Exercise 3.3 — Pair of Linear Equations in Two Variables

1. 14 marksNCERT Cl-10 Maths, Ex 3.3, Q1 (four parts)

   Solve each pair by the elimination method: (i) x + y = 5, 2x − 3y = 4; (ii) 3x + 4y = 10, 2x − 2y = 2; (iii) 3x − 5y − 4 = 0, 9x = 2y + 7; (iv) x/2 + 2y/3 = −1, x − y/3 = 3.

   Hint. Multiply one or both equations so that one variable has the same size of coefficient in both, then add or subtract to remove it.

   The single decision in elimination is which variable to remove — pick whichever needs the smaller multipliers.

   (i) x + y = 5, 2x − 3y = 4.

   Multiply the first by 3: 3x + 3y = 15. Now add the second, and the y terms cancel: 5x = 19, so x = 19/5.

   Then from x + y = 5: y = 5 − 19/5 = 6/5.

   Check in the second: 2(19/5) − 3(6/5) = 38/5 − 18/5 = 20/5 = 4 ✓

   ✦ x = 19/5, y = 6/5

   (ii) 3x + 4y = 10, 2x − 2y = 2.

   Multiply the second by 2: 4x − 4y = 4. Adding removes y: 7x = 14, so x = 2.

   Then 2(2) − 2y = 2 gives y = 1.

   ✦ x = 2, y = 1

   (iii) 3x − 5y − 4 = 0, 9x = 2y + 7.

   First put both in standard form: 3x − 5y = 4 and 9x − 2y = 7.

   Multiply the first by 3: 9x − 15y = 12. Subtract the second: (9x − 15y) − (9x − 2y) = 12 − 7, so −13y = 5 and y = −5/13.

   Substituting into 3x − 5y = 4: 3x + 25/13 = 4 ⟹ 3x = 27/13 ⟹ x = 9/13.

   ✦ x = 9/13, y = −5/13

   (iv) x/2 + 2y/3 = −1, x − y/3 = 3.

   Clear fractions first — multiply the first by 6 and the second by 3:

   3x + 4y = −6 and 3x − y = 9.

   Subtracting the second from the first: 5y = −15, so y = −3. Then 3x + 3 = 9 gives x = 2.

   ✦ x = 2, y = −3

   Where students slip. In (iii), subtracting without bracketing the whole second equation, so only the first term gets its sign flipped. Write the subtraction out in full — (9x − 15y) − (9x − 2y) — before simplifying.

   Another way. Part (i) is one of the rare cases where substitution is tidier: y = 5 − x turns the second equation into 2x − 15 + 3x = 4 in one line.

2. 25 marksNCERT Cl-10 Maths, Ex 3.3, Q2 (five word problems)

   Form the equations and solve by elimination: (i) a fraction becomes 1 if 1 is added to the numerator and 1 subtracted from the denominator, and becomes 1/2 if 1 is added to the denominator; (ii) five years ago Nuri was three times as old as Sonu, and ten years from now she will be twice as old; (iii) the digits of a two-digit number add to 9, and nine times the number equals twice the number with its digits reversed; (iv) Meena withdrew ₹2000 in ₹50 and ₹100 notes, receiving 25 notes in all; (v) a library charges a fixed amount for the first three days and a fixed amount per extra day — Saritha paid ₹27 for seven days and Susy paid ₹21 for five days.

   Hint. For the digits problem, a two-digit number with tens digit x and units digit y is 10x + y — not xy.

   (i) The fraction. Let it be x/y.

   First condition: (x + 1)/(y − 1) = 1, so x + 1 = y − 1, that is x − y = −2.

   Second condition: x/(y + 1) = 1/2, so 2x = y + 1, that is 2x − y = 1.

   Subtract the first from the second: x = 3. Then 3 − y = −2 gives y = 5.

   ✦ The fraction is 3/5.

   (ii) Nuri and Sonu. Let Nuri be x and Sonu y years old now.

   Five years ago: x − 5 = 3(y − 5), which simplifies to x − 3y = −10.

   Ten years hence: x + 10 = 2(y + 10), which simplifies to x − 2y = 10.

   Subtracting the first from the second: y = 20. Then x = 10 + 40 = 50.

   ✦ Nuri is 50 and Sonu is 20.

   (iii) The two-digit number. Let the tens digit be x and the units digit y, so the number is 10x + y and the reversed number is 10y + x.

   Digits add to 9: x + y = 9.

   Nine times the number equals twice the reverse: 9(10x + y) = 2(10y + x), so 90x + 9y = 20y + 2x, giving 88x = 11y, that is y = 8x.

   Substituting into x + y = 9: x + 8x = 9 ⟹ x = 1, y = 8.

   Check: the number is 18, and 9 × 18 = 162, while the reverse is 81 and 2 × 81 = 162 ✓

   ✦ The number is 18.

   (iv) The notes. Let x be the number of ₹50 notes and y the number of ₹100 notes.

   Total notes: x + y = 25. Total value: 50x + 100y = 2000, which simplifies to x + 2y = 40.

   Subtracting the first from the second: y = 15, so x = 10.

   Check: 10(50) + 15(100) = 500 + 1500 = 2000 ✓

   ✦ 10 notes of ₹50 and 15 of ₹100.

   (v) The library. Let the fixed charge for the first three days be ₹x and the charge per extra day ₹y.

   Seven days means four extra days: x + 4y = 27.

   Five days means two extra days: x + 2y = 21.

   Subtracting: 2y = 6, so y = 3, and then x = 15.

   Check: seven days cost 15 + 4(3) = 27 ✓ and five days cost 15 + 2(3) = 21 ✓

   ✦ Fixed charge ₹15, and ₹3 for each day beyond the first three.

   Where students slip. In (v), writing x + 7y = 27 — charging the per-day rate for all seven days. The fixed amount already covers the first three, so only four days are extra. Read carefully what the fixed charge buys.

   Another way. In (iii), once you have y = 8x you barely need the second equation: the only single digits satisfying it are x = 1, y = 8, since any larger x makes y exceed 9.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (the cross-multiplication method is no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.