Question 1
Solve the following pairs of linear equations by the elimination method and the substitution method:
(i) $x + y = 5; \quad 2x - 3y = 4$
(ii) $3x + 4y = 10; \quad 2x - 2y = 2$
(iii) $3x - 5y - 4 = 0; \quad 9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1; \quad x - \frac{y}{3} = 3$
(ii) $3x + 4y = 10; \quad 2x - 2y = 2$
(iii) $3x - 5y - 4 = 0; \quad 9x = 2y + 7$
(iv) $\frac{x}{2} + \frac{2y}{3} = -1; \quad x - \frac{y}{3} = 3$
(i) $x + y = 5$ and $2x - 3y = 4$
Multiply equation (1) by 3:
$$3x + 3y = 15 \quad \text{--- (3)}$$
Add equation (2) and equation (3):
$$(2x - 3y) + (3x + 3y) = 4 + 15 \implies 5x = 19 \implies x = \frac{19}{5}$$
Substitute $x = \frac{19}{5}$ into (1):
$$\frac{19}{5} + y = 5 \implies y = 5 - \frac{19}{5} = \frac{25 - 19}{5} = \frac{6}{5}$$
$$\therefore x = \frac{19}{5}, \quad y = \frac{6}{5}$$
(ii) $3x + 4y = 10$ and $2x - 2y = 2$
Divide equation (2) by 2:
$$x - y = 1 \implies x = y + 1 \quad \text{--- (3)}$$
Multiply (3) by 4: $4x - 4y = 4 \quad \text{--- (4)}$
Add (1) and (4): $$(3x + 4y) + (4x - 4y) = 10 + 4 \implies 7x = 14 \implies x = 2$$ Substitute $x = 2$ into (3): $$2 - y = 1 \implies y = 1$$ $$\therefore x = 2, \quad y = 1$$
Add (1) and (4): $$(3x + 4y) + (4x - 4y) = 10 + 4 \implies 7x = 14 \implies x = 2$$ Substitute $x = 2$ into (3): $$2 - y = 1 \implies y = 1$$ $$\therefore x = 2, \quad y = 1$$
(iii) $3x - 5y = 4$ and $9x - 2y = 7$
Multiply first equation by 3:
$$9x - 15y = 12 \quad \text{--- (3)}$$
Subtract (3) from $9x - 2y = 7$:
$$(9x - 2y) - (9x - 15y) = 7 - 12 \implies 13y = -5 \implies y = -\frac{5}{13}$$
Substitute $y = -\frac{5}{13}$ in first equation:
$$3x - 5\left(-\frac{5}{13}\right) = 4 \implies 3x + \frac{25}{13} = 4 \implies 3x = \frac{52 - 25}{13} = \frac{27}{13} \implies x = \frac{9}{13}$$
$$\therefore x = \frac{9}{13}, \quad y = -\frac{5}{13}$$
(iv) $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $x - \dfrac{y}{3} = 3$
Multiply both equations by their common denominators (6 and 3):
$$3x + 4y = -6 \quad \text{--- (1)}$$
$$3x - y = 9 \quad \text{--- (2)}$$
Subtract (2) from (1):
$$(3x + 4y) - (3x - y) = -6 - 9 \implies 5y = -15 \implies y = -3$$
Substitute $y = -3$ in (2):
$$3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2$$
$$\therefore x = 2, \quad y = -3$$
Answers: (i) $(\frac{19}{5}, \frac{6}{5})$; (ii) $(2, 1)$; (iii) $(\frac{9}{13}, -\frac{5}{13})$; (iv) $(2, -3)$.
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) Fraction Problem
Let the fraction be $\dfrac{x}{y}$.
Case 1: $\dfrac{x + 1}{y - 1} = 1 \implies x + 1 = y - 1 \implies x - y = -2 \quad \text{--- (1)}$
Case 2: $\dfrac{x}{y + 1} = \frac{1}{2} \implies 2x = y + 1 \implies 2x - y = 1 \quad \text{--- (2)}$
Subtract (1) from (2): $$(2x - y) - (x - y) = 1 - (-2) \implies x = 3$$ Substitute $x = 3$ in (1): $3 - y = -2 \implies y = 5$.
$$\therefore \text{The fraction is } \frac{3}{5}$$
Case 1: $\dfrac{x + 1}{y - 1} = 1 \implies x + 1 = y - 1 \implies x - y = -2 \quad \text{--- (1)}$
Case 2: $\dfrac{x}{y + 1} = \frac{1}{2} \implies 2x = y + 1 \implies 2x - y = 1 \quad \text{--- (2)}$
Subtract (1) from (2): $$(2x - y) - (x - y) = 1 - (-2) \implies x = 3$$ Substitute $x = 3$ in (1): $3 - y = -2 \implies y = 5$.
$$\therefore \text{The fraction is } \frac{3}{5}$$
(ii) Nuri and Sonu Age Problem
Let Nuri's age be $x$ years and Sonu's age be $y$ years.
5 years ago: $x - 5 = 3(y - 5) \implies x - 3y = -10 \quad \text{--- (1)}$
10 years later: $x + 10 = 2(y + 10) \implies x - 2y = 10 \quad \text{--- (2)}$
Subtract (1) from (2): $$(x - 2y) - (x - 3y) = 10 - (-10) \implies y = 20$$ From (2): $x = 2(20) + 10 = 50$.
$$\therefore \text{Nuri is } 50\text{ years old}, \quad \text{Sonu is } 20\text{ years old}$$
5 years ago: $x - 5 = 3(y - 5) \implies x - 3y = -10 \quad \text{--- (1)}$
10 years later: $x + 10 = 2(y + 10) \implies x - 2y = 10 \quad \text{--- (2)}$
Subtract (1) from (2): $$(x - 2y) - (x - 3y) = 10 - (-10) \implies y = 20$$ From (2): $x = 2(20) + 10 = 50$.
$$\therefore \text{Nuri is } 50\text{ years old}, \quad \text{Sonu is } 20\text{ years old}$$
(iii) Two-Digit Number Problem
Let tens digit be $x$ and units digit be $y$. Original number $= 10x + y$.
Reversed number $= 10y + x$.
Sum of digits: $x + y = 9 \quad \text{--- (1)}$
Condition: $9(10x + y) = 2(10y + x) \implies 90x + 9y = 20y + 2x \implies 88x - 11y = 0 \implies 8x - y = 0 \quad \text{--- (2)}$
Add (1) and (2): $$(x + y) + (8x - y) = 9 + 0 \implies 9x = 9 \implies x = 1$$ Substitute $x = 1$ in (1): $1 + y = 9 \implies y = 8$.
$$\text{Original Number} = 10(1) + 8 = 18$$ $$\therefore \text{The required number is } 18$$
Reversed number $= 10y + x$.
Sum of digits: $x + y = 9 \quad \text{--- (1)}$
Condition: $9(10x + y) = 2(10y + x) \implies 90x + 9y = 20y + 2x \implies 88x - 11y = 0 \implies 8x - y = 0 \quad \text{--- (2)}$
Add (1) and (2): $$(x + y) + (8x - y) = 9 + 0 \implies 9x = 9 \implies x = 1$$ Substitute $x = 1$ in (1): $1 + y = 9 \implies y = 8$.
$$\text{Original Number} = 10(1) + 8 = 18$$ $$\therefore \text{The required number is } 18$$
(iv) Meena Currency Notes Problem
Let the number of ₹50 notes be $x$ and ₹100 notes be $y$.
Total notes: $x + y = 25 \quad \text{--- (1)}$
Total value: $50x + 100y = 2000 \implies x + 2y = 40 \quad \text{--- (2)}$
Subtract (1) from (2): $$(x + 2y) - (x + y) = 40 - 25 \implies y = 15$$ Substitute $y = 15$ in (1): $x + 15 = 25 \implies x = 10$.
$$\therefore \text{Meena received } 10\text{ notes of } ₹50 \text{ and } 15\text{ notes of } ₹100$$
Total notes: $x + y = 25 \quad \text{--- (1)}$
Total value: $50x + 100y = 2000 \implies x + 2y = 40 \quad \text{--- (2)}$
Subtract (1) from (2): $$(x + 2y) - (x + y) = 40 - 25 \implies y = 15$$ Substitute $y = 15$ in (1): $x + 15 = 25 \implies x = 10$.
$$\therefore \text{Meena received } 10\text{ notes of } ₹50 \text{ and } 15\text{ notes of } ₹100$$
(v) Lending Library Charge
Let fixed charge for 3 days be ₹$x$ and additional charge per day be ₹$y$.
Saritha (7 days = 3 fixed + 4 extra): $x + 4y = 27 \quad \text{--- (1)}$
Susy (5 days = 3 fixed + 2 extra): $x + 2y = 21 \quad \text{--- (2)}$
Subtract (2) from (1): $$2y = 6 \implies y = 3$$ Substitute $y = 3$ in (2): $x + 2(3) = 21 \implies x + 6 = 21 \implies x = 15$.
$$\therefore \text{Fixed charge} = ₹15, \quad \text{Charge per extra day} = ₹3$$
Saritha (7 days = 3 fixed + 4 extra): $x + 4y = 27 \quad \text{--- (1)}$
Susy (5 days = 3 fixed + 2 extra): $x + 2y = 21 \quad \text{--- (2)}$
Subtract (2) from (1): $$2y = 6 \implies y = 3$$ Substitute $y = 3$ in (2): $x + 2(3) = 21 \implies x + 6 = 21 \implies x = 15$.
$$\therefore \text{Fixed charge} = ₹15, \quad \text{Charge per extra day} = ₹3$$
Solutions: (i) Fraction = $\frac{3}{5}$; (ii) Nuri = 50 yrs, Sonu = 20 yrs; (iii) Number = 18; (iv) 10 notes of ₹50, 15 notes of ₹100; (v) Fixed charge = ₹15, Extra per day = ₹3.