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NCERT Class 10 Mathematics Exercise 3.2 Solutions: Pair of Linear Equations in Two Variables

Step-by-step working for every question of Exercise 3.2 in NCERT's Class 10 Mathematics book, chapter “Pair of Linear Equations in Two Variables”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 3.2 Q1

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

   1. (i) From x − y = 4, x = y + 4. Substituting in x + y = 14: 2y + 4 = 14, so y = 5 and x = 9.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (ii) From s − t = 3, s = t + 3. Substituting in s/3 + t/2 = 6 and multiplying by 6: 2(t + 3) + 3t = 36, so 5t = 30, t = 6 and s = 9.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   3. (iii) From 3x − y = 3, y = 3x − 3. Substituting in 9x − 3y = 9: 9x − 9x + 9 = 9, which is true for every x.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   4. (iii) So the two equations describe the same line: there are infinitely many solutions, all pairs with y = 3x − 3.
   5. (iv) Multiply both equations by 10: 2x + 3y = 13 and 4x + 5y = 23. From the first, x = (13 − 3y)/2.
   6. (iv) Substituting: 2(13 − 3y) + 5y = 23, so 26 − y = 23, y = 3 and x = (13 − 9)/2 = 2.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   7. (v) From √2x + √3y = 0, x = −(√3/√2)y. Substituting in √3x − √8y = 0: −(3/√2)y − 2√2y = 0.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. (v) Multiplying by √2: −3y − 4y = 0, so −7y = 0, y = 0 and then x = 0.
   9. (vi) Clear the fractions: 3x/2 − 5y/3 = −2 becomes 9x − 10y = −12, and x/3 + y/2 = 13/6 becomes 2x + 3y = 13.
   10. (vi) From 2x + 3y = 13, x = (13 − 3y)/2. Substituting: 9(13 − 3y)/2 − 10y = −12, so 117 − 27y − 20y = −24, 47y = 141, y = 3 and x = 2.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   Answer: (i) x = 9, y = 5; (ii) s = 9, t = 6; (iii) infinitely many solutions (y = 3x − 3); (iv) x = 2, y = 3; (v) x = 0, y = 0; (vi) x = 2, y = 3

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2. Ex 3.2 Q2

   Solve 2x + 3y = 11 and 2x − 4y = −24, then find m for which y = mx + 3.

   1. Subtract the second equation from the first to eliminate x: (2x + 3y) − (2x − 4y) = 11 − (−24).
      why?

      Once one variable has the same coefficient in both equations, adding or subtracting the equations removes it, leaving one equation in one unknown.

   2. Simplify the left side to get 7y and the right side to get 35, so 7y = 35.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   3. Divide by 7 to find y = 5.
   4. Substitute y = 5 into the first equation: 2x + 3(5) = 11.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   5. Simplify to 2x + 15 = 11, then subtract 15 from both sides to get 2x = −4.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   6. Divide by 2 to find x = −2.
   7. Substitute x = −2 and y = 5 into the equation y = mx + 3.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   8. This gives 5 = m(−2) + 3, which simplifies to 2 = −2m.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   9. Divide by −2 to find m = −1.

   Answer: x = -2, y = 5, m = -1

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3. Ex 3.2 Q3

   Form the equations and solve by substitution: (i) two numbers differ by 26 and one is three times the other (ii) the larger of two supplementary angles exceeds the smaller by 18° (iii) 7 bats and 6 balls cost ₹3800, 3 bats and 5 balls cost ₹1750 (iv) taxi: ₹105 for 10 km and ₹155 for 15 km — fixed charge, charge per km, and fare for 25 km (v) a fraction becomes 9/11 when 2 is added to numerator and denominator, and 5/6 when 3 is added (vi) in 5 years Jacob will be 3 times his son's age; 5 years ago he was 7 times — present ages.

   1. (i) Let the numbers be x and y. x − y = 26 and x = 3y. Substituting x = 3y into the first equation gives 3y − y = 26, so 2y = 26 and y = 13. Then x = 3 × 13 = 39.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   2. (ii) Let the angles be x and y. x + y = 180° and x − y = 18°. Adding the equations gives 2x = 198°, so x = 99°. Then y = 180° − 99° = 81°.
   3. (iii) Let the cost of a bat be x and a ball be y. 7x + 6y = 3800 and 3x + 5y = 1750. From the second equation, x = (1750 − 5y)/3. Substituting into the first gives 7(1750 − 5y)/3 + 6y = 3800. Solving yields y = 50 and x = 500.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   4. (iv) Let the fixed charge be x and the charge per km be y. x + 10y = 105 and x + 15y = 155. Subtracting the first from the second gives 5y = 50, so y = 10. Then x = 105 − 10(10) = 5. The fare for 25 km is 5 + 25(10) = 255.
   5. (v) Let the fraction be x/y. (x + 2)/(y + 2) = 9/11 gives 11x + 22 = 9y + 18, so 11x − 9y = −4; (x + 3)/(y + 3) = 5/6 gives 6x + 18 = 5y + 15, so 6x − 5y = −3. From the second, x = (5y − 3)/6; substituting gives 11(5y − 3)/6 − 9y = −4, so 55y − 33 − 54y = −24 and y = 9, x = 7. The fraction is 7/9.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. (vi) Let Jacob's present age be x and his son's be y. x + 5 = 3(y + 5) and x − 5 = 7(y − 5). Simplifying gives x − 3y = 10 and x − 7y = −30. Subtracting the second from the first gives 4y = 40, so y = 10. Then x = 3(10 + 5) − 5 = 40.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   Answer: (i) 39 and 13; (ii) 99° and 81°; (iii) bat ₹500, ball ₹50; (iv) fixed charge ₹5, ₹10 per km, ₹255 for 25 km; (v) 7/9; (vi) Jacob 40 years, son 10 years

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