Textbook - Linear Equations in Two Variables Questions in English

(N/A) $(i)$ The equation $2x + 3y = 4.37$ can be written as $2x + 3y - 4.37 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 2$,$b = 3$,and $c = -4.37$.
$(ii)$ The equation $x - 4 = \sqrt{3}y$ can be written as $x - \sqrt{3}y - 4 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 1$,$b = -\sqrt{3}$,and $c = -4$.
$(iii)$ The equation $4 = 5x - 3y$ can be written as $5x - 3y - 4 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 5$,$b = -3$,and $c = -4$.
$(iv)$ The equation $2x = y$ can be written as $2x - y + 0 = 0$.
Comparing this with $ax + by + c = 0$,we get $a = 2$,$b = -1$,and $c = 0$.

(N/A) linear equation in two variables is expressed in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
$(i)$ $x = -5$ can be written as $1x + 0y = -5$,which is $1x + 0y + 5 = 0$.
$(ii)$ $y = 2$ can be written as $0x + 1y = 2$,which is $0x + 1y - 2 = 0$.
$(iii)$ $2x = 3$ can be written as $2x + 0y = 3$,which is $2x + 0y - 3 = 0$.
$(iv)$ $5y = 2$ can be written as $0x + 5y = 2$,which is $0x + 5y - 2 = 0$.

(A) Let the cost of a notebook be Rs. $x$.
Let the cost of a pen be Rs. $y$.
According to the given condition,the cost of a notebook is twice the cost of a pen.
Therefore,we can write the equation as:
$x = 2y$
Rearranging the terms to the standard form of a linear equation in two variables $(ax + by + c = 0)$:
$x - 2y = 0$
Thus,the required linear equation is $x - 2y = 0$.

(A) The given equation is $2x + 3y = 9.3\overline{5}$.
To express this in the standard form $ax + by + c = 0$,we subtract $9.3\overline{5}$ from both sides:
$2x + 3y - 9.3\overline{5} = 0$.
Comparing this with the general form $ax + by + c = 0$,we get:
$a = 2$
$b = 3$
$c = -9.3\overline{5}$

(A) The given equation is $x - \frac{y}{5} - 10 = 0$.
This can be rewritten in the standard form $ax + by + c = 0$ as:
$1x + (-\frac{1}{5})y + (-10) = 0$.
Comparing this with the general form $ax + by + c = 0$,we identify the coefficients as:
$a = 1$,$b = -\frac{1}{5}$,and $c = -10$.

(A) The given equation is $-2x + 3y = 6$.
To express this in the form $ax + by + c = 0$,we subtract $6$ from both sides:
$-2x + 3y - 6 = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = -2$,
$b = 3$,
$c = -6$.

(A) The given equation is $x = 3y$.
To express this in the form $ax + by + c = 0$,we subtract $3y$ from both sides:
$x - 3y = 0$
This can be written as:
$(1)x + (-3)y + 0 = 0$
Comparing this with the standard form $ax + by + c = 0$,we get:
$a = 1$
$b = -3$
$c = 0$

(N/A) The given equation is $2x = -5y$.
To express this in the form $ax + by + c = 0$,we move all terms to the left side:
$2x + 5y = 0$.
This can be written as $(2)x + (5)y + (0) = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 2$,$b = 5$,and $c = 0$.

(N/A) The given equation is $3x + 2 = 0$.
To express this in the form $ax + by + c = 0$,we can write it as:
$3x + 0y + 2 = 0$
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 3$
$b = 0$
$c = 2$

(N/A) The given equation is $y - 2 = 0$.
To express this in the form $ax + by + c = 0$,we can rewrite it as:
$(0)x + (1)y + (-2) = 0$.
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients as:
$a = 0$,$b = 1$,and $c = -2$.

(A) The given equation is $5 = 2x$.
To express this in the form $ax + by + c = 0$,we rearrange the terms:
$2x - 5 = 0$
This can be written as:
$2x + 0y + (-5) = 0$
Comparing this with the standard form $ax + by + c = 0$,we identify the coefficients:
$a = 2$
$b = 0$
$c = -5$

(N/A) To find solutions for the linear equation $x + 2y = 6$,we can assign arbitrary values to one variable and solve for the other.
$1$. Let $x = 0$. Then $0 + 2y = 6$,which gives $2y = 6$,so $y = 3$. Thus,$(0, 3)$ is a solution.
$2$. Let $y = 0$. Then $x + 2(0) = 6$,which gives $x = 6$. Thus,$(6, 0)$ is a solution.
$3$. Let $x = 2$. Then $2 + 2y = 6$,which gives $2y = 4$,so $y = 2$. Thus,$(2, 2)$ is a solution.
$4$. Let $y = 1$. Then $x + 2(1) = 6$,which gives $x + 2 = 6$,so $x = 4$. Thus,$(4, 1)$ is a solution.
Therefore,four different solutions of the equation are $(0, 3)$,$(6, 0)$,$(2, 2)$,and $(4, 1)$.

(N/A) $(i)$ For $4x + 3y = 12$,if we take $x = 0$,then $3y = 12$,which gives $y = 4$. So,$(0, 4)$ is a solution. If we take $y = 0$,then $4x = 12$,which gives $x = 3$. So,$(3, 0)$ is another solution.
$(ii)$ For $2x + 5y = 0$,if we take $x = 0$,then $5y = 0$,which gives $y = 0$. So,$(0, 0)$ is a solution. If we take $x = 1$,then $2(1) + 5y = 0$,which gives $5y = -2$,so $y = -\frac{2}{5}$. Thus,$(1, -\frac{2}{5})$ is another solution.
$(iii)$ For $3y + 4 = 0$,we can write this as $0x + 3y = -4$. For any value of $x$,$y$ remains $-\frac{4}{3}$. If we take $x = 0$,$y = -\frac{4}{3}$. If we take $x = 1$,$y = -\frac{4}{3}$. Thus,two solutions are $(0, -\frac{4}{3})$ and $(1, -\frac{4}{3})$.

(C) The equation $y = 3x + 5$ is a linear equation in two variables.
For any linear equation in two variables of the form $ax + by + c = 0$,there exist infinitely many pairs of $(x, y)$ that satisfy the equation.
This is because for every real value assigned to $x$,we can calculate a corresponding unique value for $y$ using the relation $y = 3x + 5$.
Since there are infinitely many real numbers for $x$,there are infinitely many corresponding values for $y$.
Therefore,the correct option is $(iii)$.

(N/A) To find solutions for the equation $2x + y = 7$,we can express $y$ in terms of $x$: $y = 7 - 2x$.
By substituting different values of $x$,we get corresponding values of $y$:
$1$. When $x = 0$,$y = 7 - 2(0) = 7$. Thus,the solution is $(0, 7)$.
$2$. When $x = 1$,$y = 7 - 2(1) = 5$. Thus,the solution is $(1, 5)$.
$3$. When $x = 2$,$y = 7 - 2(2) = 3$. Thus,the solution is $(2, 3)$.
$4$. When $x = 3$,$y = 7 - 2(3) = 1$. Thus,the solution is $(3, 1)$.

(N/A) To find the solutions for the linear equation $\pi x + y = 9$,we can substitute different values for $x$ and solve for $y$.
$1$. When $x = 0$:
$\pi(0) + y = 9 \Rightarrow 0 + y = 9 \Rightarrow y = 9$.
So,the solution is $(0, 9)$.
$2$. When $x = 1$:
$\pi(1) + y = 9 \Rightarrow \pi + y = 9 \Rightarrow y = 9 - \pi$.
So,the solution is $(1, 9 - \pi)$.
$3$. When $x = 2$:
$\pi(2) + y = 9 \Rightarrow 2\pi + y = 9 \Rightarrow y = 9 - 2\pi$.
So,the solution is $(2, 9 - 2\pi)$.
$4$. When $x = -1$:
$\pi(-1) + y = 9 \Rightarrow -\pi + y = 9 \Rightarrow y = 9 + \pi$.
So,the solution is $(-1, 9 + \pi)$.

(N/A) To find the solutions for the equation $x = 4y$,we can substitute different values for $y$ and find the corresponding values of $x$.
$1$. If $y = 0$,then $x = 4(0) = 0$. So,the solution is $(0, 0)$.
$2$. If $y = 1$,then $x = 4(1) = 4$. So,the solution is $(4, 1)$.
$3$. If $y = -1$,then $x = 4(-1) = -4$. So,the solution is $(-4, -1)$.
$4$. If $y = 2$,then $x = 4(2) = 8$. So,the solution is $(8, 2)$.

(B) The given point is $(0, 2)$,which means $x = 0$ and $y = 2$.
Substitute $x = 0$ and $y = 2$ into the equation $x - 2y = 4$:
$L$.$H$.$S$. $= x - 2y = 0 - 2(2) = 0 - 4 = -4$.
The $R$.$H$.$S$. of the equation is $4$.
Since $L$.$H$.$S$. $= -4$ and $R$.$H$.$S$. $= 4$,we have $L$.$H$.$S$. $\neq$ $R$.$H$.$S$.
Therefore,$(0, 2)$ is not a solution of the equation $x - 2y = 4$.

(B) The coordinate $(2, 0)$ implies $x=2$ and $y=0$.
Substituting $x=2$ and $y=0$ into the equation $x - 2y = 4$,we get:
$L$.$H$.$S$. $= 2 - 2(0) = 2 - 0 = 2$.
Since the $R$.$H$.$S$. $= 4$,we observe that $L$.$H$.$S$. $\neq$ $R$.$H$.$S$.
Therefore,$(2, 0)$ is not a solution of the given equation.

(A) Given equation: $x - 2y = 4$.
For the point $(4, 0)$,we have $x = 4$ and $y = 0$.
Substitute these values into the left-hand side ($L$.$H$.$S$.) of the equation:
$L$.$H$.$S$. $= x - 2y = 4 - 2(0) = 4 - 0 = 4$.
The right-hand side ($R$.$H$.$S$.) of the equation is $4$.
Since $L$.$H$.$S$. $=$ $R$.$H$.$S$.,the point $(4, 0)$ is a solution of the given equation.

(D) Given equation: $x - 2y = 4$.
The point is $(\sqrt{2}, 4\sqrt{2})$,which means $x = \sqrt{2}$ and $y = 4\sqrt{2}$.
Substitute these values into the left-hand side ($L$.$H$.$S$.) of the equation:
$L$.$H$.$S$. $= x - 2y = \sqrt{2} - 2(4\sqrt{2})$
$= \sqrt{2} - 8\sqrt{2}$
$= (1 - 8)\sqrt{2} = -7\sqrt{2}$.
Since the right-hand side ($R$.$H$.$S$.) is $4$,and $-7\sqrt{2} \neq 4$,the $L$.$H$.$S$. $\neq$ $R$.$H$.$S$.
Therefore,the point $(\sqrt{2}, 4\sqrt{2})$ is not a solution of the equation $x - 2y = 4$.

(B) Given equation is $x - 2y = 4$.
For the point $(1, 1)$,we have $x = 1$ and $y = 1$.
Substitute these values into the left-hand side ($L$.$H$.$S$.) of the equation:
$L$.$H$.$S$. $= x - 2y = 1 - 2(1) = 1 - 2 = -1$.
The right-hand side ($R$.$H$.$S$.) of the equation is $4$.
Since $L$.$H$.$S$. $\neq$ $R$.$H$.$S$. $(-1 \neq 4)$,the point $(1, 1)$ is not a solution of the equation $x - 2y = 4$.

(C) Given the equation is $2x + 3y = k$.
Since $x = 2$ and $y = 1$ is a solution,we substitute these values into the equation:
$2(2) + 3(1) = k$
Calculating the values:
$4 + 3 = k$
Therefore:
$k = 7$
Thus,the required value of $k$ is $7$.

(D) The point $(1, 2)$ must satisfy the linear equation $ax + by = c$. Substituting the coordinates $x = 1$ and $y = 2$,we get $a(1) + b(2) = c$,or $a + 2b = c$.
For example,if we choose $a = 1$ and $b = 1$,then $c = 1 + 2(1) = 3$. Thus,one such equation is $x + y = 3$.
Similarly,if we choose $a = -1$ and $b = 1$,then $c = -1 + 2(1) = 1$. Thus,another equation is $-x + y = 1$.
Since there are infinitely many pairs of $(a, b, c)$ that satisfy the relation $a + 2b = c$,there are infinitely many such linear equations that pass through the point $(1, 2)$.

(N/A) To draw the graph of the linear equation $x + y = 7$,we need at least two solutions of the equation.
$1$. If we put $x = 0$,then $0 + y = 7$,which gives $y = 7$. So,$(0, 7)$ is a solution.
$2$. If we put $x = 7$,then $7 + y = 7$,which gives $y = 0$. So,$(7, 0)$ is a solution.
We can represent these solutions in the following table:

$x$	$0$	$7$
$y$	$7$	$0$

Now,plot the points $(0, 7)$ and $(7, 0)$ on a Cartesian plane and join them with a straight line to obtain the graph of the equation $x + y = 7$.

(N/A) Here,the variables involved are force and acceleration. Let the force applied be $y$ units and the acceleration produced be $x$ units. From the concept of direct proportion,we can express this relationship as $y = kx$,where $k$ is a constant of proportionality. (From your study of science,you know that $k$ is actually the mass of the body.)
Since we do not know the specific value of $k$,we cannot draw a unique graph for $y = kx$. However,if we assign a specific value to $k$,we can draw the graph. Let us take $k = 3$,i.e.,we draw the line representing $y = 3x$.
To plot this,we find two solutions,for example,$(0, 0)$ and $(2, 6)$.
From the graph,you can see that when the force applied is $3$ units,the acceleration produced is $1$ unit. Also,note that $(0, 0)$ lies on the graph,which means that when the force applied is $0$ units,the acceleration produced is $0$ units.

(A-(II), B-(III), C-(IV)) In Fig. $(i)$,the points on the line are $(-1, -2)$,$(0, 0)$,$(1, 2)$. By inspection,$y = 2x$ is the equation corresponding to this graph. You can find that the $y$-coordinate in each case is double that of the $x$-coordinate.
$(b)$ In Fig. (ii),the points on the line are $(-2, 0)$,$(0, 4)$,$(1, 6)$. You know that the coordinates of the points of the graph (line) satisfy the equation $y = 2x + 4$. So,$y = 2x + 4$ is the equation corresponding to the graph in Fig. (ii).
$(c)$ In Fig. (iii),the points on the line are $(-1, -6)$,$(0, -4)$,$(1, -2)$,$(2, 0)$. By inspection,you can see that $y = 2x - 4$ is the equation corresponding to the given graph (line).

(N/A) $x + y = 4$
$\Rightarrow y = 4 - x$
If we have $x = 0$,then $y = 4 - 0 = 4$.
If we have $x = 1$,then $y = 4 - 1 = 3$.
If we have $x = 2$,then $y = 4 - 2 = 2$.
$\therefore$ We get the following table:

$x$	$0$	$1$	$2$
$y$	$4$	$3$	$2$

Plot the ordered pairs $(0, 4)$,$(1, 3)$,and $(2, 2)$ on the graph paper. Joining these points,we get a line $AB$ as shown in the graph. Thus,the line $AB$ is the required graph of $x + y = 4$.

(N/A) Given equation: $x - y = 2$
This can be rewritten as: $y = x - 2$
To draw the graph,we find at least three solutions for the equation:
If $x = 0$,then $y = 0 - 2 = -2$
If $x = 1$,then $y = 1 - 2 = -1$
If $x = 2$,then $y = 2 - 2 = 0$
We get the following table of values:

$x$	$0$	$1$	$2$
$y$	$-2$	$-1$	$0$

Plot the points $(0, -2)$,$(1, -1)$,and $(2, 0)$ on a Cartesian plane. By joining these points,we obtain a straight line $PQ$. This line $PQ$ represents the graph of the linear equation $x - y = 2$.

(N/A) $y=3x$
If $x=0$,then $y=3(0) \Rightarrow y=0$
If $x=1$,then $y=3(1) \Rightarrow y=3$
If $x=-1$,then $y=3(-1) \Rightarrow y=-3$
$\therefore$ We get the following table:

$x$	$0$	$1$	$-1$
$y$	$0$	$3$	$-3$

Plot the ordered pairs $(0, 0)$,$(1, 3)$,and $(-1, -3)$ on the graph paper. Joining these points,we get the straight line $LM$.
Thus,the line $LM$ is the required graph of $y=3x$.

(N/A) $y = 3 - 2x$
$\therefore$ If $x = 0$,then $y = 3 - 2(0) \Rightarrow y = 3$
If $x = 1$,then $y = 3 - 2(1) \Rightarrow y = 1$
If $x = 2$,then $y = 3 - 2(2) \Rightarrow y = -1$
$\therefore$ We get the following table:

$x$	$0$	$1$	$2$
$y$	$3$	$1$	$-1$

Plot the ordered pairs $(0, 3)$,$(1, 1)$,and $(2, -1)$ on the graph paper. Joining these points,we get a line $CD$.
Thus,the line $CD$ is the required graph of $3 = 2x + y$.

(A) The point $(2, 14)$ means $x = 2$ and $y = 14$.
To find equations of lines passing through this point,we can use the form $ax + by = c$ such that $a(2) + b(14) = c$.
$(i)$ If we take $a=1, b=1$,then $1(2) + 1(14) = 16$. So,$x + y = 16$ is one such equation.
$(ii)$ If we take $a=7, b=-1$,then $7(2) - 1(14) = 14 - 14 = 0$. So,$7x - y = 0$ is another such equation.
There are infinitely many such lines because an infinite number of lines can pass through a single point in a plane.

(A) The given equation of the line is $3y = ax + 7$.
Since the point $(3, 4)$ lies on the graph of this line,it must satisfy the equation.
Substituting $x = 3$ and $y = 4$ into the equation:
$3(4) = a(3) + 7$
$12 = 3a + 7$
Subtracting $7$ from both sides:
$12 - 7 = 3a$
$5 = 3a$
Dividing by $3$:
$a = 5/3$
Thus,the value of $a$ is $5/3$.

(N/A) Here,total distance covered $= x \, km$
Total taxi fare $= y$
Fare for the $1^{st} \, km = Rs. \, 8$
Remaining distance $= (x - 1) \, km$
$\therefore$ Fare for $(x - 1) \, km = Rs. \, 5(x - 1)$
Total taxi fare $= Rs. \, 8 + 5(x - 1)$
$\therefore$ According to the condition,
$y = 8 + 5(x - 1)$
$y = 8 + 5x - 5$
$y = 5x + 3$
This is the required linear equation representing the given information.
Graph: We have $y = 5x + 3$
When $x = 0, y = 5(0) + 3 = 3$
When $x = -1, y = 5(-1) + 3 = -2$
When $x = -2, y = 5(-2) + 3 = -7$
We get the following table:

$x$	$0$	$-1$	$-2$
$y$	$3$	$-2$	$-7$

Now,plotting the ordered pairs $(0, 3), (-1, -2)$ and $(-2, -7)$ on a graph paper and joining them,we get a straight line $PQ$. Thus,$PQ$ is the required graph of the linear equation $y = 5x + 3$.

(B, C) For Fig. $(i)$,the line passes through the points $(-1, 1)$,$(0, 0)$,and $(1, -1)$.
Checking the options:
For $(b)$ $x+y=0$,substituting $(-1, 1)$ gives $-1+1=0$,which is true.
Thus,the equation for Fig. $(i)$ is $x+y=0$.
For Fig. $(ii)$,the line passes through the points $(-1, 3)$,$(0, 2)$,and $(2, 0)$.
Checking the options:
For $(c)$ $y=-x+2$,substituting $(-1, 3)$ gives $3=-(-1)+2 \Rightarrow 3=1+2 \Rightarrow 3=3$,which is true.
Substituting $(0, 2)$ gives $2=-(0)+2 \Rightarrow 2=2$,which is true.
Thus,the equation for Fig. $(ii)$ is $y=-x+2$.

(N/A) Constant force is $5$ units. Let the distance travelled $= x$ units and work done $= y$ units.
Since,$\text{Work done} = \text{Force} \times \text{Displacement}$
$\Rightarrow y = 5 \times x$
$\Rightarrow y = 5x$
Drawing the graph:
We have $y = 5x$.
When $x = 0$,then $y = 5(0) = 0$.
When $x = 1$,then $y = 5(1) = 5$.
When $x = 1.5$,then $y = 5(1.5) = 7.5$.
Therefore,we get the following table:

$x$	$0$	$1$	$1.5$
$y$	$0$	$5$	$7.5$

Plotting the ordered pairs $(0, 0)$,$(1, 5)$,and $(1.5, 7.5)$ on the graph paper and joining the points,we get a straight line.
From the graph:
$(i)$ When distance travelled $= 2$ units,then $x = 2$,which gives $y = 10$.
Therefore,$\text{Work done} = 10$ units.
$(ii)$ When distance travelled $= 0$ units,then $x = 0$,which gives $y = 5(0) = 0$.
Therefore,$\text{Work done} = 0$ units.

(N/A) Let the contribution of Yamini $= ₹ x$ and the contribution of Fatima $= ₹ y$.
$\therefore$ We have $x + y = 100$
$\Rightarrow y = 100 - x$
Now,we find the coordinates of points to plot the graph:
When $x = 0, y = 100 - 0 = 100$
When $x = 50, y = 100 - 50 = 50$
When $x = 100, y = 100 - 100 = 0$
We get the following table:

$x$	$0$	$50$	$100$
$y$	$100$	$50$	$0$

For drawing the graph,plot the ordered pairs $(0, 100)$,$(50, 50)$,and $(100, 0)$ on a graph paper. Joining these points,we get a line.
Thus,the line represents the graph of $x + y = 100$.

(N/A) $(i)$ We have the equation $F = \left(\frac{9}{5}\right) C + 32$.
When $C = 0$,$F = \left(\frac{9}{5}\right) \times 0 + 32 = 32$.
When $C = -15$,$F = \frac{9}{5}(-15) + 32 = -27 + 32 = 5$.
When $C = -10$,$F = \frac{9}{5}(-10) + 32 = 9(-2) + 32 = 14$.
Table of values:

$C$	$0$	$-15$	$-10$
$F$	$32$	$5$	$14$

Plotting the points $(0, 32)$,$(-15, 5)$,and $(-10, 14)$ on a graph and joining them gives a straight line.
$(ii)$ For $C = 30$,$F = \frac{9}{5}(30) + 32 = 9(6) + 32 = 54 + 32 = 86\,^oF$.
$(iii)$ For $F = 95$,$95 = \frac{9}{5}C + 32 \implies 63 = \frac{9}{5}C \implies C = 63 \times \frac{5}{9} = 35\,^oC$.
$(iv)$ For $C = 0$,$F = 32\,^oF$. For $F = 0$,$0 = \frac{9}{5}C + 32 \implies -32 = \frac{9}{5}C \implies C = -\frac{160}{9} \approx -17.8\,^oC$.
$(v)$ Let $F = C = x$. Then $x = \frac{9}{5}x + 32 \implies x - \frac{9}{5}x = 32 \implies -\frac{4}{5}x = 32 \implies x = 32 \times (-\frac{5}{4}) = -40$. Thus,$-40\,^oC = -40\,^oF$.

(N/A) We solve $2x + 1 = x - 3$,to get
$2x - x = -3 - 1$
i.e.,$x = -4$
$(i)$ The representation of the solution on the number line is shown in Fig. $(i)$,where $x = -4$ is treated as an equation in one variable.
$(ii)$ We know that $x = -4$ can be written as
$x + 0y = -4$
which is a linear equation in the variables $x$ and $y$. This is represented by a line. Now all the values of $y$ are permissible because $0y$ is always $0$. However,$x$ must satisfy the equation $x = -4$. Hence,two solutions of the given equation are $(-4, 0)$ and $(-4, 2)$.
Note that the graph $AB$ is a line parallel to the $y$-axis and at a distance of $4$ units to the left of it (see Fig. $(ii)$).

(N/A) $(i)$ $y = 3$ [An equation in one variable]
Since $y = 3$ is an equation in one variable,it represents a unique point on the number line.
The unique solution is a point at $3$ on the number line.
$(ii)$ $y = 3$ [An equation in two variables]
We can write $y = 3$ as $0x + y = 3$.
Now,for different values of $x$,we get $y = 3$:

$X$	$1$	$2$	$3$
$Y$	$3$	$3$	$3$

Plotting the ordered pairs $(1, 3)$,$(2, 3)$,and $(3, 3)$ on a Cartesian plane,we get a line $AB$ parallel to the $x$-axis,which represents the solution of $0x + y = 3$,i.e.,$y = 3$.

(N/A) $(i)$ $2x + 9 = 0$ [An equation in one variable]
We have: $2x + 9 = 0 \Rightarrow 2x = -9 \Rightarrow x = -4.5$ or $x = -\frac{9}{2}$.
This is a linear equation in one variable '$x$' only. Its solution is the point $-\frac{9}{2}$ on the number line.
$(ii)$ $2x + 9 = 0$ [An equation in two variables]
We can write $2x + 9 = 0$ as $2x + 0y + 9 = 0$ or $2x = -9 + 0y$.
Thus,$x = \frac{-9 + 0y}{2}$.
For any value of $y$,$x$ will always be $-\frac{9}{2}$.

$x$	$-\frac{9}{2}$	$-\frac{9}{2}$	$-\frac{9}{2}$
$y$	$1$	$2$	$3$

Plotting the points $\left(-\frac{9}{2}, 1\right)$,$\left(-\frac{9}{2}, 2\right)$,and $\left(-\frac{9}{2}, 3\right)$ on a Cartesian plane and joining them,we get a vertical line parallel to the $y$-axis.