Solution of a Linear Equation Questions in English

(N/A) We know that any point lying on the $x$-axis has its ordinate (y-coordinate) equal to $0$.
Putting $y = 0$ in the equation $x + 2y = 8$,we get:
$x + 2(0) = 8 \Rightarrow x = 8$.
Thus,the point on the $x$-axis is $(8, 0)$.
We also know that any point lying on the $y$-axis has its abscissa (x-coordinate) equal to $0$.
Putting $x = 0$ in the equation $x + 2y = 8$,we get:
$0 + 2y = 8 \Rightarrow 2y = 8 \Rightarrow y = 4$.
Thus,the point on the $y$-axis is $(0, 4)$.

(B) Given the linear equation: $2x + cy = 8$.
We are looking for the value of $c$ such that $x = y$.
Substitute $y = x$ into the given equation:
$2x + c(x) = 8$
$x(2 + c) = 8$
$x = \frac{8}{2 + c}$
Since the question implies a specific value for $c$ that satisfies the condition for any solution where $x=y$,we look at the structure of the equation. If we assume a specific point like $(2, 2)$ which satisfies $x=y$:
$2(2) + c(2) = 8$
$4 + 2c = 8$
$2c = 4$
$c = 2$
Thus,for $c = 2$,the equation becomes $2x + 2y = 8$,or $x + y = 4$,which clearly allows for the solution $x = y = 2$.

(A) $y$ varies directly as $x$.
$\Rightarrow y \propto x$
$\therefore y = kx$
Substituting $y = 12$ when $x = 4$,we get:
$12 = k \times 4 \Rightarrow k = 12 \div 4 = 3$
Hence,the required linear equation is $y = 3x$.
The value of $y$ when $x = 5$ is $y = 3 \times 5 = 15$.

(N/A) To show that a point $(x, y)$ lies on the graph of the linear equation $y = 9x - 7$,we must substitute the coordinates of the point into the equation and verify if the left-hand side equals the right-hand side.
For point $A(1, 2)$:
Substitute $x = 1$ and $y = 2$ into the equation:
$2 = 9(1) - 7$
$2 = 9 - 7$
$2 = 2$
Since the left-hand side equals the right-hand side,point $A$ lies on the graph.
For point $B(-1, -16)$:
Substitute $x = -1$ and $y = -16$ into the equation:
$-16 = 9(-1) - 7$
$-16 = -9 - 7$
$-16 = -16$
Since the left-hand side equals the right-hand side,point $B$ lies on the graph.
For point $C(0, -7)$:
Substitute $x = 0$ and $y = -7$ into the equation:
$-7 = 9(0) - 7$
$-7 = 0 - 7$
$-7 = -7$
Since the left-hand side equals the right-hand side,point $C$ lies on the graph.
Thus,all three points $A$,$B$,and $C$ satisfy the equation $y = 9x - 7$ and therefore lie on its graph.

(N/A) To draw the graph of the linear equation $3x + 4y = 6$,we first find some coordinate points that satisfy the equation.
$1$. To find the $x$-intercept,set $y = 0$:
$3x + 4(0) = 6 \implies 3x = 6 \implies x = 2$. So,the point is $(2, 0)$.
$2$. To find the $y$-intercept,set $x = 0$:
$3(0) + 4y = 6 \implies 4y = 6 \implies y = 1.5$. So,the point is $(0, 1.5)$.
$3$. To find another point,set $x = -2$:
$3(-2) + 4y = 6 \implies -6 + 4y = 6 \implies 4y = 12 \implies y = 3$. So,the point is $(-2, 3)$.
Summary table:

$x$	$2$	$0$	$-2$
$y$	$0$	$1.5$	$3$

Plot these points $(2, 0)$,$(0, 1.5)$,and $(-2, 3)$ on a Cartesian plane and join them to obtain a straight line.
The graph cuts the $x$-axis at $(2, 0)$ and the $y$-axis at $(0, 1.5)$.

(D) To find the linear equation,we check which equation is satisfied by all the given points: $(-2, 2), (0, 0),$ and $(2, -2)$.
For option $D$,$x + y = 0$:
$1$. For $(-2, 2): -2 + 2 = 0$. (Satisfied)
$2$. For $(0, 0): 0 + 0 = 0$. (Satisfied)
$3$. For $(2, -2): 2 + (-2) = 0$. (Satisfied)
Since all points satisfy the equation $x + y = 0$,this is the correct linear equation.

(A) The Cartesian plane is divided into four quadrants based on the signs of the coordinates $(x, y)$.
Quadrant $I$ consists of all points $(x, y)$ where both $x > 0$ and $y > 0$.
Since a positive solution implies that both $x$ and $y$ are positive,such points must satisfy the condition for the $1^{st}$ quadrant.
Therefore,the positive solutions of the equation $ax + by + c = 0$ always lie in the $1^{st}$ quadrant.

Given equation: $2x + 5y = 20$.
Rearranging for $y$: $5y = 20 - 2x$,which gives $y = \frac{20 - 2x}{5}$.
$1$. If $x = 0$,then $y = \frac{20 - 2(0)}{5} = \frac{20}{5} = 4$. So,$(0, 4)$ is a solution.
$2$. If $x = 5$,then $y = \frac{20 - 2(5)}{5} = \frac{10}{5} = 2$. So,$(5, 2)$ is a solution.
$3$. If $x = -5$,then $y = \frac{20 - 2(-5)}{5} = \frac{30}{5} = 6$. So,$(-5, 6)$ is a solution.
$4$. If $x = 10$,then $y = \frac{20 - 2(10)}{5} = \frac{0}{5} = 0$. So,$(10, 0)$ is a solution.
Thus,four solutions of the given equation are $(0, 4), (5, 2), (-5, 6),$ and $(10, 0)$.

(N/A) Given equation: $4x - 3y = 24$.
Rearranging for $y$:
$3y = 4x - 24$
$y = \frac{4x - 24}{3}$
To find four solutions,we substitute different values for $x$:
$1$. If $x = 0$,then $y = \frac{4(0) - 24}{3} = \frac{-24}{3} = -8$. So,$(0, -8)$ is a solution.
$2$. If $x = 3$,then $y = \frac{4(3) - 24}{3} = \frac{12 - 24}{3} = \frac{-12}{3} = -4$. So,$(3, -4)$ is a solution.
$3$. If $x = 6$,then $y = \frac{4(6) - 24}{3} = \frac{24 - 24}{3} = \frac{0}{3} = 0$. So,$(6, 0)$ is a solution.
$4$. If $x = 9$,then $y = \frac{4(9) - 24}{3} = \frac{36 - 24}{3} = \frac{12}{3} = 4$. So,$(9, 4)$ is a solution.
Thus,four solutions of the given equation are $(0, -8), (3, -4), (6, 0),$ and $(9, 4)$.

(N/A) Given equation: $3x + 5y = 0$.
Rearranging for $y$:
$5y = -3x$
$y = \frac{-3x}{5}$
To find solutions,we substitute different values for $x$:
$1$. If $x = 0$,then $y = \frac{-3(0)}{5} = 0$. So,$(0, 0)$ is a solution.
$2$. If $x = 5$,then $y = \frac{-3(5)}{5} = -3$. So,$(5, -3)$ is a solution.
$3$. If $x = -5$,then $y = \frac{-3(-5)}{5} = 3$. So,$(-5, 3)$ is a solution.
$4$. If $x = 10$,then $y = \frac{-3(10)}{5} = -6$. So,$(10, -6)$ is a solution.
Thus,four solutions of the given equation are $(0, 0), (5, -3), (-5, 3),$ and $(10, -6)$.

(N/A) Given equation: $2x - 12 = 0$
Step $1$: Simplify the equation to solve for $x$.
$2x = 12$
$x = 6$
Step $2$: Express the equation in the form $ax + by + c = 0$.
$1x + 0y - 6 = 0$
Step $3$: Since the coefficient of $y$ is $0$,the value of $x$ will always be $6$ regardless of the value of $y$. We can choose any four arbitrary values for $y$ to find the corresponding solutions $(x, y)$.
Let $y = 0$,then $x = 6$. Solution: $(6, 0)$
Let $y = 1$,then $x = 6$. Solution: $(6, 1)$
Let $y = 2$,then $x = 6$. Solution: $(6, 2)$
Let $y = 3$,then $x = 6$. Solution: $(6, 3)$
Thus,four solutions are $(6, 0), (6, 1), (6, 2),$ and $(6, 3)$.

(N/A) To find the solutions for the equation $2x - 5y = 10$,we can substitute different values for $x$ and solve for $y$,or vice versa.
$1$. If $x = 0$,then $2(0) - 5y = 10 \implies -5y = 10 \implies y = -2$. So,$(0, -2)$ is a solution.
$2$. If $y = 0$,then $2x - 5(0) = 10 \implies 2x = 10 \implies x = 5$. So,$(5, 0)$ is a solution.
$3$. If $x = -5$,then $2(-5) - 5y = 10 \implies -10 - 5y = 10 \implies -5y = 20 \implies y = -4$. So,$(-5, -4)$ is a solution.
$4$. If $y = 2$,then $2x - 5(2) = 10 \implies 2x - 10 = 10 \implies 2x = 20 \implies x = 10$. So,$(10, 2)$ is a solution.
Thus,four solutions are $(0, -2), (5, 0), (-5, -4), (10, 2)$.

(N/A) To find the solutions for the linear equation $2x + 3y = 7$,we can express $y$ in terms of $x$:
$3y = 7 - 2x$
$y = \frac{7 - 2x}{3}$
Now,we substitute different values of $x$ to find corresponding values of $y$:
$1$. If $x = -1$,then $y = \frac{7 - 2(-1)}{3} = \frac{7 + 2}{3} = \frac{9}{3} = 3$. Solution: $(-1, 3)$.
$2$. If $x = 2$,then $y = \frac{7 - 2(2)}{3} = \frac{7 - 4}{3} = \frac{3}{3} = 1$. Solution: $(2, 1)$.
$3$. If $x = 5$,then $y = \frac{7 - 2(5)}{3} = \frac{7 - 10}{3} = \frac{-3}{3} = -1$. Solution: $(5, -1)$.
$4$. If $x = 8$,then $y = \frac{7 - 2(8)}{3} = \frac{7 - 16}{3} = \frac{-9}{3} = -3$. Solution: $(8, -3)$.
Thus,four solutions are $(-1, 3), (2, 1), (5, -1), (8, -3)$.

(N/A) To find the solutions for the linear equation $5x + 3y = 16$,we can assign arbitrary values to $x$ and solve for $y$ using the formula $y = \frac{16 - 5x}{3}$.
$1$. If $x = 2$,then $y = \frac{16 - 5(2)}{3} = \frac{16 - 10}{3} = \frac{6}{3} = 2$. Solution: $(2, 2)$.
$2$. If $x = 5$,then $y = \frac{16 - 5(5)}{3} = \frac{16 - 25}{3} = \frac{-9}{3} = -3$. Solution: $(5, -3)$.
$3$. If $x = -1$,then $y = \frac{16 - 5(-1)}{3} = \frac{16 + 5}{3} = \frac{21}{3} = 7$. Solution: $(-1, 7)$.
$4$. If $x = -4$,then $y = \frac{16 - 5(-4)}{3} = \frac{16 + 20}{3} = \frac{36}{3} = 12$. Solution: $(-4, 12)$.

To find the solutions for the equation $3x + y = 11$,we can express $y$ in terms of $x$:
$y = 11 - 3x$
Now,we substitute different values of $x$ to find the corresponding values of $y$:
$1$. If $x = 0$,then $y = 11 - 3(0) = 11$. So,the solution is $(0, 11)$.
$2$. If $x = 1$,then $y = 11 - 3(1) = 8$. So,the solution is $(1, 8)$.
$3$. If $x = 2$,then $y = 11 - 3(2) = 5$. So,the solution is $(2, 5)$.
$4$. If $x = -1$,then $y = 11 - 3(-1) = 11 + 3 = 14$. So,the solution is $(-1, 14)$.
Thus,four solutions are $(0, 11), (1, 8), (2, 5), (-1, 14)$.

(N/A) The given equation is $4y - 11 = 0$,which can be written as $0x + 4y = 11$.
This is a linear equation in two variables $x$ and $y$.
Since the coefficient of $x$ is $0$,the value of $y$ will always be $\frac{11}{4}$ regardless of the value of $x$.
We can choose any four arbitrary values for $x$ to find the corresponding solutions.
Let $x = 0$,then $y = \frac{11}{4}$. Solution: $(0, 2.75)$.
Let $x = 1$,then $y = \frac{11}{4}$. Solution: $(1, 2.75)$.
Let $x = 2$,then $y = \frac{11}{4}$. Solution: $(2, 2.75)$.
Let $x = 3$,then $y = \frac{11}{4}$. Solution: $(3, 2.75)$.
Thus,four solutions are $(0, 2.75), (1, 2.75), (2, 2.75), (3, 2.75)$.

(N/A) The given equation is $3x - 24 = 0$.
This can be simplified as $3x = 24$, which gives $x = 8$.
Since the equation is in the form $x = 8$, it can be written as a linear equation in two variables as $1x + 0y = 8$.
For any value of $y$, the value of $x$ will always be $8$.
Therefore, four possible solutions are $(8, 0), (8, 1), (8, 2), \text{ and } (8, 3)$.

(N/A) To check if a point $(x, y)$ is a solution to the equation $3x - 2y = 12$,we substitute the values of $x$ and $y$ into the equation and check if the left-hand side $(LHS)$ equals the right-hand side ($RHS$ = $12$).
$(1) (0, -6): 3(0) - 2(-6) = 0 + 12 = 12$. Since $LHS$ = $RHS$,$(0, -6)$ is a solution.
$(2) (2, 3): 3(2) - 2(3) = 6 - 6 = 0 \neq 12$. Not a solution.
$(3) (2, -3): 3(2) - 2(-3) = 6 + 6 = 12$. Since $LHS$ = $RHS$,$(2, -3)$ is a solution.
$(4) (-4, 0): 3(-4) - 2(0) = -12 - 0 = -12 \neq 12$. Not a solution.
$(5) (-2, -9): 3(-2) - 2(-9) = -6 + 18 = 12$. Since $LHS$ = $RHS$,$(-2, -9)$ is a solution.
$(6) (6, 4): 3(6) - 2(4) = 18 - 8 = 10 \neq 12$. Not a solution.
Therefore,$(1), (3),$ and $(5)$ are solutions,while $(2), (4),$ and $(6)$ are not.

(A) Given the equation is $2x + 5y = k$.
Since $(3,5)$ is a solution of the equation,we substitute $x = 3$ and $y = 5$ into the equation.
$2(3) + 5(5) = k$
$6 + 25 = k$
$k = 31$
Therefore,the value of $k$ is $31$.

(A) Given that $(2, -3)$ is a solution of the equation $3x + ky = 18$.
This means that when we substitute $x = 2$ and $y = -3$ into the equation,it must satisfy the equality.
Substituting the values: $3(2) + k(-3) = 18$.
$6 - 3k = 18$.
Subtract $6$ from both sides: $-3k = 18 - 6$.
$-3k = 12$.
Divide by $-3$: $k = 12 / -3$.
Therefore,$k = -4$.

(B) Given the equation is $kx + 4y = 33$.
Since $(5, 2)$ is a solution,we substitute $x = 5$ and $y = 2$ into the equation.
$k(5) + 4(2) = 33$
$5k + 8 = 33$
$5k = 33 - 8$
$5k = 25$
$k = \frac{25}{5}$
$k = 5$.

(A) Given the equation is $2x + 5y = 17$.
Since $(k, 3)$ is a solution,we substitute $x = k$ and $y = 3$ into the equation.
$2(k) + 5(3) = 17$
$2k + 15 = 17$
$2k = 17 - 15$
$2k = 2$
$k = 1$.

(A) Step $1$: Substitute $(x, y) = (2, 3)$ into the equation $7x - 3y = a$.
$7(2) - 3(3) = a$
$14 - 9 = a$
$a = 5$.
Step $2$: Substitute $a = 5$ into the point $(a, a+1)$ to get $(5, 6)$.
Step $3$: Substitute $(x, y) = (5, 6)$ into the equation $2x + y = b$.
$2(5) + 6 = b$
$10 + 6 = b$
$b = 16$.
Thus,the values are $a = 5$ and $b = 16$.

The equation of a line passing through a point $(x_1, y_1)$ can be represented in the form $a(x - x_1) + b(y - y_1) = 0$. Since there are infinitely many lines passing through a single point $(2, 3)$,we can choose different values for the coefficients $a$ and $b$ to generate them.
$1$. For $a=1, b=1$: $(x - 2) + (y - 3) = 0 \implies x + y = 5$.
$2$. For $a=3, b=-2$: $3(x - 2) - 2(y - 3) = 0 \implies 3x - 6 - 2y + 6 = 0 \implies 3x - 2y = 0$.
$3$. For $a=5, b=-3$: $5(x - 2) - 3(y - 3) = 0 \implies 5x - 10 - 3y + 9 = 0 \implies 5x - 3y = 1$.
$4$. For $a=2, b=3$: $2(x - 2) + 3(y - 3) = 0 \implies 2x - 4 + 3y - 9 = 0 \implies 2x + 3y = 13$.

(C) Since every point on the graph of a linear equation is a solution of the equation,the coordinates $(x, y) = (3, 5)$ must satisfy the equation $ax + y = 20$.
Substituting $x = 3$ and $y = 5$ into the equation:
$a(3) + 5 = 20$
$3a + 5 = 20$
Subtracting $5$ from both sides:
$3a = 20 - 5$
$3a = 15$
Dividing by $3$:
$a = \frac{15}{3}$
$a = 5$
Thus,the value of $a$ is $5$.

(N/A) Given equation: $2x - 3y = 0$
$\therefore 3y = 2x$
$\therefore y = \frac{2}{3}x$
To draw the graph,we find at least two solutions for the equation:
$1$. For $x = 0$,we get $y = \frac{2}{3}(0) = 0$. So,the point is $(0, 0)$.
$2$. For $x = 3$,we get $y = \frac{2}{3}(3) = 2$. So,the point is $(3, 2)$.
$3$. For $x = -3$,we get $y = \frac{2}{3}(-3) = -2$. So,the point is $(-3, -2)$.
We can represent these solutions in the tabular form as below:

$x$	$0$	$3$	$-3$
$y$	$0$	$2$	$-2$

Plot these points on a Cartesian plane and join them to obtain the straight line representing the equation $2x - 3y = 0$.

(N/A) $4x - 3y = 12$
$\therefore 4x - 12 = 3y$
$\therefore y = \frac{4x - 12}{3}$
For $x = 0$,we get:
$y = \frac{4(0) - 12}{3} = \frac{-12}{3} = -4$; i.e.,$y = -4$
For $x = 3$,we get:
$y = \frac{4(3) - 12}{3} = \frac{0}{3} = 0$; i.e.,$y = 0$
For $x = 6$,we get:
$y = \frac{4(6) - 12}{3} = \frac{12}{3} = 4$; i.e.,$y = 4$
We can represent these solutions in the tabular form as below:

$x$	$0$	$3$	$6$
$y$	$-4$	$0$	$4$

Plot the points $(0, -4)$,$(3, 0)$,and $(6, 4)$ on a Cartesian plane and join them to obtain the graph of the line $4x - 3y = 12$.

$A$ linear equation in two variables passing through a point $(x_0, y_0)$ can be represented in the form $a(x - x_0) + b(y - y_0) = 0$.
Given the point $(3, 5)$,we can choose different values for $a$ and $b$ to find various lines:
$1$. For $a=1, b=1$: $(x - 3) + (y - 5) = 0 \implies x + y = 8$.
$2$. For $a=-1, b=1$: $-(x - 3) + (y - 5) = 0 \implies -x + 3 + y - 5 = 0 \implies y - x = 2$.
$3$. For $a=1, b=2$: $(x - 3) + 2(y - 5) = 0 \implies x - 3 + 2y - 10 = 0 \implies x + 2y = 13$.
$4$. For $a=2, b=1$: $2(x - 3) + (y - 5) = 0 \implies 2x - 6 + y - 5 = 0 \implies 2x + y = 11$.

The general equation of a line passing through a point $(x_1, y_1)$ is given by $(y - y_1) = m(x - x_1)$,where $m$ is the slope.
For the point $(-2, 4)$,the equation is $(y - 4) = m(x + 2)$.
By choosing different values for the slope $m$,we can find infinitely many lines:
$1$. If $m = -1$,then $(y - 4) = -1(x + 2) \implies y - 4 = -x - 2 \implies x + y = 2$.
$2$. If $m = 1$,then $(y - 4) = 1(x + 2) \implies y - 4 = x + 2 \implies x - y = -6$.
$3$. If $m = -2$,then $(y - 4) = -2(x + 2) \implies y - 4 = -2x - 4 \implies 2x + y = 0$.
$4$. If $m = -0.5$,then $(y - 4) = -0.5(x + 2) \implies y - 4 = -0.5x - 1 \implies 0.5x + y = 3 \implies x + 2y = 6$.

(B) The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the intercept form: $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $x$-intercept and $y$-intercept respectively.
Given points are $(3,0)$ and $(0,4)$.
Here,the $x$-intercept $a = 3$ and the $y$-intercept $b = 4$.
Substituting these values into the intercept form equation:
$\frac{x}{3} + \frac{y}{4} = 1$
To simplify,multiply the entire equation by the least common multiple of $3$ and $4$,which is $12$:
$12 \times (\frac{x}{3}) + 12 \times (\frac{y}{4}) = 12 \times 1$
$4x + 3y = 12$
Thus,the correct equation is $4x + 3y = 12$.

(C) The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Substituting the given points $(-2, 0)$ and $(0, 3)$:
$m = \frac{3 - 0}{0 - (-2)} = \frac{3}{2}$.
The equation of a line in slope-intercept form is $y = mx + c$,where $c$ is the $y$-intercept.
Since the line passes through $(0, 3)$,the $y$-intercept $c = 3$.
Thus,the equation is $y = \frac{3}{2}x + 3$.
Multiplying by $2$,we get $2y = 3x + 6$,which rearranges to $3x - 2y = -6$.

(C) Given that the point $(2, 3)$ lies on the graph of the equation $3x + ay = 18$.
This means that the coordinates $x = 2$ and $y = 3$ must satisfy the given equation.
Substituting $x = 2$ and $y = 3$ into the equation:
$3(2) + a(3) = 18$
$6 + 3a = 18$
Subtracting $6$ from both sides:
$3a = 18 - 6$
$3a = 12$
Dividing by $3$:
$a = 4$.

(A) Given the equation is $ax + 3y = 7$.
Since the point $(-2, 5)$ lies on the graph,it must satisfy the equation.
Substitute $x = -2$ and $y = 5$ into the equation:
$a(-2) + 3(5) = 7$
$-2a + 15 = 7$
$-2a = 7 - 15$
$-2a = -8$
$a = \frac{-8}{-2}$
$a = 4$

(A) Given that the point $(a, a-2)$ lies on the graph of the equation $3x + 5y = 30$.
This means that the coordinates of the point must satisfy the equation.
Substitute $x = a$ and $y = a - 2$ into the equation:
$3(a) + 5(a - 2) = 30$
$3a + 5a - 10 = 30$
$8a - 10 = 30$
$8a = 30 + 10$
$8a = 40$
$a = 40 / 8$
$a = 5$

(C) To determine the correct equation,we check which equation is satisfied by the points shown on the graph. The points on the line are $(0, 2)$,$(2, 1)$,and $(4, 0)$.
Testing the points in option $(C)$,$x + 2y = 4$:
For $(0, 2)$: $0 + 2(2) = 4$,which is $4 = 4$ (True).
For $(2, 1)$: $2 + 2(1) = 4$,which is $4 = 4$ (True).
For $(4, 0)$: $4 + 2(0) = 4$,which is $4 = 4$ (True).
Since all points satisfy the equation $x + 2y = 4$,this is the correct equation.

(D) To determine the correct equation,we check the points lying on the line shown in the graph.
The points marked on the line are $(0, 0)$,$(2, 4)$,and $(-2, -4)$.
Now,we test these points in the given equations:
For option $D$,$y = 2x$:
$1$. For $(0, 0)$: $0 = 2(0) \implies 0 = 0$ (True)
$2$. For $(2, 4)$: $4 = 2(2) \implies 4 = 4$ (True)
$3$. For $(-2, -4)$: $-4 = 2(-2) \implies -4 = -4$ (True)
Since all points satisfy the equation $y = 2x$,this is the correct equation.

(N/A) Let the number of students be $x$ and the total cost be $y$.
The fixed cost for the bus is ₹ $200$.
The cost of snacks per student is ₹ $30$.
Therefore,the total cost $y$ is given by the linear equation: $y = 30x + 200$.
To find the total cost for $40$ students,substitute $x = 40$ into the equation:
$y = 30(40) + 200$
$y = 1200 + 200$
$y = 1400$.
Thus,the total cost for $40$ students is ₹ $1400$.

(N/A) line parallel to the $x$-axis is given by the equation $y = k$,where $k$ is the distance from the $x$-axis.
Since the line is $3$ units below the $x$-axis,the value of $y$ is negative.
Therefore,the equation of the line is $y = -3$.
The graph is a horizontal straight line passing through the point $(0, -3)$ on the $y$-axis.

(B) The statement is False.
$A$ linear equation in two variables,typically represented as $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a, b \neq 0$,has infinitely many solutions.
For any value assigned to one variable,there exists a corresponding value for the other variable that satisfies the equation. Therefore,it does not have a unique solution.

(A) The given equation is $3x = 4y$,which can be rewritten as $y = \frac{3}{4}x$.
This is a linear equation of the form $y = mx$,where $m = \frac{3}{4}$.
Any linear equation of the form $y = mx$ (where $c = 0$) represents a straight line that passes through the origin $(0, 0)$.
Substituting $x = 0$ in the equation $3x = 4y$,we get $3(0) = 4y$,which implies $4y = 0$,so $y = 0$.
Since the point $(0, 0)$ satisfies the equation,the graph passes through the origin.
Therefore,the statement is True.

(B) To determine if the statement is true or false,we substitute the coordinates $(x, y) = (5, -2)$ into the given equation $5x + 2y = k$.
Substituting $x = 5$ and $y = -2$:
$5(5) + 2(-2) = k$
$25 - 4 = k$
$21 = k$
Since the calculated value of $k$ is $21$ and not $0$,the statement is False.

(B) The given equation is $3x = 15$.
Dividing both sides by $3$,we get $x = 5$.
In a Cartesian plane,the equation $x = a$ represents a vertical line parallel to the $y$-axis,passing through the point $(a, 0)$.
Therefore,the graph of $x = 5$ is a vertical line parallel to the $y$-axis,not the $x$-axis.
Thus,the statement is False.

(B) The statement is False.
To find the intersection point with the $y$-axis,we set $x = 0$ in the equation $5x - 3y = 30$.
Substituting $x = 0$: $5(0) - 3y = 30 \implies -3y = 30 \implies y = -10$.
Thus,the graph intersects the $y$-axis at the point $(0, -10)$.
Conversely,if we check the point $(6, 0)$ by substituting $x = 6$ and $y = 0$ into the equation: $5(6) - 3(0) = 30 - 0 = 30$. This point lies on the $x$-axis,not the $y$-axis.

(A) To find a solution for the linear equation $3x + 5y = 30$,we can substitute the given options into the equation.
For option $A$ $(10, 0)$,substitute $x = 10$ and $y = 0$:
$3(10) + 5(0) = 30 + 0 = 30$.
Since the left-hand side equals the right-hand side,$(10, 0)$ is a solution.
For option $B$ $(0, 10)$,substitute $x = 0$ and $y = 10$:
$3(0) + 5(10) = 0 + 50 = 50 \neq 30$.
For option $C$ $(3, 5)$,substitute $x = 3$ and $y = 5$:
$3(3) + 5(5) = 9 + 25 = 34 \neq 30$.
For option $D$ $(0, 0)$,substitute $x = 0$ and $y = 0$:
$3(0) + 5(0) = 0 \neq 30$.
Therefore,the correct solution is $(10, 0)$.

(B) To check if a point $(x, y)$ is a solution to the equation $2x + 7y = 28$,we substitute the values of $x$ and $y$ into the equation and check if the left-hand side equals the right-hand side $(28)$.
For $(14, 0)$: $2(14) + 7(0) = 28 + 0 = 28$. (This is a solution).
For $(2, 7)$: $2(2) + 7(7) = 4 + 49 = 53 \neq 28$. (This is not a solution).
For $(0, 4)$: $2(0) + 7(4) = 0 + 28 = 28$. (This is a solution).
For $(7, 2)$: $2(7) + 7(2) = 14 + 14 = 28$. (This is a solution).
Therefore,$(2, 7)$ is not a solution of the given equation.

(C) Given the equation is $x + 3y = k$.
Since $(5, 2)$ is a solution to the equation,we substitute $x = 5$ and $y = 2$ into the equation.
$5 + 3(2) = k$
$5 + 6 = k$
$k = 11$
Therefore,the value of $k$ is $11$.

(D) Given the equation: $3x - 2y = 12$
To express $y$ in terms of $x$,we isolate $y$:
Subtract $3x$ from both sides: $-2y = 12 - 3x$
Multiply both sides by $-1$: $2y = 3x - 12$
Divide by $2$: $y = \frac{3x - 12}{2}$
Thus,the correct option is $D$.

(A) linear equation of the form $ax + by = k$ represents a line passing through the origin $(0, 0)$ if and only if the constant term $k$ is equal to $0$.
Substituting the coordinates of the origin $(x = 0, y = 0)$ into the given equation $3x + 7y = k$:
$3(0) + 7(0) = k$
$0 + 0 = k$
$k = 0$
Therefore,the value of $k$ is $0$.

(B) Given the equation is $3x + ay = 7$.
Since the graph passes through the point $(3, -1)$,this point must satisfy the equation.
Substitute $x = 3$ and $y = -1$ into the equation:
$3(3) + a(-1) = 7$
$9 - a = 7$
$-a = 7 - 9$
$-a = -2$
$a = 2$
Therefore,the value of $a$ is $2$.

(C) To find the points through which the graph of the equation $3x - 2y = 12$ passes,we test the given coordinates by substituting them into the equation.
For point $(4, 0)$:
$3(4) - 2(0) = 12 - 0 = 12$. This point satisfies the equation.
For point $(0, -6)$:
$3(0) - 2(-6) = 0 + 12 = 12$. This point also satisfies the equation.
Thus,the graph passes through $(4, 0)$ and $(0, -6)$.