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 linear equation in two variables, $2x + 3y - 8 = 0$.

To Find: Another linear equation in two variables, $a_2x + b_2y + c_2 = 0$, such that the pair of linear equations represents coincident lines.

Theoretical Background:

For a pair of linear equations in two variables given by:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

The lines are coincident if and only if the ratios of their coefficients are equal, satisfying the condition:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Step 1: Identify the coefficients of the given equation.

The given equation is $2x + 3y - 8 = 0$.

Comparing this with the standard form $a_1x + b_1y + c_1 = 0$, we have:

$a_1 = 2$

$b_1 = 3$

$c_1 = -8$

Step 2: Apply the condition for coincident lines.

To obtain a coincident line, we can multiply the entire equation by a non-zero constant $k$. Let us choose $k = 2$ for simplicity.

The new coefficients will be:

$a_2 = k \cdot a_1 = 2 \cdot 2 = 4$

$b_2 = k \cdot b_1 = 2 \cdot 3 = 6$

$c_2 = k \cdot c_1 = 2 \cdot (-8) = -16$

Step 3: Formulate the new equation.

Substituting the values of $a_2, b_2,$ and $c_2$ into the standard form $a_2x + b_2y + c_2 = 0$:

$4x + 6y - 16 = 0$

Step 4: Verification of the condition.

Check the ratios:

$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$

[Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.]

Final Answer: One such linear equation is 4x + 6y - 16 = 0.

Given: A pair of linear equations in two variables:

(1) $3x - y = 3$

(2) $9x - 3y = 9$

To Find: The solution $(x, y)$ for the given pair of linear equations using the substitution method.

Step 1: Express one variable in terms of the other using Equation (1).

From Equation (1):

$3x - y = 3$

Add $y$ to both sides:

$3x = 3 + y$

Subtract $3$ from both sides:

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

Step 2: Substitute the expression for $y$ into Equation (2).

Equation (2) is $9x - 3y = 9$.

Substitute $y = 3x - 3$ into Equation (2):

$9x - 3(3x - 3) = 9$

Step 3: Solve the resulting equation.

Distribute the $-3$ across the terms inside the parentheses:

$9x - 9x + 9 = 9$

Combine the $x$ terms ($9x - 9x = 0$):

$0 + 9 = 9$

$9 = 9$

Step 4: Interpret the result.

[Since the variable $x$ has been eliminated and we have arrived at a true statement ($9 = 9$), this indicates that the two equations are dependent and represent the same line.]

Specifically, if we divide Equation (2) by $3$:

$\frac{9x}{3} - \frac{3y}{3} = \frac{9}{3}$

$3x - y = 3$

This is identical to Equation (1). Therefore, the pair of linear equations has infinitely many solutions.

Step 5: General form of the solution.

Since the equations are coincident, any value of $x$ will yield a corresponding value of $y$ that satisfies both equations. We can express the solution set as:

$y = 3x - 3$ for any real number $x$.

Final Answer: The pair of linear equations has infinitely many solutions, represented by the relation $y = 3x - 3$.

Given: A pair of linear equations in two variables:

Equation 1: $5x - 4y + 8 = 0$

Equation 2: $7x + 6y - 9 = 0$

To Find: Determine whether the lines representing these equations intersect at a point, are parallel, or are coincident by comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.

Step 1: Identify the coefficients of the given equations.

The standard form of a linear equation in two variables is $ax + by + c = 0$.

For Equation 1 ($5x - 4y + 8 = 0$):

$a_1 = 5$

$b_1 = -4$

$c_1 = 8$

For Equation 2 ($7x + 6y - 9 = 0$):

$a_2 = 7$

$b_2 = 6$

$c_2 = -9$

Step 2: Calculate the ratios of the coefficients.

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{5}{7}$

Ratio of $y$-coefficients: $\frac{b_1}{b_2} = \frac{-4}{6} = -\frac{2}{3}$

Ratio of constants: $\frac{c_1}{c_2} = \frac{8}{-9} = -\frac{8}{9}$

Step 3: Compare the ratios and apply the geometric conditions.

We observe that $\frac{a_1}{a_2} = \frac{5}{7}$ and $\frac{b_1}{b_2} = -\frac{2}{3}$.

Since $\frac{5}{7} \neq -\frac{2}{3}$, it follows that $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Theoretical Justification:

According to the theory of linear equations in two variables:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point (consistent system).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (infinitely many solutions).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel (no solution).

Step 4: Conclusion.

Since the condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ is satisfied, the lines representing the given pair of linear equations intersect at a single point.

Final Answer: The lines representing the equations $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$ intersect at a point.

Given:

1. The cost of 5 pencils and 7 pens together is ₹ 50.

2. The cost of 7 pencils and 5 pens together is ₹ 46.

To Find:

The cost of one pencil and the cost of one pen.

Step 1: Defining Variables

Let the cost of one pencil be $x$ and the cost of one pen be $y$.

Step 2: Formulating the Equations

Based on the given conditions, we can write the following system of linear equations:

Equation 1: $5x + 7y = 50$

Equation 2: $7x + 5y = 46$

Step 3: Finding Coordinates for Graphical Representation

To plot these lines, we find at least two points for each equation.

For Equation 1: $5x + 7y = 50 \implies y = \frac{50 - 5x}{7}$

For Equation 2: $7x + 5y = 46 \implies y = \frac{46 - 7x}{5}$

Step 4: Visual Representation

Step 5: Solving the System

From the table of values, we observe that the point $(3, 5)$ satisfies both equations.

Verification for Equation 1: $5(3) + 7(5) = 15 + 35 = 50$ (Correct)

Verification for Equation 2: $7(3) + 5(5) = 21 + 25 = 46$ (Correct)

Step 6: Conclusion

Since the lines intersect at the point $(3, 5)$, the solution to the system is $x = 3$ and $y = 5$.

Final Answer: The cost of one pencil is ₹ 3 and the cost of one pen is ₹ 5.