). (Slope-intercept form, pages 98 and 99.)

Pairs of equations

Question 5. Solve by substitution: y = 2x − 17 and 2x + 3y = 261.

Answer: x = 39, y = 61. Substitute y = 2x − 17 into the second equation: 2x + 3(2x − 17) = 261, so 2x + 6x − 51 = 261, so 8x = 312 and x = 39. Then y = 2 × 39 − 17 = 78 − 17 = 61. Check: 2 × 39 + 3 × 61 = 78 + 183 = 261. (Substitution method, page 106.)

Question 6. A nursery sells rose plants at ₹x each and tulsi plants at ₹y each. One school buys 44 rose and 17 tulsi plants for ₹3242. Another buys 33 rose and 51 tulsi plants for ₹3885. Find the price of each plant by elimination.

Answer: a rose plant costs ₹59 and a tulsi plant ₹38. The equations are 44x + 17y = 3242 and 33x + 51y = 3885. To eliminate x, multiply the first by 3 and the second by 4, so both have 132x: 132x + 51y = 9726 and 132x + 204y = 15540. Subtract the first from the second: 153y = 5814, so y = 38. Then 44x + 17 × 38 = 3242, so 44x = 3242 − 646 = 2596 and x = 59. Check: 33 × 59 + 51 × 38 = 1947 + 1938 = 3885. (Elimination method, page 108.)

Question 7. Compare each line with 41x + 53y = 47. Does it meet that line in one point, run parallel to it, or coincide with it? (a) 123x + 159y = 141 (b) 123x + 159y = 142 (c) 53x + 41y = 47

Answer: (a) coincides; (b) parallel; (c) meets it in one point. Write each as a₁/a₂, b₁/b₂, c₁/c₂ against 41x + 53y = 47.

(Ratios of the coefficients, pages 110 to 112; the three cases on a graph, page 113.)

Question 8. The line x + y = 123 passes through (0, 123) and (123, 0). The line x − y = 41 passes through (41, 0) and (0, −41). Where do the two lines cross, and what does that point tell you?

Answer: at (82, 41), the unique solution of the pair. Adding the equations: 2x = 164, so x = 82; then y = 123 − 82 = 41. Check: 82 + 41 = 123 and 82 − 41 = 41. The point (82, 41) lies on both lines, so it is the one solution of the pair; drawn on a graph, the two lines meet there and nowhere else. (Graphical method, pages 112 to 114.)

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Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 84.

The methods are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.