Solution:

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:
The given equations are:
$1. x + y = 5$
$2. 2x - 3y = 4$
Multiplying the first equation by 2 to eliminate $x$:
$2(x + y) = 2(5)$
$2x + 2y = 10$
Now subtract equation 2 from the modified equation 1:
$(2x + 2y) - (2x - 3y) = 10 - 4$
$2x - 2x + 2y + 3y = 6$
$5y = 6$
$y = \dfrac{6}{5}$
Substitute $y = \dfrac{6}{5}$ into $x + y = 5$:
$x + \dfrac{6}{5} = 5$
$x = 5 - \dfrac{6}{5}$
$x = \dfrac{25}{5} - \dfrac{6}{5}$
$x = \dfrac{19}{5}$
Substitution Method:
From $x + y = 5$, express $x$ in terms of $y$:
$x = 5 - y$
Substitute this into $2x - 3y = 4$:
$2(5 - y) - 3y = 4$
$10 - 2y - 3y = 4$
$10 - 5y = 4$
$-5y = 4 - 10$
$-5y = -6$
$y = \dfrac{-6}{-5}$
$y = \dfrac{6}{5}$
Now substitute $y = \dfrac{6}{5}$ back into $x = 5 - y$:
$x = 5 - \dfrac{6}{5}$
$x = \dfrac{25}{5} - \dfrac{6}{5}$
$x = \dfrac{19}{5}$

(ii) $3x + 4y = 10$ and $2x - 2y = 2$

Elimination Method:
The given equations are:
$1. 3x + 4y = 10$
$2. 2x - 2y = 2$
To eliminate $x$, multiply the second equation by 3:
$3(2x - 2y) = 3(2)$
$6x - 6y = 6$
Now subtract equation 1 from the modified equation 2:
$(6x - 6y) - (3x + 4y) = 6 - 10$
$6x - 3x - 6y - 4y = -4$
$3x - 10y = -4$
We now have the system of equations:
$3x + 4y = 10$
$3x - 10y = -4$
Subtract the second equation from the first:
$(3x + 4y) - (3x - 10y) = 10 - (-4)$
$3x - 3x + 4y + 10y = 10 + 4$
$14y = 14$
$y = 1$
Now substitute $y = 1$ into $3x + 4y = 10$:
$3x + 4(1) = 10$
$3x + 4 = 10$
$3x = 6$
$x = 2$
Substitution Method:
From the second equation $2x - 2y = 2$, express $x$ in terms of $y$:
$2x = 2y + 2$
$x = y + 1$
Substitute $x = y + 1$ into $3x + 4y = 10$:
$3(y + 1) + 4y = 10$
$3y + 3 + 4y = 10$
$7y + 3 = 10$
$7y = 7$
$y = 1$
Substitute $y = 1$ into $x = y + 1$:
$x = 1 + 1$
$x = 2$

(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$

Elimination Method:
The given equations are:
$1. 3x - 5y - 4 = 0$
$2. 9x = 2y + 7$
Rewrite equation 1 as $3x - 5y = 4$:
$1. 3x - 5y = 4$
Rewrite equation 2 as $9x - 2y = 7$:
$2. 9x - 2y = 7$
Multiply the first equation by 3 to eliminate $x$:
$3(3x - 5y) = 3(4)$
$9x - 15y = 12$
Now subtract equation 2 from the modified equation 1:
$(9x - 15y) - (9x - 2y) = 12 - 7$
$9x - 9x - 15y + 2y = 5$
$-13y = 5$
$y = \dfrac{5}{-13}$
Substitute $y = \dfrac{5}{-13}$ into $3x - 5y = 4$:
$3x - 5\left(\dfrac{5}{-13}\right) = 4$
$3x + \dfrac{25}{13} = 4$
$3x = 4 - \dfrac{25}{13}$
$3x = \dfrac{52}{13} - \dfrac{25}{13}$
$3x = \dfrac{27}{13}$
$x = \dfrac{9}{13}$
Substitution Method:
From $3x - 5y = 4$, express $x$ in terms of $y$:
$3x = 5y + 4$
$x = \dfrac{5y + 4}{3}$
Substitute this into $9x - 2y = 7$:
$9\left(\dfrac{5y + 4}{3}\right) - 2y = 7$
$3(5y + 4) - 2y = 7$
$15y + 12 - 2y = 7$
$13y + 12 = 7$
$13y = -5$
$y = \dfrac{-5}{13}$
Substitute $y = \dfrac{-5}{13}$ into $x = \dfrac{5y + 4}{3}$:
$x = \dfrac{5\left(\dfrac{-5}{13}\right) + 4}{3}$
$x = \dfrac{\dfrac{-25}{13} + 4}{3}$
$x = \dfrac{\dfrac{-25 + 52}{13}}{3}$
$x = \dfrac{\dfrac{27}{13}}{3}$
$x = \dfrac{9}{13}$

Elimination Method:

The given system of equations is: 1) $\dfrac{x}{2} + \dfrac{2y}{3} = -1$ 2) $x + \dfrac{y}{3} = 3$ Step 1: Multiply the first equation by 6 to eliminate the fractions. $\left(6 \times \dfrac{x}{2}\right) + \left(6 \times \dfrac{2y}{3}\right) = 6 \times (-1)$ This simplifies to: $3x + 4y = -6$ Step 2: Multiply the second equation by 3 to eliminate the fraction. $\left(3 \times x\right) + \left(3 \times \dfrac{y}{3}\right) = 3 \times 3$ This simplifies to: $3x + y = 9$ Now, the system of equations is: 1) $3x + 4y = -6$ 2) $3x + y = 9$ Step 3: Subtract the second equation from the first equation to eliminate $x$. $\left(3x + 4y\right) - \left(3x + y\right) = -6 - 9$ This simplifies to: $3x - 3x + 4y - y = -15$ $3y = -15$ Step 4: Solve for $y$: $y = \dfrac{-15}{3} = -5$ Step 5: Substitute $y = -5$ into the second equation $3x + y = 9$. $3x + (-5) = 9$ $3x - 5 = 9$ $3x = 14$ $x = \dfrac{14}{3}$ Thus, the solution is: $x = \dfrac{14}{3}$, $y = -5$

Solution:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

Let the fraction be $\dfrac{x}{y}$, where $x$ is the numerator and $y$ is the denominator. According to the given condition: 1. $\dfrac{x+1}{y-1} = 1$ 2. $\dfrac{x}{y+1} = \dfrac{1}{2}$ Solving these equations will give the values of $x$ and $y$.

(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 present age be $x$ and Sonu's present age be $y$. According to the given conditions: 1. $x - 5 = 3(y - 5)$ 2. $x + 10 = 2(y + 10)$ Solving these equations will give the values of $x$ and $y$.

(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 the two-digit number be $10x + y$, where $x$ is the tens digit and $y$ is the ones digit. According to the given conditions: 1. $x + y = 9$ 2. $9(10x + y) = 2(10y + x)$ Solving these equations will give the values of $x$ and $y$.

(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.

Let the number of Rs. 50 notes be $x$ and the number of Rs. 100 notes be $y$. According to the given conditions: 1. $x + y = 25$ 2. $50x + 100y = 2000$ Solving these equations will give the values of $x$ and $y$.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge be $x$ and the charge for each extra day be $y$. According to the given conditions: 1. $x + 4y = 27$ 2. $x + 2y = 21$ Solving these equations will give the values of $x$ and $y$.