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NCERT Class 10 Mathematics Exercise 3.3 Solutions: Pair of Linear Equations in Two Variables

Step-by-step working for every question of Exercise 3.3 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 3.3 Q1

   Solve by elimination and by substitution: (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.

   1. (i) Multiply the first equation by 3 to get 3x + 3y = 15.
   2. (i) Add this to the second equation (2x − 3y = 4) to eliminate y, giving 5x = 19, so x = 19/5.
      why?

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

   3. (i) Substitute x = 19/5 into x + y = 5 to find y = 5 − 19/5 = 6/5.
      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. (ii) Multiply the second equation by 2 to get 4x − 4y = 4.
   5. (ii) Add this to the first equation (3x + 4y = 10) to eliminate y, giving 7x = 14, so x = 2.
      why?

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

   6. (ii) Substitute x = 2 into 2x − 2y = 2 to find 4 − 2y = 2, so y = 1.
      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. (iii) Rewrite the second equation as 9x − 2y = 7.
   8. (iii) Multiply the first equation (3x − 5y = 4) by 3 to get 9x − 15y = 12.
   9. (iii) Subtract the modified first equation from the second to eliminate x, giving 13y = −5, so y = −5/13.
      why?

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

   10. (iii) Substitute y = −5/13 into 3x − 5y = 4 to find 3x + 25/13 = 4, so x = 9/13.
       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.

   11. (iv) Multiply the first equation by 6 to get 3x + 4y = −6.
   12. (iv) Multiply the second equation by 3 to get 3x − y = 9.
   13. (iv) Subtract the second modified equation from the first to eliminate x, giving 5y = −15, so y = −3.
       why?

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

   14. (iv) Substitute y = −3 into x − y/3 = 3 to find x + 1 = 3, so 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 = 19/5, y = 6/5; (ii) x = 2, y = 1; (iii) x = 9/13, y = -5/13; (iv) x = 2, y = -3

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2. Ex 3.3 Q2

   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 1/2 if 1 is added to the denominator only (ii) 5 years ago Nuri was thrice Sonu's age; in 10 years she will be twice his age (iii) a two-digit number's digits add to 9, and 9 times the number is twice the reversed number (iv) Meena got 25 notes of ₹50 and ₹100 for ₹2000 (v) a library charges a fixed amount for 3 days plus a daily charge: ₹27 for 7 days, ₹21 for 5 days.

   1. (i) Let the fraction be x/y. Adding 1 to the numerator and subtracting 1 from the denominator gives 1: x + 1 = y − 1, so x − y = −2.
   2. (i) Adding 1 to the denominator gives 1/2: 2x = y + 1, so 2x − y = 1. Subtracting the first equation from this: x = 3, then y = 5. The fraction is 3/5.
   3. (ii) Let Nuri's age be x and Sonu's be y. Five years ago: x − 5 = 3(y − 5), so x − 3y = −10. In ten years: x + 10 = 2(y + 10), so x − 2y = 10.
   4. (ii) Subtracting the first from the second: y = 20, then x = 10 + 2 × 20 = 50.
   5. (iii) Let the tens digit be x and the units digit y, so the number is 10x + y and x + y = 9.
   6. (iii) 9(10x + y) = 2(10y + x) gives 88x − 11y = 0, so y = 8x. Then x + 8x = 9 gives x = 1, y = 8, and the number is 18.
   7. (iv) Let x notes be of ₹50 and y of ₹100: x + y = 25 and 50x + 100y = 2000, i.e. x + 2y = 40.
   8. (iv) Subtracting x + y = 25 from x + 2y = 40: y = 15, then x = 10.
   9. (v) Let the fixed charge (first 3 days) be ₹x and the charge per extra day ₹y. 7 days = 4 extra days: x + 4y = 27; 5 days = 2 extra days: x + 2y = 21.
   10. (v) Subtracting: 2y = 6, so y = 3, and x = 21 − 6 = 15.

   Answer: (i) 3/5; (ii) Nuri 50 years, Sonu 20 years; (iii) 18; (iv) 10 notes of ₹50, 15 notes of ₹100; (v) fixed charge ₹15, ₹3 per extra day

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