Exercise 3.2 — Pair of Linear Equations in Two Variables

1. 16 marksNCERT Cl-10 Maths, Ex 3.2, Q1 (six parts)

   Solve each pair by substitution: (i) x + y = 14, x − y = 4; (ii) s − t = 3, s/3 + t/2 = 6; (iii) 3x − y = 3, 9x − 3y = 9; (iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3; (v) √2x + √3y = 0, √3x − √8y = 0; (vi) (3/2)x − (5/3)y = −2, x/3 + y/2 = 13/6.

   Hint. Pick the equation where one variable already has coefficient 1 (or is easiest to isolate) — that choice decides how messy the algebra gets.

   (i) x + y = 14, x − y = 4.

   From the second, x = y + 4. Substituting into the first: (y + 4) + y = 14, so 2y = 10 and y = 5. Then x = 9.

   ✦ x = 9, y = 5

   (ii) s − t = 3, s/3 + t/2 = 6.

   Clear the fractions in the second by multiplying by 6: 2s + 3t = 36. From the first, s = t + 3. Substituting: 2(t + 3) + 3t = 36 ⟹ 5t = 30 ⟹ t = 6, and s = 9.

   ✦ s = 9, t = 6

   (iii) 3x − y = 3, 9x − 3y = 9.

   From the first, y = 3x − 3. Substituting into the second: 9x − 3(3x − 3) = 9x − 9x + 9 = 9, which reduces to 9 = 9.

   That is a true statement carrying no information about x — the signal that the second equation is just the first tripled.

   ✦ Infinitely many solutions: every point on 3x − y = 3.

   (iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3.

   Multiply both by 10 first: 2x + 3y = 13 and 4x + 5y = 23. From the first, x = (13 − 3y)/2. Substituting: 2(13 − 3y) + 5y = 23 ⟹ 26 − y = 23 ⟹ y = 3, then x = 2.

   ✦ x = 2, y = 3

   (v) √2x + √3y = 0, √3x − √8y = 0.

   Both constants are zero, so (0, 0) is clearly a solution — the question is whether it is the only one. From the first, x = −√3y/√2. Substituting into the second:

   √3(−√3y/√2) − √8y = −3y/√2 − 2√2y = 0.

   Since −3/√2 − 2√2 ≠ 0, the only possibility is y = 0, and then x = 0.

   ✦ x = 0, y = 0

   (vi) (3/2)x − (5/3)y = −2, x/3 + y/2 = 13/6.

   Clear fractions: multiply the first by 6 to get 9x − 10y = −12; multiply the second by 6 to get 2x + 3y = 13. From the second, x = (13 − 3y)/2. Substituting:

   9(13 − 3y)/2 − 10y = −12 ⟹ 117 − 27y − 20y = −24 ⟹ 47y = 141 ⟹ y = 3, and then x = 2.

   ✦ x = 2, y = 3

   Where students slip. In (iii), writing 'no solution' when the variables cancel. Check what is left behind: 9 = 9 is true, so there are infinitely many solutions. A false statement like 9 = 5 would mean none.

   Another way. For (i), adding the equations immediately gives 2x = 18. Substitution is asked for here, but noticing the shortcut is a good way to check your answer.

2. 23 marksNCERT Cl-10 Maths, Ex 3.2, Q2

   Solve 2x + 3y = 11 and 2x − 4y = −24, then use your answer to find the value of m for which y = mx + 3.

   Hint. Solve the pair first. The last part is just substituting the point you found into y = mx + 3 and solving for m.

   Step 1 — From the first equation, 2x = 11 − 3y, so x = (11 − 3y)/2.

   Step 2 — Substitute into the second: (11 − 3y) − 4y = −24, since 2x is exactly 11 − 3y.

   Step 3 — Simplify: 11 − 7y = −24 ⟹ −7y = −35 ⟹ y = 5.

   Step 4 — Back-substitute: 2x + 15 = 11 ⟹ 2x = −4 ⟹ x = −2.

   Step 5 — Now use the point (−2, 5) in y = mx + 3: 5 = m(−2) + 3, so −2m = 2 and m = −1.

   Step 6 — Check: with m = −1, y = −x + 3 gives y = 2 + 3 = 5 at x = −2 ✓

   ✦ x = −2, y = 5, and m = −1.

   Where students slip. Substituting only x, or only y, into y = mx + 3. You need the whole point — the equation has two unknowns filled in and one, m, to find.

   Another way. Subtracting the second equation from the first eliminates x immediately: 7y = 35.

3. 36 marksNCERT Cl-10 Maths, Ex 3.2, Q3 (six word problems)

   Form a pair of linear equations and solve by substitution: (i) the difference between two numbers is 26 and the larger is three times the smaller; (ii) the larger of two supplementary angles exceeds the smaller by 18°; (iii) a coach buys 7 bats and 6 balls for ₹3800, and later 3 bats and 5 balls for ₹1750; (iv) a taxi charges a fixed amount plus a rate per km — 10 km costs ₹105 and 15 km costs ₹155; find the charge for 25 km; (v) a fraction becomes 9/11 when 2 is added to both numerator and denominator, and 5/6 when 3 is added to both; (vi) five years from now Jacob will be three times as old as his son, and five years ago he was seven times as old.

   Hint. In every part, name the two unknowns in one sentence before writing any equation. Most errors here come from unclear definitions, not from the algebra.

   (i) Two numbers. Let them be x (larger) and y. Then x − y = 26 and x = 3y. Substituting: 3y − y = 26 ⟹ y = 13, so x = 39.

   ✦ 39 and 13

   (ii) Supplementary angles. Supplementary means they add to 180°. Let them be x (larger) and y. Then x + y = 180 and x − y = 18. From the second, x = y + 18, so (y + 18) + y = 180 ⟹ 2y = 162 ⟹ y = 81, and x = 99.

   ✦ 99° and 81°

   (iii) Bats and balls. Let a bat cost ₹x and a ball ₹y. Then 7x + 6y = 3800 and 3x + 5y = 1750.

   From the second, x = (1750 − 5y)/3. Substituting into the first:

   7(1750 − 5y)/3 + 6y = 3800 ⟹ 12250 − 35y + 18y = 11400 ⟹ −17y = −850 ⟹ y = 50, then x = 500.

   ✦ A bat costs ₹500 and a ball ₹50.

   (iv) Taxi fare. Let the fixed charge be ₹x and the rate ₹y per km. Then x + 10y = 105 and x + 15y = 155.

   From the first, x = 105 − 10y. Substituting: 105 − 10y + 15y = 155 ⟹ 5y = 50 ⟹ y = 10, then x = 5.

   For 25 km: 5 + 25(10) = 255.

   ✦ Fixed charge ₹5, rate ₹10/km, and 25 km costs ₹255.

   (v) The fraction. Let it be x/y. Adding 2 to both parts gives (x + 2)/(y + 2) = 9/11, so 11x + 22 = 9y + 18, that is 11x − 9y = −4.

   Adding 3 gives (x + 3)/(y + 3) = 5/6, so 6x + 18 = 5y + 15, that is 6x − 5y = −3.

   From the second, x = (5y − 3)/6. Substituting into the first: 11(5y − 3)/6 − 9y = −4 ⟹ 55y − 33 − 54y = −24 ⟹ y = 9, then x = 7.

   ✦ The fraction is 7/9.

   (vi) Jacob and his son. Let Jacob be x years old now and his son y.

   In five years: x + 5 = 3(y + 5), which simplifies to x − 3y = 10.

   Five years ago: x − 5 = 7(y − 5), which simplifies to x − 7y = −30.

   From the first, x = 3y + 10. Substituting: 3y + 10 − 7y = −30 ⟹ −4y = −40 ⟹ y = 10, then x = 40.

   ✦ Jacob is 40 and his son is 10.

   Where students slip. In (vi), writing the age relations as x + 5 = 3y + 5 — applying the 'five years' only to Jacob. Both people age, so the son's future age is y + 5 too. The same trap appears in (v): the 2 is added to numerator *and* denominator.

   Another way. For (v), a useful check on the final answer: 7/9 with 2 added to both parts gives 9/11 ✓ and with 3 added gives 10/12 = 5/6 ✓. Always test word-problem answers against the original sentences rather than the equations you derived.

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.