NCERT Solutions Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables

#### **Q1. Solve the following pair of linear equations by the elimination method and the substitution method.** **(i)** $x + y = 5$ and $2x – 3y = 4$ **(ii)** $3x + 4y = 10$ and $2x – 2y = 2$ **(iii)** $3x – 5y – 4 = 0$ and $9x = 2y + 7$ **(iv)** $\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$ #### **A1. Solution:** **(i) Solution:** **Elimination Method:** 1. $x + y = 5 \dots(1)$ 2. $2x - 3y = 4 \dots(2)$ Multiply eq(1) by 2: $2x + 2y = 10 \dots(3)$ Subtract eq(2) from eq(3): $(2x + 2y) - (2x - 3y) = 10 - 4$ $5y = 6 \Rightarrow y = \frac{6}{5}$ Substitute $y = \frac{6}{5}$ in eq(1): $x + \frac{6}{5} = 5$ $x = 5 - \frac{6}{5} = \frac{25-6}{5} = \frac{19}{5}$ **Answer: $x = \frac{19}{5}, y = \frac{6}{5}$** **(ii) Solution:** **Elimination Method:** 1. $3x + 4y = 10 \dots(1)$ 2. $2x - 2y = 2 \dots(2)$ Multiply eq(2) by 2: $4x - 4y = 4 \dots(3)$ Add eq(1) and eq(3): $(3x + 4y) + (4x - 4y) = 10 + 4$ $7x = 14 \Rightarrow x = 2$ Substitute $x=2$ in eq(2): $2(2) - 2y = 2$ $4 - 2y = 2$ $-2y = -2 \Rightarrow y = 1$ **Answer: $x = 2, y = 1$** **(iii) Solution:** **Elimination Method:** Rearrange equations: 1. $3x - 5y = 4 \dots(1)$ 2. $9x - 2y = 7 \dots(2)$ Multiply eq(1) by 3: $9x - 15y = 12 \dots(3)$ Subtract eq(3) from eq(2): $(9x - 2y) - (9x - 15y) = 7 - 12$ $13y = -5 \Rightarrow y = -\frac{5}{13}$ Substitute $y = -\frac{5}{13}$ in eq(1): $3x - 5(-\frac{5}{13}) = 4$ $3x + \frac{25}{13} = 4$ $3x = 4 - \frac{25}{13} = \frac{52-25}{13} = \frac{27}{13}$ $x = \frac{9}{13}$ **Answer: $x = \frac{9}{13}, y = -\frac{5}{13}$** **(iv) Solution:** **Elimination Method:** Simplify equations: 1. $\frac{x}{2} + \frac{2y}{3} = -1 \Rightarrow 3x + 4y = -6 \dots(1)$ 2. $x - \frac{y}{3} = 3 \Rightarrow 3x - y = 9 \dots(2)$ Subtract eq(2) from eq(1): $(3x + 4y) - (3x - y) = -6 - 9$ $5y = -15 \Rightarrow y = -3$ Substitute $y = -3$ in eq(2): $3x - (-3) = 9$ $3x + 3 = 9$ $3x = 6 \Rightarrow x = 2$ **Answer: $x = 2, y = -3$** #### **Q2. 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. #### **A2. Solution:** **(i) Solution:** Let the fraction be $\frac{x}{y}$. 1. $\frac{x+1}{y-1} = 1 \Rightarrow x + 1 = y - 1 \Rightarrow x - y = -2 \dots(1)$ 2. $\frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x - y = 1 \dots(2)$ Subtract eq(1) from eq(2): $(2x - y) - (x - y) = 1 - (-2)$ $x = 3$ Put $x=3$ in eq(1): $3 - y = -2 \Rightarrow -y = -5 \Rightarrow y = 5$ **Answer: The fraction is $\frac{3}{5}$** **(ii) Solution:** Let Nuri's age = $x$ and Sonu's age = $y$. 1. 5 years ago: $(x-5) = 3(y-5) \Rightarrow x - 3y = -10 \dots(1)$ 2. 10 years later: $(x+10) = 2(y+10) \Rightarrow x - 2y = 10 \dots(2)$ Subtract eq(1) from eq(2): $(x - 2y) - (x - 3y) = 10 - (-10)$ $y = 20$ Put $y=20$ in eq(2): $x - 2(20) = 10 \Rightarrow x - 40 = 10 \Rightarrow x = 50$ **Answer: Nuri is 50 years old, Sonu is 20 years old.** **(iii) Solution:** Let units digit = $y$ and tens digit = $x$. The number is $10x + y$. 1. Sum of digits: $x + y = 9 \dots(1)$ 2. Reversing condition: $9(10x + y) = 2(10y + x)$ $90x + 9y = 20y + 2x$ $88x - 11y = 0$ Divide by 11: $8x - y = 0 \dots(2)$ Add eq(1) and eq(2): $(x + y) + (8x - y) = 9 + 0$ $9x = 9 \Rightarrow x = 1$ Put $x=1$ in eq(1): $1 + y = 9 \Rightarrow y = 8$ Number = $10(1) + 8 = 18$ **Answer: The number is 18.** **(iv) Solution:** Let number of ₹50 notes = $x$ and ₹100 notes = $y$. 1. Total notes: $x + y = 25 \dots(1)$ 2. Total value: $50x + 100y = 2000$ Divide by 50: $x + 2y = 40 \dots(2)$ Subtract eq(1) from eq(2): $(x + 2y) - (x + y) = 40 - 25$ $y = 15$ Put $y=15$ in eq(1): $x + 15 = 25 \Rightarrow x = 10$ **Answer: 10 notes of ₹50 and 15 notes of ₹100.** **(v) Solution:** Let fixed charge (for 3 days) = $x$ and charge per extra day = $y$. 1. Saritha (7 days = 3 fixed + 4 extra): $x + 4y = 27 \dots(1)$ 2. Susy (5 days = 3 fixed + 2 extra): $x + 2y = 21 \dots(2)$ Subtract eq(2) from eq(1): $2y = 6 \Rightarrow y = 3$ Put $y=3$ in eq(2): $x + 2(3) = 21$ $x + 6 = 21 \Rightarrow x = 15$ **Answer: Fixed charge = ₹15, Charge per extra day = ₹3.**