Given: A pair of linear equations in two variables:

Equation 1: $2x - 3y = 8$

Equation 2: $4x - 6y = 9$

To Find: Determine whether the given pair of linear equations is consistent or inconsistent by comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.

Step 1: Standardizing the Equations

The standard form of a linear equation in two variables is $ax + by + c = 0$. We rewrite the given equations in this form:

Equation 1: $2x - 3y - 8 = 0$

Equation 2: $4x - 6y - 9 = 0$

Step 2: Identifying Coefficients

Comparing these with the general forms $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we identify:

For Equation 1: $a_1 = 2$, $b_1 = -3$, $c_1 = -8$

For Equation 2: $a_2 = 4$, $b_2 = -6$, $c_2 = -9$

Step 3: Calculating the Ratios

We calculate the ratios of the coefficients:

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$

Ratio of $y$-coefficients: $\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$

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

Step 4: Comparing the Ratios and Applying the Consistency Condition

We observe the following relationship between the ratios:

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

Since $\frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$, the condition for parallel lines (no solution) is satisfied.

[Theorem: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and the system of equations has no solution, making it inconsistent.]

Final Answer: Since the ratios satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the given pair of linear equations is inconsistent.

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: Two linear equations in two variables:
1) $x - y + 1 = 0$
2) $3x + 2y - 12 = 0$

To Find: The coordinates of the vertices of the triangle formed by these two lines and the x-axis, and to represent the region graphically.

Step 1: Finding coordinates for the first equation $x - y + 1 = 0$
Rearranging the equation: $y = x + 1$.
If $x = -1$, then $y = 0$. Point: $(-1, 0)$
If $x = 0$, then $y = 1$. Point: $(0, 1)$
If $x = 1$, then $y = 2$. Point: $(1, 2)$

Step 2: Finding coordinates for the second equation $3x + 2y - 12 = 0$
Rearranging the equation: $2y = 12 - 3x \implies y = \frac{12 - 3x}{2}$.
If $x = 0$, then $y = 6$. Point: $(0, 6)$
If $x = 2$, then $y = 3$. Point: $(2, 3)$
If $x = 4$, then $y = 0$. Point: $(4, 0)$

Step 3: Determining the intersection point of the two lines
We solve the system of equations:
(i) $x - y = -1 \implies y = x + 1$
(ii) $3x + 2y = 12$
Substitute (i) into (ii):
$3x + 2(x + 1) = 12$
$3x + 2x + 2 = 12$
$5x = 10 \implies x = 2$
Substitute $x = 2$ into $y = x + 1$:
$y = 2 + 1 = 3$
The intersection point is $(2, 3)$.

Step 4: Identifying the vertices of the triangle
The triangle is formed by the two lines and the x-axis ($y=0$).
- The first line $x - y + 1 = 0$ intersects the x-axis at $y=0 \implies x = -1$. Vertex: $(-1, 0)$.
- The second line $3x + 2y - 12 = 0$ intersects the x-axis at $y=0 \implies 3x = 12 \implies x = 4$. Vertex: $(4, 0)$.
- The two lines intersect at $(2, 3)$. Vertex: $(2, 3)$.

Final Answer: The vertices of the triangle formed are $(-1, 0)$, $(4, 0)$, and $(2, 3)$.

Given:

• Total number of students in the Mathematics quiz = $10$.
• The number of girls is $4$ more than the number of boys.

To Find:

• The number of boys and the number of girls who took part in the quiz using the graphical method.

Step 1: Defining Variables

Let the number of girls be $x$.

Let the number of boys be $y$.

Step 2: Formulating the Linear Equations

Based on the problem statement:

Equation 1: The total number of students is $10$.

$x + y = 10$ --- (i)

Equation 2: The number of girls is $4$ more than the number of boys.

$x = y + 4$ or $x - y = 4$ --- (ii)

Step 3: Determining Coordinates for Graphical Representation

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

For Equation (i): $x + y = 10 \implies y = 10 - x$

For Equation (ii): $x - y = 4 \implies y = x - 4$

Step 4: Graphical Representation

Step 5: Solving the Equations

By observing the graph, the two lines intersect at the point $(7, 3)$.

Verification by substitution:

Substitute $x = 7$ and $y = 3$ into Equation (i): $7 + 3 = 10$ (Correct).

Substitute $x = 7$ and $y = 3$ into Equation (ii): $7 - 3 = 4$ (Correct).

Final Answer: The number of girls is 7 and the number of boys is 3.

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.

Given: A pair of linear equations in two variables:

Equation 1: $3x + 2y = 5$

Equation 2: $2x - 3y = 7$

To Find: Determine whether the given pair of linear equations is consistent or inconsistent by comparing the ratios $\frac{a_1}{a_2}$, $\frac{b_1}{b_2}$, and $\frac{c_1}{c_2}$.

Theoretical Background:

For a pair of linear equations of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point, and the system is consistent.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, and the system is consistent (dependent).
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel, and the system is inconsistent.

Step 1: Standardizing the Equations

Rewrite the equations in the standard form $ax + by + c = 0$:

Equation 1: $3x + 2y - 5 = 0$

Equation 2: $2x - 3y - 7 = 0$

Step 2: Identifying Coefficients

Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

For Equation 1: $a_1 = 3$, $b_1 = 2$, $c_1 = -5$

For Equation 2: $a_2 = 2$, $b_2 = -3$, $c_2 = -7$

Step 3: Calculating the Ratios

Ratio of coefficients of $x$: $\frac{a_1}{a_2} = \frac{3}{2}$

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

Ratio of constant terms: $\frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7}$

Step 4: Comparing the Ratios

Observe the ratios $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$:

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

Step 5: Conclusion

[Since the condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ is satisfied, the lines intersect at a single point, implying the system has a unique solution.]

Therefore, the pair of linear equations is consistent.

Final Answer: The pair of linear equations is consistent.

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.