Exercise 3.2 Solutions
Pair of Linear Equations in Two Variables
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x − y = 4
From x - y = 4, we get x = 4 + y.
Substitute into the first equation: (4 + y) + y = 14 => 4 + 2y = 14 => 2y = 10 => y = 5.
Substitute y back: x = 4 + 5 = 9.
Answer: x = 9, y = 5.
(ii) s − t = 3; s/3 + t/2 = 6
From s - t = 3, we get s = 3 + t.
Substitute into the second equation: (3 + t)/3 + t/2 = 6.
Multiply by 6: 2(3 + t) + 3t = 36 => 6 + 2t + 3t = 36 => 5t = 30 => t = 6.
Substitute t back: s = 3 + 6 = 9.
Answer: s = 9, t = 6.
(iii) 3x − y = 3; 9x − 3y = 9
From 3x - y = 3, we get y = 3x - 3.
Substitute into the second equation: 9x - 3(3x - 3) = 9 => 9x - 9x + 9 = 9 => 9 = 9.
This is a true statement, which means the lines are coincident.
Answer: Infinitely many solutions.
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
Multiply by 10: 2x + 3y = 13; 4x + 5y = 23.
From 2x + 3y = 13, we get x = (13 - 3y)/2.
Substitute: 4((13 - 3y)/2) + 5y = 23 => 2(13 - 3y) + 5y = 23 => 26 - 6y + 5y = 23 => 26 - y = 23 => y = 3.
Substitute y back: x = (13 - 3*3)/2 = 4/2 = 2.
Answer: x = 2, y = 3.
(v) √2x + √3y = 0; √3x − √8y = 0
From √2x + √3y = 0, we get x = -(√3/√2)y.
Substitute: √3(-(√3/√2)y) - √8y = 0 => (-3/√2)y - 2√2y = 0.
y(-3/√2 - 2√2) = 0. Since the term in the bracket is not zero, y = 0.
Substitute y back: x = -(√3/√2)(0) = 0.
Answer: x = 0, y = 0.
(vi) (3x/2) − (5y/3) = −2; x/3 + y/2 = 13/6
Simplify: 9x - 10y = -12; 2x + 3y = 13.
From 2x + 3y = 13, we get x = (13 - 3y)/2.
Substitute: 9((13 - 3y)/2) - 10y = -12. Multiply by 2: 9(13 - 3y) - 20y = -24.
117 - 27y - 20y = -24 => 117 - 47y = -24 => 141 = 47y => y = 3.
Substitute y back: x = (13 - 3*3)/2 = 4/2 = 2.
Answer: x = 2, y = 3.
2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.
Step 1: Solve the system of equations.
From 2x + 3y = 11, we get 2x = 11 - 3y.
Substitute into the second equation: (11 - 3y) - 4y = -24 => 11 - 7y = -24 => 35 = 7y => y = 5.
Substitute y back: 2x = 11 - 3(5) = 11 - 15 = -4 => x = -2.
Step 2: Find the value of ‘m’.
Substitute x = -2 and y = 5 into y = mx + 3:
5 = m(-2) + 3 => 5 - 3 = -2m => 2 = -2m => m = -1.
Answer: x = -2, y = 5, and m = -1.
3. Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Let numbers be x and y (x > y). Equations: x - y = 26 and x = 3y.
Substitute x = 3y into the first equation: 3y - y = 26 => 2y = 26 => y = 13.
x = 3(13) = 39. Answer: The numbers are 39 and 13.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let angles be x and y (x > y). Equations: x + y = 180 and x = y + 18.
Substitute: (y + 18) + y = 180 => 2y = 162 => y = 81.
x = 81 + 18 = 99. Answer: The angles are 99° and 81°.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Let cost of a bat be x, a ball be y. Equations: 7x + 6y = 3800 and 3x + 5y = 1750.
From the second eq: x = (1750 - 5y)/3.
Substitute: 7((1750 - 5y)/3) + 6y = 3800. Multiply by 3: 7(1750 - 5y) + 18y = 11400.
12250 - 35y + 18y = 11400 => 12250 - 17y = 11400 => 850 = 17y => y = 50.
x = (1750 - 5*50)/3 = 1500/3 = 500. Answer: Bat = ₹500, Ball = ₹50.
(iv) The taxi charges… What are the fixed charges and the charge per km? How much for 25 km?
Let fixed charge be x, charge per km be y. Equations: x + 10y = 105 and x + 15y = 155.
From the first eq: x = 105 - 10y.
Substitute: (105 - 10y) + 15y = 155 => 105 + 5y = 155 => 5y = 50 => y = 10.
x = 105 - 10(10) = 5. Fixed charge is ₹5, per km charge is ₹10.
Charge for 25 km = x + 25y = 5 + 25(10) = 255. Answer: Fixed charge = ₹5, Charge/km = ₹10, Cost for 25km = ₹255.
(v) A fraction becomes 9/11 if 2 is added to both… it becomes 5/6 if 3 is added to both… Find the fraction.
Let fraction be x/y. Equations: 11(x+2) = 9(y+2) => 11x - 9y = -4 and 6(x+3) = 5(y+3) => 6x - 5y = -3.
From the second eq: x = (5y - 3)/6.
Substitute: 11((5y - 3)/6) - 9y = -4. Multiply by 6: 11(5y - 3) - 54y = -24.
55y - 33 - 54y = -24 => y - 33 = -24 => y = 9.
x = (5*9 - 3)/6 = 42/6 = 7. Answer: The fraction is 7/9.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Let Jacob’s age be x, son’s be y. Equations: x+5 = 3(y+5) => x - 3y = 10 and x-5 = 7(y-5) => x - 7y = -30.
From the first eq: x = 10 + 3y.
Substitute: (10 + 3y) - 7y = -30 => 10 - 4y = -30 => 40 = 4y => y = 10.
x = 10 + 3(10) = 40. Answer: Jacob is 40, his son is 10.
