Exercise 3.3 Solutions
Pair of Linear Equations in Two Variables
1. Solve the following pair of linear equations by the elimination method and the substitution method.
(i) x + y = 5 and 2x − 3y = 4
Elimination Method
Multiply eq (1) by 3: 3x + 3y = 15.
Add this to eq (2): (3x + 3y) + (2x - 3y) = 15 + 4 => 5x = 19 => x = 19/5.
Substitute x in eq (1): 19/5 + y = 5 => y = 5 - 19/5 = (25-19)/5 = 6/5.
Substitution Method
From eq (1), y = 5 - x. Substitute into eq (2): 2x - 3(5-x) = 4 => 2x - 15 + 3x = 4 => 5x = 19 => x = 19/5.
y = 5 - 19/5 = 6/5.
Answer: x = 19/5, y = 6/5.
(ii) 3x + 4y = 10 and 2x − 2y = 2
Elimination Method
Multiply eq (2) by 2: 4x - 4y = 4.
Add this to eq (1): (3x + 4y) + (4x - 4y) = 10 + 4 => 7x = 14 => x = 2.
Substitute x in eq (1): 3(2) + 4y = 10 => 6 + 4y = 10 => 4y = 4 => y = 1.
Answer: x = 2, y = 1.
(iii) 3x − 5y − 4 = 0 and 9x = 2y + 7
Elimination Method
Rearrange: 3x - 5y = 4 and 9x - 2y = 7.
Multiply eq (1) by 3: 9x - 15y = 12.
Subtract eq (2): (9x - 15y) - (9x - 2y) = 12 - 7 => -13y = 5 => y = -5/13.
Substitute y in eq (1): 3x - 5(-5/13) = 4 => 3x + 25/13 = 4 => 3x = 4 - 25/13 = 27/13 => x = 9/13.
Answer: x = 9/13, y = -5/13.
(iv) x/2 + 2y/3 = −1 and x − y/3 = 3
Elimination Method
Simplify: 3x + 4y = -6 and 3x - y = 9.
Subtract eq (2) from eq (1): (3x + 4y) - (3x - y) = -6 - 9 => 5y = -15 => y = -3.
Substitute y in eq (2): 3x - (-3) = 9 => 3x + 3 = 9 => 3x = 6 => x = 2.
Answer: x = 2, y = -3.
2. Form the pair of linear equations and find their solutions by the elimination method.
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let fraction be x/y. Equations: (x+1)/(y-1) = 1 => x-y = -2. And x/(y+1) = 1/2 => 2x-y = 1.
Subtract first eq from second: (2x - y) - (x - y) = 1 - (-2) => x = 3.
Substitute x: 3 - y = -2 => y = 5. Answer: Fraction is 3/5.
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Let Nuri’s age be x, Sonu’s be y. Equations: x-5 = 3(y-5) => x-3y = -10. And x+10 = 2(y+10) => x-2y = 10.
Subtract first eq from second: (x - 2y) - (x - 3y) = 10 - (-10) => y = 20.
Substitute y: x - 2(20) = 10 => x = 50. Answer: Nuri is 50, Sonu is 20.
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let tens digit be x, units be y. Number = 10x+y. Equations: x+y = 9. And 9(10x+y) = 2(10y+x) => 88x - 11y = 0 => 8x - y = 0.
Add the two equations: (x+y) + (8x-y) = 9 + 0 => 9x = 9 => x = 1.
Substitute x: 1 + y = 9 => y = 8. Answer: The number is 18.
(iv) Meena went to a bank to withdraw ₹ 2000… She got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
Let number of ₹50 notes be x, ₹100 notes be y. Equations: x+y = 25. And 50x + 100y = 2000 => x + 2y = 40.
Subtract first eq from second: (x + 2y) - (x + y) = 40 - 25 => y = 15.
Substitute y: x + 15 = 25 => x = 10. Answer: 10 notes of ₹50 and 15 notes of ₹100.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter… Find the fixed charge and the charge for each extra day.
Let fixed charge be x, extra day charge be y. Saritha (7 days = 3 fixed + 4 extra): x + 4y = 27. Susy (5 days = 3 fixed + 2 extra): x + 2y = 21.
Subtract second eq from first: (x + 4y) - (x + 2y) = 27 - 21 => 2y = 6 => y = 3.
Substitute y: x + 2(3) = 21 => x = 15. Answer: Fixed charge is ₹15, extra day charge is ₹3.
