Solution of a Linear Equation Questions in English

(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 $x = 3y$.
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) 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 $(1, 2)$.

(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) 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 $x - y = 2$.

(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 $5/3$.

(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) 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. $(ii)$).

(A) Given the linear equation $3x - y = x - 1$.
Rearranging the terms to bring variables to one side: $3x - x - y = -1$.
This simplifies to $2x - y = -1$,or $y = 2x + 1$.
This is a linear equation in two variables ($x$ and $y$).
$A$ linear equation in two variables of the form $ax + by + c = 0$ always has infinitely many solutions because for every value of $x$,there exists a corresponding value of $y$ that satisfies the equation.
Therefore,the equation has infinitely many solutions.

(D) The given equation $2x - 5y = 7$ is a linear equation in two variables ($x$ and $y$).
For any linear equation in two variables of the form $ax + by = c$,there exist infinitely many pairs of $(x, y)$ that satisfy the equation.
By assigning any arbitrary real value to $x$,we can find a corresponding value for $y$,and vice versa.
Therefore,the equation has infinitely many solutions.

(A) The linear equation $2x + 5y = 7$ represents a line in the Cartesian plane,which has infinitely many solutions in the set of real numbers.
However,if we restrict $x$ and $y$ to natural numbers (i.e.,$x, y \in \{1, 2, 3, ...\}$),we can test for solutions:
If $x = 1$,then $2(1) + 5y = 7 \implies 5y = 5 \implies y = 1$.
If $x = 2$,then $2(2) + 5y = 7 \implies 5y = 3 \implies y = 3/5$ (not a natural number).
If $x \ge 2$,$y$ will not be a natural number.
Thus,$(1, 1)$ is the only solution in natural numbers.

(B) For the linear equation $2x + 5y = 7$,we look for solutions where $x$ and $y$ are natural numbers (i.e.,$x, y \in \{1, 2, 3, ...\}$).
If $x = 1$,then $2(1) + 5y = 7$,which gives $5y = 5$,so $y = 1$.
If $x = 2$,then $2(2) + 5y = 7$,which gives $4 + 5y = 7$,so $5y = 3$,or $y = 3/5$,which is not a natural number.
If $x \ge 2$,$5y$ would be $7 - 2x \le 3$,meaning $y$ would not be a natural number.
Thus,the only pair of natural numbers that satisfies the equation is $(1, 1)$,making it a unique solution.

(C) To find the value of $k,$ we substitute the given solution $(x, y) = (2, 0)$ into the linear equation $2x + 3y = k.$
Substituting $x = 2$ and $y = 0$ in the equation:
$2(2) + 3(0) = k$
$4 + 0 = k$
$k = 4$
Therefore,the value of $k$ is $4.$

(D) The given linear equation is $2x + 0y + 9 = 0$.
This simplifies to $2x = -9$.
Dividing by $2$,we get $x = -9/2$.
Since the coefficient of $y$ is $0$,the value of $y$ can be any real number $m$.
Therefore,any solution of the equation is of the form $(-9/2, m)$,where $m$ is any real number.

(A) The graph of the linear equation $2x + 3y = 6$ cuts the $y$-axis at the point where the $x$-coordinate is $0$.
By substituting $x = 0$ into the equation $2x + 3y = 6$,we get:
$2(0) + 3y = 6$
$0 + 3y = 6$
$3y = 6$
$y = 6 / 3 = 2$
Therefore,the point where the graph cuts the $y$-axis is $(0, 2)$.

(B) linear equation in two variables is of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation $x = 7$,we can rewrite it as $x - 7 = 0$.
To express this in two variables $x$ and $y$,we include the variable $y$ with a coefficient of $0$.
Thus,the equation becomes $1 \cdot x + 0 \cdot y = 7$.

(D) For any point on the line $y=x$,the $x$-coordinate and the $y$-coordinate must be equal.
If we let the $x$-coordinate be $a$,then the $y$-coordinate must also be $a$.
Therefore,any point on the line $y=x$ is of the form $(a, a)$.

(B) The given equation $y=6$ does not contain the variable $x$.
This implies that for any value of $x$,the value of $y$ remains constant at $6$.
Therefore,the graph of $y=6$ is a horizontal line that is parallel to the $x$-axis.
Since the $y$-coordinate is always $6$,the line is at a constant distance of $6$ units from the origin along the $y$-axis.

(C) To check if $(x=5, y=2)$ is a solution,substitute the values into each equation:
For option $A$: $x + 2y = 5 + 2(2) = 5 + 4 = 9 \neq 7$.
For option $B$: $5x + 2y = 5(5) + 2(2) = 25 + 4 = 29 \neq 7$.
For option $C$: $x + y = 5 + 2 = 7$. This matches the equation.
For option $D$: $5x + y = 5(5) + 2 = 25 + 2 = 27 \neq 7$.
Therefore,the correct equation is $x + y = 7$.

(D) The graph of the linear equation $2x + 3y = 6$ is a line that meets the $x$-axis at the point where $y = 0$.
Substituting $y = 0$ into the equation $2x + 3y = 6$,we get:
$2x + 3(0) = 6$
$2x = 6$
$x = 6 / 2 = 3$
Therefore,the line meets the $x$-axis at the point $(3, 0)$.

(A) The given linear equation is $y = x$.
For a point $(x, y)$ to lie on the graph of this equation,the $x$-coordinate must be equal to the $y$-coordinate.
Checking the options:
$A$: $(1, 1)$ where $x = 1$ and $y = 1$. Since $1 = 1$,the point $(1, 1)$ satisfies the equation $y = x$.
$B$: $(0, 3/2)$ where $x = 0$ and $y = 1.5$. Since $0 \neq 1.5$,it does not lie on the line.
$C$: $(3/2, -3/2)$ where $x = 1.5$ and $y = -1.5$. Since $1.5 \neq -1.5$,it does not lie on the line.
$D$: $(-1/2, 1/2)$ where $x = -0.5$ and $y = 0.5$. Since $-0.5 \neq 0.5$,it does not lie on the line.
Therefore,the graph passes through the point $(1, 1)$.

(B) If we multiply or divide both sides of a linear equation with a non-zero number,the equality holds true. Since the relationship between the variables remains balanced,the solution of the linear equation remains the same.

(C) There are infinitely many linear equations that can be satisfied by the values $x = 1$ and $y = 2$.
$A$ linear equation in two variables is of the form $ax + by + c = 0$. Since we have a specific point $(1, 2)$,we can construct infinitely many lines passing through this point by varying the coefficients $a, b,$ and $c$.
For example:
$1$. $x + y = 3$ (since $1 + 2 = 3$)
$2$. $y = 2x$ (since $2 = 2(1)$)
$3$. $y - x = 1$ (since $2 - 1 = 1$)
$4$. $2y - x = 3$ (since $2(2) - 1 = 3$)
Since we can choose infinitely many combinations of $a, b,$ and $c$ such that $a(1) + b(2) + c = 0$,there are infinitely many such linear equations.

(A) point of the form $(a, -a)$ has an $x$-coordinate equal to $a$ and a $y$-coordinate equal to $-a$.
To find the line on which this point lies,we can observe the relationship between the coordinates:
$x = a$
$y = -a$
Adding these two equations gives:
$x + y = a + (-a)$
$x + y = 0$
Therefore,the point $(a, -a)$ always satisfies the equation $x + y = 0$.

(A) The statement is False.
To verify if the points $(x, y)$ satisfy the equation $2x + 2 = y$,we substitute the $x$-values from the table into the equation and check if the resulting $y$-value matches the table.
For $x = 0$: $2(0) + 2 = 2$. (Matches $y = 2$)
For $x = 1$: $2(1) + 2 = 4$. (Matches $y = 4$)
For $x = 2$: $2(2) + 2 = 6$. (Matches $y = 6$)
For $x = 3$: $2(3) + 2 = 8$. (Matches $y = 8$)
For $x = 4$: $2(4) + 2 = 10$. (Matches $y = 10$)
Wait,checking the equation $2x + 2 = y$ against the table values:
If $x=0, y=2(0)+2=2$.
If $x=1, y=2(1)+2=4$.
If $x=2, y=2(2)+2=6$.
If $x=3, y=2(3)+2=8$.
If $x=4, y=2(4)+2=10$.
All points satisfy the equation $2x + 2 = y$. Therefore,the statement is True.

(TRUE) To check if the point $(0, 3)$ lies on the graph of the linear equation $3x + 4y = 12$,we substitute $x = 0$ and $y = 3$ into the equation.
Substituting the values:
$3(0) + 4(3) = 0 + 12 = 12$.
Since the left-hand side equals the right-hand side $(12 = 12)$,the point $(0, 3)$ satisfies the equation.
Therefore,the given statement is True.

(FALSE) To check if the graph of the linear equation $x + 2y = 7$ passes through the point $(0, 7)$,we substitute $x = 0$ and $y = 7$ into the equation.
Substituting the values:
$0 + 2(7) = 7$
$0 + 14 = 7$
$14 = 7$
Since $14 \neq 7$,the point $(0, 7)$ does not satisfy the equation.
Therefore,the statement is False.

(TRUE) We know that the graph of the equation $x = a$ is a line parallel to the $y$-axis and at a distance of $|a|$ units from it.
If $a > 0$,the line lies to the right of the $y$-axis.
In the given graph,the line is parallel to the $y$-axis and passes through the point $(3, 0)$ on the $x$-axis,which is at a distance of $3$ units to the right of the $y$-axis.
Therefore,the given statement is True.

(B) To determine if the points are solutions,we substitute the values of $x$ and $y$ into the equation $x-y+2=0$.
For $(0, 2)$: $0 - 2 + 2 = 0$. (True)
For $(1, 3)$: $1 - 3 + 2 = 0$. (True)
For $(2, 4)$: $2 - 4 + 2 = 0$. (True)
For $(3, -5)$: $3 - (-5) + 2 = 3 + 5 + 2 = 10 \neq 0$. (False)
For $(4, 6)$: $4 - 6 + 2 = 0$. (True)
Since the point $(3, -5)$ does not satisfy the equation,the statement is False.

(B) The given statement is $False$.
By definition,the graph of a linear equation in two variables is the collection of all points $(x, y)$ that satisfy the equation.
Therefore,every point lying on the line represented by the linear equation is a solution to that equation.

(N/A) The graph of the linear equation $3x + 4y = 12$ cuts the $x$-axis at the point where $y = 0$.
Putting $y = 0$ in the equation,we get $3x + 4(0) = 12$,which simplifies to $3x = 12$,so $x = 4$.
Thus,the point on the $x$-axis is $(4, 0)$.
The graph of the linear equation $3x + 4y = 12$ cuts the $y$-axis at the point where $x = 0$.
Putting $x = 0$ in the equation,we get $3(0) + 4y = 12$,which simplifies to $4y = 12$,so $y = 3$.
Thus,the point on the $y$-axis is $(0, 3)$.

(A) The line parallel to the $y$-axis at a distance of $2$ units from the origin in the positive direction of the $x$-axis is given by the equation $x=2$.
To find the point of intersection,we substitute $x=2$ into the given linear equation $x+y=5$.
$2+y=5$
$y=5-2$
$y=3$
Therefore,the point of intersection is $(2, 3)$.

(C) Let the point be $(x, y)$.
According to the problem,the $x$-coordinate is $\frac{5}{2}$ times its ordinate $(y)$,so $x = \frac{5}{2}y$.
Substitute $x = \frac{5}{2}y$ into the equation $2x + 5y = 20$:
$2(\frac{5}{2}y) + 5y = 20$
$5y + 5y = 20$
$10y = 20$
$y = 2$
Now,find the $x$-coordinate using $x = \frac{5}{2}y$:
$x = \frac{5}{2}(2) = 5$
Thus,the required point is $(5, 2)$.

(N/A) Any straight line parallel to the $x$-axis is given by the equation $y=k$,where $k$ is the distance of the line from the $x$-axis.
Here,the line is $4$ units above the $x$-axis,so $k=4$.
Therefore,the equation of the line is $y=4$.
To draw the graph of this equation,plot the points $(1,4)$,$(2,4)$,and $(3,4)$ on the Cartesian plane and join them with a straight line. This line represents all points where the $y$-coordinate is $4$.

(A) Let the abscissa of the point be $x$ and the ordinate be $y$.
According to the problem,the ordinate is $1 \frac{1}{2}$ times the abscissa,so $y = 1 \frac{1}{2} x = \frac{3}{2} x$.
Substitute $y = \frac{3}{2} x$ into the given linear equation $2x + 5y = 19$:
$2x + 5(\frac{3}{2} x) = 19$
$2x + \frac{15}{2} x = 19$
Multiply the entire equation by $2$ to clear the fraction:
$4x + 15x = 38$
$19x = 38$
$x = 2$
Now,find the ordinate $y$ using $y = \frac{3}{2} x$:
$y = \frac{3}{2} (2) = 3$
Thus,the point is $(2, 3)$.

(N/A) The equation of a line parallel to the $x$-axis at a distance $a$ units below it is given by $y = -a$.
Here,the distance is $3$ units below the $x$-axis,so $a = 3$.
Therefore,the equation of the line is $y = -3$.
This line is parallel to the $x$-axis and passes through every point where the $y$-coordinate is $-3$,such as $(0, -3)$,$(1, -3)$,$(-1, -3)$,etc.
The graph of the equation $y = -3$ is a horizontal line passing through the point $(0, -3)$ on the $y$-axis,as shown in the figure.

(A) The point $(3,4)$ lies on the graph of $3y = ax + 7$.
Substituting $x = 3$ and $y = 4$ in the given equation $3y = ax + 7$,we get:
$3 \times 4 = a \times 3 + 7$
$12 = 3a + 7$
$3a = 12 - 7$
$3a = 5$
$a = \frac{5}{3}$