Mix Examples - Linear Equations in Two Variables Questions in English

(N/A) To find the linear equation $ax + by = c$,we use the two given points $(6, -2)$ and $(-6, 6)$.
For $(6, -2)$: $6a - 2b = c$
For $(-6, 6)$: $-6a + 6b = c$
Adding the two equations: $4b = 2c \implies c = 2b$.
Substituting $c = 2b$ into the first equation: $6a - 2b = 2b \implies 6a = 4b \implies a = \frac{2}{3}b$.
Let $b = 3$,then $a = 2$ and $c = 6$. The equation is $2x + 3y = 6$.
Plot the points $(6, -2)$ and $(-6, 6)$ on a Cartesian plane and join them with a straight line.
$(i)$ To find the $x$-intercept,set $y = 0$: $2x + 3(0) = 6 \implies 2x = 6 \implies x = 3$. The graph cuts the $x$-axis at $(3, 0)$.
$(ii)$ To find the $y$-intercept,set $x = 0$: $2(0) + 3y = 6 \implies 3y = 6 \implies y = 2$. The graph cuts the $y$-axis at $(0, 2)$.

(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 $(0, 1.5)$.

(N/A) $C = \frac{5F - 160}{9}$
$(i)$ Putting $F = 86^{\circ}$,we get $C = \frac{5(86) - 160}{9} = \frac{430 - 160}{9} = \frac{270}{9} = 30^{\circ}$.
Hence,the temperature in Celsius is $30^{\circ} C$.
$(ii)$ Putting $C = 35^{\circ}$,we get $35 = \frac{5F - 160}{9} \Rightarrow 315 = 5F - 160$.
$\Rightarrow 5F = 315 + 160 = 475$.
$\therefore F = \frac{475}{5} = 95^{\circ}$.
Hence,the temperature in Fahrenheit is $95^{\circ} F$.
$(iii)$ Putting $C = 0^{\circ}$,we get $0 = \frac{5F - 160}{9} \Rightarrow 0 = 5F - 160$.
$\Rightarrow 5F = 160 \Rightarrow F = \frac{160}{5} = 32^{\circ}$.
Now,putting $F = 0^{\circ}$,we get $C = \frac{5(0) - 160}{9} = -\frac{160}{9} \approx -17.78^{\circ} C$.
If the temperature is $0^{\circ} C$,the temperature in Fahrenheit is $32^{\circ} F$,and if the temperature is $0^{\circ} F$,the temperature in Celsius is $-\frac{160}{9}^{\circ} C$.
$(iv)$ Putting $C = F$ in the given relation,we get $F = \frac{5F - 160}{9} \Rightarrow 9F = 5F - 160$.
$\Rightarrow 4F = -160 \Rightarrow F = -40^{\circ}$.
Hence,the numerical value of the temperature which is the same in both scales is $-40$.

(N/A) The given linear equation is $y = \frac{9}{5}(x - 273) + 32$.
$(i)$ Given $x = 313^{\circ} K$:
$y = \frac{9}{5}(313 - 273) + 32$
$y = \frac{9}{5}(40) + 32$
$y = 9 \times 8 + 32 = 72 + 32 = 104^{\circ} F$.
$(ii)$ Given $y = 158^{\circ} F$:
$158 = \frac{9}{5}(x - 273) + 32$
$158 - 32 = \frac{9}{5}(x - 273)$
$126 = \frac{9}{5}(x - 273)$
$x - 273 = 126 \times \frac{5}{9}$
$x - 273 = 14 \times 5 = 70$
$x = 273 + 70 = 343^{\circ} K$.

(N/A) Let $y$ be the force and $x$ be the acceleration. Since force is directly proportional to acceleration,we have $y \propto x$,which implies $y = mx$,where $m$ is the constant mass.
Given $m = 6 \,kg$,the linear equation is $y = 6x$.
To draw the graph,we find some points satisfying the equation:

$x$ (Acceleration)	$0$	$1$	$2$
$y$ (Force)	$0$	$6$	$12$

Plotting the points $(0,0)$,$(1,6)$,and $(2,12)$ on a graph and joining them gives a straight line.
From the graph:
$(i)$ When acceleration $x = 5 \,m/s^2$,the corresponding force $y = 30 \,N$.
$(ii)$ When acceleration $x = 6 \,m/s^2$,the corresponding force $y = 36 \,N$.

(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 $5$.

(D) To find the correct equation,we check which equation is satisfied by the points $(0,6)$,$(2,3)$,and $(4,0)$ shown on the graph.
$1$. For $x+y=4$: Substituting $(4,0)$ gives $4+0=4$ (True),but $(0,6)$ gives $0+6=6 \neq 4$ (False).
$2$. For $x+y=5$: Substituting $(2,3)$ gives $2+3=5$ (True),but $(4,0)$ gives $4+0=4 \neq 5$ (False).
$3$. For $x+y=6$: Substituting $(0,6)$ gives $0+6=6$ (True),but $(4,0)$ gives $4+0=4 \neq 6$ (False).
$4$. For $3x+2y=12$:
- For $(4,0)$: $3(4)+2(0) = 12+0 = 12$ (True).
- For $(2,3)$: $3(2)+2(3) = 6+6 = 12$ (True).
- For $(0,6)$: $3(0)+2(6) = 0+12 = 12$ (True).
Since all three points satisfy the equation $3x+2y=12$,this is the correct equation.

(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) We solve the equation $2y + 1 = y + 4$:
$\therefore 2y - y = 4 - 1$
$\therefore y = 3$
$(1)$ Representation on the number line:
Since $y = 3$ is an equation in one variable,it represents a unique point on the number line at the position $3$.
$(2)$ Representation on the Cartesian plane:
We know that $y = 3$ can be written as a linear equation in two variables: $0x + 1y = 3$.
For any value of $x$,the value of $y$ remains $3$. Thus,we can choose points like $(0, 3)$,$(1, 3)$,and $(3, 3)$. Plotting these points on the Cartesian plane and joining them results in a line parallel to the $x$-axis passing through $y = 3$.

$A$ 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.

(A) To find the intersection points with the coordinate axes:
- For the $x$-axis,set $y = 0$ and solve for $x$.
- For the $y$-axis,set $x = 0$ and solve for $y$.

Equation	Intersection at $x$-axis $(y=0)$	Intersection at $y$-axis $(x=0)$
$1. 3x + 5y = 15$	$(5, 0)$	$(0, 3)$
$2. 5x - 2y = 10$	$(2, 0)$	$(0, -5)$
$3. 4x + 3y = -12$	$(-3, 0)$	$(0, -4)$
$4. 3x - 7y = 21$	$(7, 0)$	$(0, -3)$
$5. x - y = 0$	$(0, 0)$	$(0, 0)$
$6. 2x - 3y = 0$	$(0, 0)$	$(0, 0)$
$7. x - y = -5$	$(-5, 0)$	$(0, 5)$
$8. 5x - 3y = 15$	$(3, 0)$	$(0, -5)$

(N/A) Step $1$: Solve the equation for $x$.
$3x - 1 = x + 7$
$3x - x = 7 + 1$
$2x = 8$
$x = 4$
Step $2$: Representation on the number line.
On the number line,mark the point $4$ as a solid dot.
Step $3$: Representation on the Cartesian plane.
In the Cartesian plane,the equation $x = 4$ represents a vertical line passing through $x = 4$ on the $x$-axis,parallel to the $y$-axis.

(Y = -3) Step $1$: Solve the equation for $y$.
$5y + 10 = 3y + 4$
$5y - 3y = 4 - 10$
$2y = -6$
$y = -3$
Step $2$: Representation on the number line.
On a horizontal number line,mark the point at $-3$.
Step $3$: Representation on the Cartesian plane.
The equation $y = -3$ represents a horizontal line parallel to the $x$-axis,passing through the point $(0, -3)$.

(D) Let the total distance covered be $x$ $km$ and the total fare be ₹ $y$.
For the first $1$ $km$,the fare is ₹ $10$.
For the remaining distance $(x - 1)$ $km$,the fare is $3(x - 1)$.
Therefore,the total fare $y = 10 + 3(x - 1)$.
Simplifying the equation: $y = 10 + 3x - 3$,which gives $y = 3x + 7$.
To draw the graph,we find points:
If $x = 1, y = 10$.
If $x = 2, y = 13$.
If $x = 3, y = 16$.
Plotting these points on a graph and joining them gives a straight line.
For a journey of $4$ $km$,substitute $x = 4$ in the equation: $y = 3(4) + 7 = 12 + 7 = 19$.
Thus,the total fare for $4$ $km$ is ₹ $19$.

(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.

Let the coordinates of the point be $(x, y)$.
According to the problem,the sum of the coordinates is $10$,so the linear equation is $x + y = 10$.
To draw the graph,we find at least two solutions for the equation:
If $x = 0$,then $0 + y = 10 \implies y = 10$. So,the point is $(0, 10)$.
If $x = 10$,then $10 + y = 10 \implies y = 0$. So,the point is $(10, 0)$.
If $x = 5$,then $5 + y = 10 \implies y = 5$. So,the point is $(5, 5)$.
Plot these points $(0, 10)$,$(10, 0)$,and $(5, 5)$ on a Cartesian plane and join them with a straight line to obtain the graph of the equation $x + y = 10$.

(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.