Given:

A lending library charges a fixed amount for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for 7 days. Susy paid ₹21 for a book kept for 5 days.

To Find:

The fixed charge for the first three days and the additional charge for each extra day.

Step 1: Defining Variables

Let the fixed charge for the first three days be $x$ (in ₹).
Let the additional charge for each extra day be $y$ (in ₹).

Step 2: Formulating the Equations

For Saritha: She kept the book for 7 days. This includes 3 fixed days and 4 extra days ($7 - 3 = 4$).
The equation is: $x + 4y = 27$ --- (Equation 1)

For Susy: She kept the book for 5 days. This includes 3 fixed days and 2 extra days ($5 - 3 = 2$).
The equation is: $x + 2y = 21$ --- (Equation 2)

Step 3: Solving by Elimination Method

To eliminate $x$, we subtract Equation 2 from Equation 1:

$(x + 4y) - (x + 2y) = 27 - 21$

$x - x + 4y - 2y = 6$

$2y = 6$

$y = \frac{6}{2}$

$y = 3$

[Since the additional charge per day is ₹3]

Step 4: Finding the value of $x$

Substitute the value of $y = 3$ into Equation 2:

$x + 2(3) = 21$

$x + 6 = 21$

$x = 21 - 6$

$x = 15$

[Since the fixed charge for the first three days is ₹15]

Step 5: Verification

Check with Equation 1: $15 + 4(3) = 15 + 12 = 27$. (Matches the given condition for Saritha).

Final Answer: The fixed charge for the first three days is ₹15 and the charge for each extra day is ₹3.

Given:

A fraction where the numerator and denominator are unknown. Let the numerator be $x$ and the denominator be $y$. The fraction is represented as $\frac{x}{y}$.

Condition 1: If we add 1 to the numerator and subtract 1 from the denominator, the fraction reduces to 1.

Condition 2: If we only add 1 to the denominator, the fraction becomes $\frac{1}{2}$.

To Find:

The value of the fraction $\frac{x}{y}$ using the elimination method.

Step 1: Formulating the Linear Equations

Based on Condition 1: $\frac{x + 1}{y - 1} = 1$

Multiplying both sides by $(y - 1)$: $x + 1 = y - 1$

Rearranging the terms to standard form $ax + by = c$: $x - y = -2$ --- (Equation 1)

Based on Condition 2: $\frac{x}{y + 1} = \frac{1}{2}$

Using cross-multiplication: $2x = 1(y + 1)$

Rearranging the terms: $2x - y = 1$ --- (Equation 2)

Step 2: Applying the Elimination Method

We have the system of equations:

(1) $x - y = -2$

(2) $2x - y = 1$

To eliminate the variable $y$, we subtract Equation 1 from Equation 2:

$(2x - y) - (x - y) = 1 - (-2)$

$2x - y - x + y = 1 + 2$ [Distributing the negative sign]

$x = 3$ [Combining like terms]

Step 3: Solving for the second variable

Substitute $x = 3$ into Equation 1:

$3 - y = -2$

$-y = -2 - 3$ [Subtracting 3 from both sides]

$-y = -5$

$y = 5$ [Multiplying both sides by -1]

Step 4: Verification

Check with Condition 1: $\frac{3+1}{5-1} = \frac{4}{4} = 1$. (Correct)

Check with Condition 2: $\frac{3}{5+1} = \frac{3}{6} = \frac{1}{2}$. (Correct)

Final Answer: The fraction is $\frac{3}{5}$.

Given: A pair of linear equations in two variables:

(1) $x + y = 5$

(2) $2x - 3y = 4$

To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.

Method 1: Substitution Method

Step 1: Express one variable in terms of the other.
From equation (1):
$x + y = 5$
$x = 5 - y$ --- (Equation 3)

Step 2: Substitute the expression into the second equation.
Substitute $x = 5 - y$ into equation (2):
$2(5 - y) - 3y = 4$ [Substituting $x$ from Eq 3 into Eq 2]

Step 3: Solve for $y$.
$10 - 2y - 3y = 4$ [Distributive property]
$10 - 5y = 4$ [Combining like terms]
$-5y = 4 - 10$ [Transposing 10 to the RHS]
$-5y = -6$
$y = \frac{-6}{-5}$
$y = \frac{6}{5}$

Step 4: Solve for $x$.
Substitute $y = \frac{6}{5}$ into equation (3):
$x = 5 - \frac{6}{5}$
$x = \frac{25 - 6}{5}$ [Finding a common denominator]
$x = \frac{19}{5}$

Method 2: Elimination Method

Step 1: Make the coefficients of one variable equal.
To eliminate $y$, multiply equation (1) by $3$ so that the coefficient of $y$ in both equations is numerically equal:
$3(x + y) = 3(5)$
$3x + 3y = 15$ --- (Equation 4)

Step 2: Add the equations to eliminate the variable.
Add equation (4) and equation (2):
$(3x + 3y) + (2x - 3y) = 15 + 4$
$3x + 2x + 3y - 3y = 19$ [Grouping like terms]
$5x = 19$
$x = \frac{19}{5}$

Step 3: Substitute the value of $x$ to find $y$.
Substitute $x = \frac{19}{5}$ into equation (1):
$\frac{19}{5} + y = 5$
$y = 5 - \frac{19}{5}$
$y = \frac{25 - 19}{5}$
$y = \frac{6}{5}$

Final Answer: The solution to the system of equations is $x = \frac{19}{5}$ and $y = \frac{6}{5}$.

Given:

1. A two-digit number where the sum of its digits is $9$.

2. Nine times the original number is equal to twice the number obtained by reversing the digits.

To Find:

The original two-digit number.

Step 1: Defining the Variables

Let the digit at the tens place be $x$ and the digit at the units place be $y$.

Since it is a two-digit number, the value of the number can be expressed as:

Original Number $= 10x + y$

When the digits are reversed, the new tens digit becomes $y$ and the new units digit becomes $x$.

Reversed Number $= 10y + x$

Step 2: Formulating the Equations

According to the first condition, the sum of the digits is $9$:

$x + y = 9$ --- (Equation 1)

According to the second condition, nine times the original number is twice the reversed number:

$9(10x + y) = 2(10y + x)$

$90x + 9y = 20y + 2x$

$90x - 2x + 9y - 20y = 0$

$88x - 11y = 0$

Dividing the entire equation by $11$ to simplify:

$8x - y = 0$ --- (Equation 2)

Step 3: Solving by the Elimination Method

We have the system of equations:

(1) $x + y = 9$

(2) $8x - y = 0$

To eliminate $y$, we add Equation 1 and Equation 2:

$(x + y) + (8x - y) = 9 + 0$

$x + 8x + y - y = 9$

$9x = 9$

$x = \frac{9}{9}$

$x = 1$

Step 4: Finding the value of $y$

Substitute $x = 1$ into Equation 1:

$1 + y = 9$

$y = 9 - 1$

$y = 8$

Step 5: Determining the Number

The tens digit $x = 1$ and the units digit $y = 8$.

Original Number $= 10x + y = 10(1) + 8 = 18$.

Verification:

Sum of digits: $1 + 8 = 9$ (Satisfied).

Nine times the number: $9 \times 18 = 162$.

Twice the reversed number: $2 \times 81 = 162$.

Since $162 = 162$, the solution is correct.

Final Answer: The two-digit number is 18.

Given: A pair of linear equations in two variables:

(1) $3x + 4y = 10$

(2) $2x - 2y = 2$

To Find: The values of $x$ and $y$ using both the Elimination Method and the Substitution Method.

Method 1: Elimination Method

Step 1: Align the coefficients. To eliminate $y$, we multiply equation (2) by $2$ so that the coefficients of $y$ in both equations have the same magnitude but opposite signs.

Equation (2) $\times 2$: $2(2x - 2y) = 2(2) \implies 4x - 4y = 4$ --- (3)

Step 2: Add the equations. Add equation (1) and equation (3) to eliminate $y$.

$(3x + 4y) + (4x - 4y) = 10 + 4$

$3x + 4x + 4y - 4y = 14$

$7x = 14$

Step 3: Solve for $x$.

$x = \frac{14}{7} = 2$

Step 4: Substitute $x$ into equation (2) to find $y$.

$2(2) - 2y = 2$

$4 - 2y = 2$

$-2y = 2 - 4$

$-2y = -2$

$y = 1$

Method 2: Substitution Method

Step 1: Express one variable in terms of the other. From equation (2):

$2x - 2y = 2$

Divide by 2: $x - y = 1 \implies x = y + 1$ --- (4)

Step 2: Substitute equation (4) into equation (1).

$3(y + 1) + 4y = 10$

Step 3: Solve for $y$.

$3y + 3 + 4y = 10$ [Using the Distributive Property]

$7y + 3 = 10$

$7y = 10 - 3$

$7y = 7$

$y = 1$

Step 4: Substitute $y = 1$ back into equation (4) to find $x$.

$x = 1 + 1$

$x = 2$

Verification:

Substitute $x=2, y=1$ into equation (1): $3(2) + 4(1) = 6 + 4 = 10$ (Correct).

Substitute $x=2, y=1$ into equation (2): $2(2) - 2(1) = 4 - 2 = 2$ (Correct).

Final Answer: $x = 2$ and $y = 1$

Given:

1. Five years ago, Nuri's age was thrice the age of Sonu.

2. Ten years later, Nuri's age will be twice the age of Sonu.

To Find:

The present ages of Nuri and Sonu.

Step 1: Defining Variables

Let the present age of Nuri be $x$ years.

Let the present age of Sonu be $y$ years.

Step 2: Formulating the Equations

Condition 1: Five years ago

Nuri's age was $(x - 5)$ and Sonu's age was $(y - 5)$.

According to the problem: $(x - 5) = 3(y - 5)$

$x - 5 = 3y - 15$

$x - 3y = -15 + 5$

$x - 3y = -10$ --- (Equation 1)

Condition 2: Ten years later

Nuri's age will be $(x + 10)$ and Sonu's age will be $(y + 10)$.

According to the problem: $(x + 10) = 2(y + 10)$

$x + 10 = 2y + 20$

$x - 2y = 20 - 10$

$x - 2y = 10$ --- (Equation 2)

Step 3: Solving by Elimination Method

We have the system of equations:

(1) $x - 3y = -10$

(2) $x - 2y = 10$

To eliminate $x$, subtract Equation (1) from Equation (2):

$(x - 2y) - (x - 3y) = 10 - (-10)$

$x - 2y - x + 3y = 10 + 10$

$y = 20$

Step 4: Finding the value of $x$

Substitute $y = 20$ into Equation (2):

$x - 2(20) = 10$

$x - 40 = 10$

$x = 10 + 40$

$x = 50$

Step 5: Verification

Five years ago: Nuri was $50 - 5 = 45$, Sonu was $20 - 5 = 15$. Since $45 = 3 \times 15$, the first condition is satisfied.

Ten years later: Nuri will be $50 + 10 = 60$, Sonu will be $20 + 10 = 30$. Since $60 = 2 \times 30$, the second condition is satisfied.

Final Answer: The present age of Nuri is 50 years and the present age of Sonu is 20 years.

Given:

• Total amount withdrawn by Meena = ₹ 2000
• Types of notes received = ₹ 50 notes and ₹ 100 notes
• Total number of notes received = 25

To Find:

The number of ₹ 50 notes and the number of ₹ 100 notes received by Meena.

Step 1: Defining Variables

Let $x$ be the number of ₹ 50 notes.

Let $y$ be the number of ₹ 100 notes.

Step 2: Formulating the Equations

Based on the total number of notes:

$x + y = 25$ --- (Equation 1)

Based on the total value of the notes:

$50x + 100y = 2000$ --- (Equation 2)

To simplify Equation 2, we divide the entire equation by 50:

$\frac{50x}{50} + \frac{100y}{50} = \frac{2000}{50}$

$x + 2y = 40$ --- (Equation 3)

Step 3: Solving by Elimination Method

We have the system of equations:

(1) $x + y = 25$

(3) $x + 2y = 40$

Subtract Equation 1 from Equation 3 to eliminate the variable $x$:

$(x + 2y) - (x + y) = 40 - 25$

$x - x + 2y - y = 15$

$y = 15$

Step 4: Finding the value of $x$

Substitute the value of $y = 15$ into Equation 1:

$x + 15 = 25$

$x = 25 - 15$

$x = 10$

Step 5: Verification

Check the total number of notes: $10 + 15 = 25$ (Correct).

Check the total value: $50(10) + 100(15) = 500 + 1500 = 2000$ (Correct).

Final Answer:

Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Given: A pair of linear equations in two variables:

(i) $3x - 5y - 4 = 0$

(ii) $9x = 2y + 7$

To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.

Part 1: Substitution Method

Step 1: Standardize the equations.

Equation (i): $3x - 5y = 4$ --- (1)

Equation (ii): $9x - 2y = 7$ --- (2)

Step 2: Express one variable in terms of the other.

From equation (1), isolate $3x$:

$3x = 5y + 4$

$x = \frac{5y + 4}{3}$ --- (3)

Step 3: Substitute equation (3) into equation (2).

$9\left(\frac{5y + 4}{3}\right) - 2y = 7$

[Since $9/3 = 3$, we simplify the expression]

$3(5y + 4) - 2y = 7$

$15y + 12 - 2y = 7$ [Distributive property]

$13y + 12 = 7$ [Combining like terms]

$13y = 7 - 12$

$13y = -5$

$y = -\frac{5}{13}$

Step 4: Solve for $x$ by substituting $y$ into equation (3).

$x = \frac{5(-\frac{5}{13}) + 4}{3}$

$x = \frac{-\frac{25}{13} + 4}{3}$

$x = \frac{\frac{-25 + 52}{13}}{3}$ [Finding common denominator]

$x = \frac{27}{13 \times 3} = \frac{9}{13}$

Part 2: Elimination Method

Step 1: Align the equations.

(1) $3x - 5y = 4$

(2) $9x - 2y = 7$

Step 2: Make the coefficients of $x$ equal.

Multiply equation (1) by 3 to match the $x$-coefficient of equation (2):

$3(3x - 5y) = 3(4)$

$9x - 15y = 12$ --- (4)

Step 3: Eliminate $x$ by subtracting equation (4) from equation (2).

$(9x - 2y) - (9x - 15y) = 7 - 12$

$9x - 2y - 9x + 15y = -5$

$13y = -5$

$y = -\frac{5}{13}$

Step 4: Substitute $y$ into equation (1) to find $x$.

$3x - 5(-\frac{5}{13}) = 4$

$3x + \frac{25}{13} = 4$

$3x = 4 - \frac{25}{13}$

$3x = \frac{52 - 25}{13}$

$3x = \frac{27}{13}$

$x = \frac{27}{13 \times 3} = \frac{9}{13}$

Final Answer: $x = \frac{9}{13}, y = -\frac{5}{13}$

Given: A pair of linear equations in two variables:

(1) $\frac{x}{2} + \frac{2y}{3} = -1$

(2) $x - \frac{y}{3} = 3$

To Find: The values of $x$ and $y$ using both the Substitution Method and the Elimination Method.

Part 1: Substitution Method

Step 1: Simplify the given equations.

For equation (1): $\frac{x}{2} + \frac{2y}{3} = -1$. Multiplying the entire equation by the Least Common Multiple (LCM) of 2 and 3, which is 6:

$6 \cdot (\frac{x}{2}) + 6 \cdot (\frac{2y}{3}) = 6 \cdot (-1)$

$3x + 4y = -6$ --- (Equation 3)

For equation (2): $x - \frac{y}{3} = 3$. Multiplying the entire equation by 3:

$3 \cdot (x) - 3 \cdot (\frac{y}{3}) = 3 \cdot (3)$

$3x - y = 9$ --- (Equation 4)

Step 2: Express one variable in terms of the other.

From Equation (4), we can isolate $y$:

$y = 3x - 9$ --- (Equation 5)

Step 3: Substitute Equation (5) into Equation (3).

$3x + 4(3x - 9) = -6$

$3x + 12x - 36 = -6$ [Distributive property]

$15x - 36 = -6$ [Combining like terms]

$15x = 30$ [Adding 36 to both sides]

$x = \frac{30}{15} = 2$

Step 4: Find the value of $y$.

Substitute $x = 2$ into Equation (5):

$y = 3(2) - 9$

$y = 6 - 9 = -3$

Part 2: Elimination Method

Step 1: Align the simplified equations.

(3) $3x + 4y = -6$

(4) $3x - y = 9$

Step 2: Eliminate one variable.

Since the coefficients of $x$ are identical ($3$), we subtract Equation (4) from Equation (3):

$(3x + 4y) - (3x - y) = -6 - 9$

$3x + 4y - 3x + y = -15$ [Distributing the negative sign]

$5y = -15$ [Combining like terms]

$y = \frac{-15}{5} = -3$

Step 3: Solve for the remaining variable.

Substitute $y = -3$ into Equation (4):

$3x - (-3) = 9$

$3x + 3 = 9$

$3x = 9 - 3$

$3x = 6$

$x = \frac{6}{3} = 2$

Final Answer: The solution to the system of equations is $x = 2$ and $y = -3$.