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: 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.

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

To Find: Another linear equation in two variables such that the pair of equations represents parallel lines.

Theoretical Background:

For a pair of linear equations in two variables of the form:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

The lines are parallel if and only if the ratios of the coefficients of $x$ and $y$ are equal, but the ratio of the constant terms is different. Mathematically, the condition is:

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

Step 1: Identify the coefficients of the given equation.

Given equation: $2x + 3y - 8 = 0$

Here, $a_1 = 2$, $b_1 = 3$, and $c_1 = -8$.

Step 2: Determine the coefficients for the second equation.

To satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$, we can multiply $a_1$ and $b_1$ by a common constant $k$. Let us choose $k = 2$.

$a_2 = 2 \times 2 = 4$

$b_2 = 3 \times 2 = 6$

Now, we must ensure $\frac{c_1}{c_2} \neq \frac{a_1}{a_2}$. Since $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, we must choose $c_2$ such that $\frac{-8}{c_2} \neq \frac{1}{2}$.

If we choose $c_2 = -9$, then $\frac{-8}{-9} = \frac{8}{9}$, which is clearly not equal to $\frac{1}{2}$.

Step 3: Formulate the second equation.

Using the coefficients $a_2 = 4$, $b_2 = 6$, and $c_2 = -9$, the equation is:

$4x + 6y - 9 = 0$

Verification:

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

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

$\frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$

Since $\frac{1}{2} = \frac{1}{2} \neq \frac{8}{9}$, the condition for parallel lines is satisfied.

Final Answer: One such linear equation is 4x + 6y - 9 = 0. (Note: Many such equations are possible by choosing different values for k and c2).

Given: A pair of linear equations in two variables:

Equation 1: $9x + 3y + 12 = 0$

Equation 2: $18x + 6y + 24 = 0$

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 linear equations.

The general form of a linear equation in two variables is $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

For Equation 1 ($9x + 3y + 12 = 0$):

$a_1 = 9$, $b_1 = 3$, $c_1 = 12$

For Equation 2 ($18x + 6y + 24 = 0$):

$a_2 = 18$, $b_2 = 6$, $c_2 = 24$

Step 2: Calculate the ratios of the coefficients.

Ratio of coefficients of $x$: $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$

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

Ratio of constant terms: $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$

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

We observe that:

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

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

$\frac{c_1}{c_2} = \frac{1}{2}$

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

Step 4: Theoretical Justification.

[According to the algebraic conditions for lines in a plane]:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a single point.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (they lie on top of each other).

Since our calculated ratios satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, we conclude that the lines are coincident.

Final Answer: The lines representing the given pair of linear equations are coincident.

Given: A pair of linear equations in two variables:

Equation 1: $\frac{4}{3}x + 2y = 8$

Equation 2: $2x + 3y = 12$

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 $ax + by + c = 0$ or $ax + by = c$. We will express both equations in the form $ax + by = c$.

Equation 1: $\frac{4}{3}x + 2y = 8$

Equation 2: $2x + 3y = 12$

Step 2: Identifying Coefficients

Comparing the given equations with the general form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

For Equation 1: $a_1 = \frac{4}{3}$, $b_1 = 2$, $c_1 = 8$

For Equation 2: $a_2 = 2$, $b_2 = 3$, $c_2 = 12$

Step 3: Calculating the Ratios

Now, we calculate the ratios of the coefficients:

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{4}{3 \times 2} = \frac{4}{6} = \frac{2}{3}$

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

Ratio of constants: $\frac{c_1}{c_2} = \frac{8}{12} = \frac{8 \div 4}{12 \div 4} = \frac{2}{3}$

Step 4: Comparing the Ratios and Determining Consistency

We observe that:

$\frac{a_1}{a_2} = \frac{2}{3}$

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

$\frac{c_1}{c_2} = \frac{2}{3}$

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

[Justification: According to the algebraic condition for consistency, if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, meaning they have infinitely many solutions. A system of linear equations is defined as consistent if it has at least one solution.]

Since the ratios are equal, the pair of linear equations represents coincident lines and possesses infinitely many solutions.

Final Answer: The pair of linear equations is consistent.

Given: A pair of linear equations in two variables:

Equation 1: $5x - 3y = 11$

Equation 2: $-10x + 6y = -22$

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 $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$. We rewrite the given equations in this form:

Equation 1: $5x - 3y - 11 = 0$

Equation 2: $-10x + 6y + 22 = 0$

Step 2: Identifying the Coefficients

By comparing the equations with the standard form, we identify the coefficients:

For Equation 1: $a_1 = 5$, $b_1 = -3$, $c_1 = -11$

For Equation 2: $a_2 = -10$, $b_2 = 6$, $c_2 = 22$

Step 3: Calculating the Ratios

We calculate the ratios of the coefficients as follows:

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2}$

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

Ratio of constants: $\frac{c_1}{c_2} = \frac{-11}{22} = -\frac{1}{2}$

Step 4: Comparing the Ratios and Determining Consistency

We observe that:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$

[Since the ratios are equal, the lines represented by these equations are coincident.]

[According to the algebraic condition for consistency: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system has infinitely many solutions and is considered consistent (specifically, dependent 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 intersecting lines.

Theoretical Context:

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 intersecting if and only if the ratio of the coefficients of $x$ and $y$ are not equal, i.e.,

$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

Step 1: Identify the coefficients of the given equation.

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

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

$a_1 = 2$

$b_1 = 3$

$c_1 = -8$

Step 2: Formulate the condition for intersecting lines.

We need to choose $a_2$ and $b_2$ such that:

$\frac{2}{a_2} \neq \frac{3}{b_2}$

This implies $2b_2 \neq 3a_2$.

Step 3: Select arbitrary values for $a_2$ and $b_2$ satisfying the condition.

Let us choose $a_2 = 3$ and $b_2 = 2$.

Checking the condition: $\frac{a_1}{a_2} = \frac{2}{3}$ and $\frac{b_1}{b_2} = \frac{3}{2}$.

Since $\frac{2}{3} \neq \frac{3}{2}$, the condition for intersecting lines is satisfied.

Step 4: Write the final equation.

We can choose any constant $c_2$. Let $c_2 = -7$.

Substituting these values into the general form $a_2x + b_2y + c_2 = 0$, we get:

$3x + 2y - 7 = 0$

Verification:

For the pair $2x + 3y - 8 = 0$ and $3x + 2y - 7 = 0$:

$\frac{a_1}{a_2} = \frac{2}{3}$

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

Since $\frac{2}{3} \neq \frac{3}{2}$, the lines intersect at a unique point.

Final Answer: One such equation is $3x + 2y - 7 = 0$ (Note: Many other solutions are possible, such as $x + y - 1 = 0$).

Given: A pair of linear equations in two variables:

Equation 1: $6x - 3y + 10 = 0$

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

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 linear equations.

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

For Equation 1 ($6x - 3y + 10 = 0$):

$a_1 = 6$, $b_1 = -3$, $c_1 = 10$

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

$a_2 = 2$, $b_2 = -1$, $c_2 = 9$

Step 2: Calculate the ratios of the coefficients.

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{6}{2} = 3$

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

Ratio of constant terms: $\frac{c_1}{c_2} = \frac{10}{9}$

Step 3: Compare the ratios and apply the geometric interpretation criteria.

We observe the following relationship between the calculated ratios:

$\frac{a_1}{a_2} = 3$

$\frac{b_1}{b_2} = 3$

$\frac{c_1}{c_2} = \frac{10}{9}$

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

Justification: According to the algebraic conditions for the graphical representation of a pair of linear equations:

• If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a single point.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.
• If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.

[Since our calculated ratios satisfy the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines do not meet at any point.]

Final Answer: The lines representing the given pair of linear equations are parallel.

Given:

1. The garden is rectangular in shape.

2. The length of the garden is $4\text{ m}$ more than its width.

3. The half-perimeter of the garden is $36\text{ m}$.

To Find:

The dimensions (length and width) of the rectangular garden.

Step 1: Defining Variables

Let the width of the rectangular garden be $w$ meters.

Let the length of the rectangular garden be $l$ meters.

Step 2: Formulating the Equations

Based on the given conditions:

Condition 1: The length is $4\text{ m}$ more than the width.

$l = w + 4$ --- (Equation 1)

Condition 2: The half-perimeter of the rectangle is $36\text{ m}$.

The formula for the perimeter of a rectangle is $P = 2(l + w)$.

Therefore, the half-perimeter is $\frac{P}{2} = l + w$.

$l + w = 36$ --- (Equation 2)

Step 3: Solving the System of Equations

We have a system of two linear equations in two variables:

1) $l - w = 4$

2) $l + w = 36$

Substitute Equation 1 into Equation 2 [Using the Substitution Method]:

$(w + 4) + w = 36$

$2w + 4 = 36$

Subtract $4$ from both sides [Addition Property of Equality]:

$2w = 36 - 4$

$2w = 32$

Divide both sides by $2$ [Division Property of Equality]:

$w = \frac{32}{2}$

$w = 16$

Step 4: Finding the Length

Substitute the value of $w = 16$ back into Equation 1:

$l = 16 + 4$

$l = 20$

Step 5: Verification

Check the conditions:

Length ($20$) is $4$ more than width ($16$): $20 - 16 = 4$ (Correct).

Half-perimeter: $l + w = 20 + 16 = 36$ (Correct).

Final Answer: The dimensions of the garden are length = 20 m and width = 16 m.

Given: A pair of linear equations in two variables:

Equation 1: $x - y = 8$

Equation 2: $3x - 3y = 16$

To Find: Determine whether the given pair of linear equations is consistent or inconsistent. If consistent, obtain the solution graphically.

Step 1: Standardizing the Equations

A pair of linear equations in the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ can be analyzed by comparing the ratios of their coefficients.

Rewrite the equations in standard form:

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

Equation 2: $3x - 3y - 16 = 0$

Step 2: Comparing Ratios of Coefficients

Identify the coefficients:

$a_1 = 1, b_1 = -1, c_1 = -8$

$a_2 = 3, b_2 = -3, c_2 = -16$

Calculate the ratios:

$\frac{a_1}{a_2} = \frac{1}{3}$

$\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$

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

Step 3: Analyzing Consistency

[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 has no solution (inconsistent).]

Since $\frac{1}{3} = \frac{1}{3} \neq \frac{1}{2}$, the condition for inconsistency is satisfied.

Visual Representation (Conceptual):

Step 4: Conclusion

Because the ratios of the coefficients of $x$ and $y$ are equal, but not equal to the ratio of the constant terms, the lines represented by these equations are parallel to each other. Parallel lines never intersect, meaning there is no common point that satisfies both equations simultaneously.

Final Answer: The pair of linear equations is inconsistent.

Given: A pair of linear equations in two variables:

Equation (1): $2x - 2y - 2 = 0$

Equation (2): $4x - 4y - 5 = 0$

To Find: Determine whether the given pair of linear equations is consistent or inconsistent. If consistent, obtain the solution graphically.

Step 1: Comparing the coefficients to determine consistency.

For a pair of linear equations of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the consistency is determined by the ratios of the coefficients:

From Equation (1): $a_1 = 2, b_1 = -2, c_1 = -2$

From Equation (2): $a_2 = 4, b_2 = -4, c_2 = -5$

Calculating the ratios:

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

$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$

Step 2: Analyzing the ratios.

We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (since $\frac{1}{2} \neq \frac{2}{5}$).

[According to the algebraic condition for lines, if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and do not intersect.]

Since the lines do not intersect, there is no common solution for the pair of equations.

Step 3: Conclusion on consistency.

A pair of linear equations is called consistent if it has at least one solution, and inconsistent if it has no solution.

Since the pair of equations $2x - 2y - 2 = 0$ and $4x - 4y - 5 = 0$ has no solution, the pair is inconsistent.

Final Answer: The pair of linear equations is inconsistent because the lines represented by the equations are parallel and have no point of intersection.

Given: A pair of linear equations in two variables:
Equation 1: $2x + y - 6 = 0$
Equation 2: $4x - 2y - 4 = 0$

To Find: Determine if the system is consistent or inconsistent. If consistent, find the solution graphically.

Step 1: Comparing coefficients to check for consistency.
For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the consistency is determined by the ratios of coefficients:
$a_1 = 2, b_1 = 1, c_1 = -6$
$a_2 = 4, b_2 = -2, c_2 = -4$

Calculate the ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a unique point. Therefore, the system is consistent.

Step 2: Finding coordinates for graphical representation.
For Equation 1 ($y = 6 - 2x$):
If $x = 0, y = 6 - 2(0) = 6 \implies (0, 6)$
If $x = 3, y = 6 - 2(3) = 0 \implies (3, 0)$

For Equation 2 ($2y = 4x - 4 \implies y = 2x - 2$):
If $x = 0, y = 2(0) - 2 = -2 \implies (0, -2)$
If $x = 1, y = 2(1) - 2 = 0 \implies (1, 0)$

Step 3: Visual Representation

Step 4: Solving the equations algebraically to verify the intersection point.
Multiply Equation 1 by 2:
$2(2x + y - 6) = 0 \implies 4x + 2y - 12 = 0$ (Equation 3)
Add Equation 3 and Equation 2:
$(4x + 2y - 12) + (4x - 2y - 4) = 0$
$8x - 16 = 0$
$8x = 16 \implies x = 2$
Substitute $x = 2$ into Equation 1:
$2(2) + y - 6 = 0$
$4 + y - 6 = 0$
$y - 2 = 0 \implies y = 2$

Final Answer: The system is consistent, and the solution is $x = 2, y = 2$.

Given: A pair of linear equations in two variables:
Equation 1: $x + y = 5$
Equation 2: $2x + 2y = 10$

To Find: Determine if the system is consistent or inconsistent. If consistent, find the solution graphically.

Step 1: Comparing the coefficients to check for consistency.
For a pair of linear equations of the form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, the consistency is determined by the ratios of the coefficients:
Equation 1: $1x + 1y - 5 = 0 \implies a_1 = 1, b_1 = 1, c_1 = -5$
Equation 2: $2x + 2y - 10 = 0 \implies a_2 = 2, b_2 = 2, c_2 = -10$

Calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$

[Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident. A system with coincident lines has infinitely many solutions and is therefore consistent.]

Step 2: Graphical Representation.
To plot the lines, we find coordinates for each equation.

Step 3: Interpretation of the Graph.
Since the equations represent the same line, every point on the line $x + y = 5$ is a solution to the system. We can express the solution set as:
$y = 5 - x$, where $x$ is any real number.

Final Answer: The system is consistent and has infinitely many solutions. The solution set is defined by the line $x + y = 5$.

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: A pair of linear equations in two variables:

Equation 1: $\frac{3}{2}x + \frac{5}{3}y = 7$

Equation 2: $9x - 10y = 14$

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

A linear equation in two variables is represented in the standard form $ax + by + c = 0$ or $ax + by = c$. We will write both equations in the form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$.

Equation 1: $\frac{3}{2}x + \frac{5}{3}y = 7$

Equation 2: $9x - 10y = 14$

Step 2: Identifying Coefficients

Comparing the given equations with $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

For Equation 1: $a_1 = \frac{3}{2}$, $b_1 = \frac{5}{3}$, $c_1 = 7$

For Equation 2: $a_2 = 9$, $b_2 = -10$, $c_2 = 14$

Step 3: Calculating the Ratios

Now, we calculate the ratios of the coefficients:

Ratio of $x$-coefficients: $\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{3}{2 \times 9} = \frac{3}{18} = \frac{1}{6}$

Ratio of $y$-coefficients: $\frac{b_1}{b_2} = \frac{5/3}{-10} = \frac{5}{3 \times (-10)} = \frac{5}{-30} = -\frac{1}{6}$

Ratio of constants: $\frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$

Step 4: Comparing the Ratios

We observe the values obtained:

$\frac{a_1}{a_2} = \frac{1}{6}$

$\frac{b_1}{b_2} = -\frac{1}{6}$

Since $\frac{1}{6} \neq -\frac{1}{6}$, it follows that $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.

Step 5: Applying the Consistency Condition

[Theorem: For a pair of linear equations $a_1x + b_1y + c_1 = 0$ 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.]

Because the condition $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ is satisfied, the pair of linear equations has a unique solution.

Final Answer: The pair of linear equations is consistent.