This page provides comprehensive Pair of Linear Equations – Exercise 3.3. Free NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Exercise 3-3. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
NCERT Exercise Practice questions with detailed step-by-step solutions
(i)
$x + y = 5$ ...(1)
$2x - 3y = 4$ ...(2)
Multiply (1) by 3: $3x + 3y = 15$ ...(3)
Add (2) and (3):
$(2x - 3y) + (3x + 3y) = 4 + 15 \Rightarrow 5x = 19 \Rightarrow x = \frac{19}{5}$.
Substitute $x$ in (1): $\frac{19}{5} + y = 5 \Rightarrow y = 5 - \frac{19}{5} = \frac{6}{5}$.
✅ $x = \frac{19}{5}, y = \frac{6}{5}$(ii)
$3x + 4y = 10$ ...(1)
$2x - 2y = 2 \Rightarrow x - y = 1$ ...(2)
Multiply (2) by 4: $4x - 4y = 4$ ...(3)
Add (1) and (3):
$7x = 14 \Rightarrow x = 2$.
Put $x=2$ in (2): $2 - y = 1 \Rightarrow y = 1$.
✅ $x = 2, y = 1$(iii)
$3x - 5y = 4$ ...(1)
$9x - 2y = 7$ ...(2)
Multiply (1) by 3: $9x - 15y = 12$ ...(3)
Subtract (3) from (2):
$(9x - 2y) - (9x - 15y) = 7 - 12 \Rightarrow 13y = -5 \Rightarrow y = -\frac{5}{13}$.
Substitute $y$ in (1): $3x - 5(-\frac{5}{13}) = 4 \Rightarrow 3x + \frac{25}{13} = 4$
$3x = 4 - \frac{25}{13} = \frac{52 - 25}{13} = \frac{27}{13} \Rightarrow x = \frac{9}{13}$.
✅ $x = \frac{9}{13}, y = -\frac{5}{13}$(iv)
Simplify equations:
$\frac{x}{2} + \frac{2y}{3} = -1 \Rightarrow 3x + 4y = -6$ ...(1)
$x - \frac{y}{3} = 3 \Rightarrow 3x - y = 9$ ...(2)
Subtract (2) from (1):
$(3x + 4y) - (3x - y) = -6 - 9 \Rightarrow 5y = -15 \Rightarrow y = -3$.
Substitute $y=-3$ in (2): $3x - (-3) = 9 \Rightarrow 3x = 6 \Rightarrow x = 2$.
✅ $x = 2, y = -3$
Question 2
Form the pair of linear equations in the following problems, and find their solutions (if they exist) 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 $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?
(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?
(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.
(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
(i) Let fraction be $\frac{x}{y}$.
$\frac{x+1}{y-1} = 1 \Rightarrow x + 1 = y - 1 \Rightarrow x - y = -2$ ...(1)
$\frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x - y = 1$ ...(2)
Subtract (1) from (2):
$(2x - y) - (x - y) = 1 - (-2) \Rightarrow x = 3$.
Put $x=3$ in (1): $3 - y = -2 \Rightarrow y = 5$.
✅ Fraction is $\frac{3}{5}$.(ii) Let Nuri's age = $x$, Sonu's age = $y$.
5 years ago: $x - 5 = 3(y - 5) \Rightarrow x - 3y = -10$ ...(1)
10 years later: $x + 10 = 2(y + 10) \Rightarrow x - 2y = 10$ ...(2)
Subtract (1) from (2):
$(x - 2y) - (x - 3y) = 10 - (-10) \Rightarrow y = 20$.
Put $y=20$ in (2): $x - 40 = 10 \Rightarrow x = 50$.
✅ Nuri is 50 years old, Sonu is 20 years old.(iii) Let number be $10x + y$.
Sum of digits: $x + y = 9$ ...(1)
Condition: $9(10x + y) = 2(10y + x)$
$90x + 9y = 20y + 2x \Rightarrow 88x - 11y = 0 \Rightarrow 8x - y = 0 \Rightarrow y = 8x$ ...(2)
Substitute (2) in (1): $x + 8x = 9 \Rightarrow 9x = 9 \Rightarrow x = 1$.
$y = 8(1) = 8$.
✅ The number is 18.(iv) Let ₹ 50 notes = $x$, ₹ 100 notes = $y$.
Total notes: $x + y = 25$ ...(1)
Total value: $50x + 100y = 2000 \Rightarrow x + 2y = 40$ ...(2)
Subtract (1) from (2):
$(x + 2y) - (x + y) = 40 - 25 \Rightarrow y = 15$.
Put $y=15$ in (1): $x + 15 = 25 \Rightarrow x = 10$.
✅ 10 notes of ₹ 50 and 15 notes of ₹ 100.(v) Let fixed charge (for 3 days) = $x$, charge per day = $y$.
Saritha (7 days = 3 fixed + 4 extra): $x + 4y = 27$ ...(1)
Susy (5 days = 3 fixed + 2 extra): $x + 2y = 21$ ...(2)
Subtract (2) from (1):
$2y = 6 \Rightarrow y = 3$.
Put $y=3$ in (2): $x + 6 = 21 \Rightarrow x = 15$.
✅ Fixed charge = ₹ 15, Charge per extra day = ₹ 3.