Graphical Method of Solution Questions in English

(D) For the equation $3x + 4y = 6$:
$4y = 6 - 3x$
$y = \frac{6 - 3x}{4}$
If $x = -2$,$y = \frac{6 - 3(-2)}{4} = \frac{12}{4} = 3$.
If $x = 2$,$y = \frac{6 - 3(2)}{4} = \frac{0}{4} = 0$.

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

Plot the points $(-2, 3)$ and $(2, 0)$ and join them to draw the line.
For the equation $3x + 4y = 19$:
$4y = 19 - 3x$
$y = \frac{19 - 3x}{4}$
If $x = 5$,$y = \frac{19 - 3(5)}{4} = \frac{4}{4} = 1$.
If $x = 1$,$y = \frac{19 - 3(1)}{4} = \frac{16}{4} = 4$.

$x$	$5$	$1$
$y$	$1$	$4$

Plot the points $(5, 1)$ and $(1, 4)$ and join them to draw the line.
Since the lines are parallel and do not intersect,the system of equations has no solution. The solution set is $\varnothing$.

(N/A) Here,dividing each term of $3x+3y=15$ by $3$,we get the equation $x+y=5$.
Thus,both the equations of the pair are identical.
Hence,we say that both the lines are the same.
So,they are coincident. It means that there are infinitely many solutions. To draw the graph,we make the following table:
$x+y=5$
$\therefore y=5-x$
For $x=0, y=5-0=5$
For $x=5, y=5-5=0$
and
$3x+3y=15$
$\therefore 3y=15-3x$
$\therefore y=\frac{15-3x}{3}$
$\therefore y=5-x$
$\therefore$ Both the tables are the same.

$x$	$0$	$5$
$y$	$5$	$0$

$\therefore$ Plot the ordered pairs $(0, 5)$ and $(5, 0)$ of the solution set of $x+y=5$ (or $3x+3y=15$,i.e.,$x+y=5$) on the graph paper and draw the line by joining them.
Here,the graphs of both the equations are the same. Also,we can see that there are infinite points on the line,and they all make the solution set. Thus,the solution set of the pair of linear equations is ${(x, y) \mid x+y=5, x, y \in R}$.

(A) Given equations are:
$1) \, y + x = 5 \implies y = 5 - x$
$2) \, y - x = 9 \implies y = 9 + x$
To plot the graph, we find points for each line:
For $y = 5 - x$: If $x = 0, y = 5$; If $x = 5, y = 0$. Points are $(0, 5)$ and $(5, 0)$.
For $y = 9 + x$: If $x = 0, y = 9$; If $x = -9, y = 0$. Points are $(0, 9)$ and $(-9, 0)$.
Solving algebraically for intersection:
Adding the two equations: $(y + x) + (y - x) = 5 + 9 \implies 2y = 14 \implies y = 7$.
Substituting $y = 7$ in $y + x = 5$: $7 + x = 5 \implies x = -2$.
The point of intersection is $(-2, 7)$.

(D) Given equations are:
$1) 3x + 6y = 4$
$2) 2x + 4y = \frac{8}{3}$
Multiply equation $(2)$ by $\frac{3}{2}$:
$\frac{3}{2}(2x + 4y) = \frac{3}{2} \times \frac{8}{3}$
$3x + 6y = 4$
Since both equations represent the same line,they are coincident lines.
Therefore,the system has infinitely many solutions. The solution set is $\{(x, y) \mid 3x + 6y = 4; x, y \in R \}$.

(A) Given equations are:
$1) 2x + y = 7 \implies y = 7 - 2x$
For $x = 0, y = 7$. For $x = 2, y = 3$. For $x = 3, y = 1$.
$2) x - 2y = 6 \implies x = 6 + 2y$
For $y = 0, x = 6$. For $y = -1, x = 4$. For $y = -2, x = 2$.
Plotting these points on a graph,the two lines intersect at the point $(4, -1)$.
Thus,the solution is $x = 4$ and $y = -1$.

(D) Given equations are:
$1) y = 2$
$2) x + 4y = 10$
Step $1$: The first equation $y = 2$ represents a horizontal line passing through all points where the $y$-coordinate is $2$.
Step $2$: For the second equation $x + 4y = 10$,substitute $y = 2$ into the equation:
$x + 4(2) = 10$
$x + 8 = 10$
$x = 10 - 8$
$x = 2$
Step $3$: The point of intersection of the two lines is $(2, 2)$.
Graphically,the line $y = 2$ is a horizontal line,and $x + 4y = 10$ is a line passing through $(10, 0)$ and $(2, 2)$. Both lines intersect at the point $(2, 2)$.

(A) Given equations are:
$1) x + y = 5$
$2) 5x - 2y = 4$
For equation $(1)$,if $x = 0$,then $y = 5$. If $y = 0$,then $x = 5$. Points are $(0, 5)$ and $(5, 0)$.
For equation $(2)$,if $x = 0$,then $-2y = 4 \implies y = -2$. If $y = 0$,then $5x = 4 \implies x = 0.8$. Points are $(0, -2)$ and $(0.8, 0)$.
By plotting these lines on a graph,the point of intersection is $(2, 3)$.
Verification: $2 + 3 = 5$ (Correct) and $5(2) - 2(3) = 10 - 6 = 4$ (Correct).
Thus,the solution is $(2, 3)$.

(B) Given equations are:
$1) \, 2x - 3y = 5$
$2) \, 4x - 6y = -13$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = -3, c_1 = -5$
$a_2 = 4, b_2 = -6, c_2 = 13$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-5}{13}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines represented by these equations are parallel to each other.
Parallel lines do not intersect at any point,which means the system of equations has no solution $(\varnothing)$.

(C) Given equations are:
$1) 4x + y = 7$
$2) 16x + 4y = 28$
Divide equation $(2)$ by $4$:
$4x + y = 7$
Since both equations are identical,they represent the same line on the graph.
For any value of $x$,we can find a corresponding value of $y$ such that $y = 7 - 4x$.
Therefore,the system has infinitely many solutions.
The solution set is ${(x, y) \mid 4x + y = 7; x, y \in R}$.

(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:
$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$. If $y = 0$,then $x = -4$. The line passes through $(0, -2)$ and $(-4, 0)$.
$2$. For the second equation $3x + 4y = -6$,if $x = 0$,then $y = -1.5$. If $y = 0$,then $x = -2$. The line passes through $(0, -1.5)$ and $(-2, 0)$.
$3$. Solving the equations algebraically: Multiply the first equation by $2$: $2x + 4y = -8$. Subtract this from the second equation: $(3x + 4y) - (2x + 4y) = -6 - (-8)$,which gives $x = 2$.
$4$. Substitute $x = 2$ into $x + 2y = -4$: $2 + 2y = -4 \implies 2y = -6 \implies y = -3$.
$5$. The intersection point is $(2, -3)$.

(A) To solve the system of equations $x + y = 8$ and $x - y = 2$ graphically:
$1$. For the equation $x + y = 8$,we find two points: If $x = 0, y = 8$; if $x = 8, y = 0$. The line passes through $(0, 8)$ and $(8, 0)$.
$2$. For the equation $x - y = 2$,we find two points: If $x = 0, y = -2$; if $x = 2, y = 0$. The line passes through $(0, -2)$ and $(2, 0)$.
$3$. Plotting these two lines on a Cartesian plane,they intersect at the point $(5, 3)$.
$4$. Therefore,the solution is $x = 5$ and $y = 3$.

(B) Given the equations:
$1) \ 2x + 3y = 12$
$2) \ 2x + 3y = 6$
Comparing these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have:
$a_1 = 2, b_1 = 3, c_1 = -12$
$a_2 = 2, b_2 = 3, c_2 = -6$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{2} = 1$
$\frac{b_1}{b_2} = \frac{3}{3} = 1$
$\frac{c_1}{c_2} = \frac{-12}{-6} = 2$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other.
Parallel lines do not intersect at any point,which means the system of equations has no solution.
Therefore,the solution set is $\varnothing$.

(C) $1$. For the equation $x+y=10$,if $x=0, y=10$; if $y=0, x=10$. The line passes through $(0,10)$ and $(10,0)$.
$2$. For the equation $x-y=4$,if $x=0, y=-4$; if $y=0, x=4$. The line passes through $(0,-4)$ and $(4,0)$.
$3$. Solving the equations simultaneously: Adding $(x+y=10)$ and $(x-y=4)$ gives $2x=14$,so $x=7$. Substituting $x=7$ into $x+y=10$ gives $y=3$. The intersection point is $(7,3)$.
$4$. The triangle is formed by the intersection of the two lines $(7,3)$ and their respective intersections with the $X$-axis,which are $(4,0)$ and $(10,0)$.
$5$. Thus,the vertices of the triangle are $(7,3), (4,0),$ and $(10,0)$.

(D) $1$. For the equation $x + 3y = 6$: If $x = 0$,$3y = 6 \implies y = 2$. If $y = 0$,$x = 6$. Points are $(0, 2)$ and $(6, 0)$.
$2$. For the equation $2x - 3y = 12$: If $x = 0$,$-3y = 12 \implies y = -4$. If $y = 0$,$2x = 12 \implies x = 6$. Points are $(0, -4)$ and $(6, 0)$.
$3$. Solving the system: Adding the two equations,$(x + 3y) + (2x - 3y) = 6 + 12 \implies 3x = 18 \implies x = 6$. Substituting $x = 6$ into $x + 3y = 6$,we get $6 + 3y = 6 \implies y = 0$. The intersection point is $(6, 0)$.
$4$. The triangle is formed by the lines and the $Y$-axis. The vertices are the intersection of the two lines $(6, 0)$,the $Y$-intercept of the first line $(0, 2)$,and the $Y$-intercept of the second line $(0, -4)$.
$5$. Thus,the vertices are $(6, 0), (0, 2), (0, -4)$.

$(A)$ Let the number of boys be $x$ and the number of girls be $y$.
According to the problem,the total number of students is $6$,so $x + y = 6$.
The number of girls is $2$ less than the number of boys,so $y = x - 2$,which can be written as $x - y = 2$.
Solving the system of equations:
$1) x + y = 6$
$2) x - y = 2$
Adding the two equations: $(x + y) + (x - y) = 6 + 2 \rightarrow 2x = 8 \rightarrow x = 4$.
Substituting $x = 4$ into the first equation: $4 + y = 6 \rightarrow y = 2$.
Thus,there are $4$ boys and $2$ girls.

(A) To check if the pair of equations is consistent,we compare the ratios of the coefficients:
$a_1 = 2, b_1 = 4, c_1 = -9$
$a_2 = 3, b_2 = 6, c_2 = -\frac{27}{2}$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{3}$
$\frac{b_1}{b_2} = \frac{4}{6} = \frac{2}{3}$
$\frac{c_1}{c_2} = \frac{-9}{-27/2} = \frac{9 \times 2}{27} = \frac{18}{27} = \frac{2}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$,the lines are coincident.
Coincident lines have infinitely many solutions,therefore the system is consistent.

(D) To find the solution,we determine the points for each line:
For $2x + 3y = 8$:
If $x = 1, 2(1) + 3y = 8 \implies 3y = 6 \implies y = 2$. Point: $(1, 2)$.
If $x = 4, 2(4) + 3y = 8 \implies 3y = 0 \implies y = 0$. Point: $(4, 0)$.
For $x + 2y = 5$:
If $x = 1, 1 + 2y = 5 \implies 2y = 4 \implies y = 2$. Point: $(1, 2)$.
If $x = 3, 3 + 2y = 5 \implies 2y = 2 \implies y = 1$. Point: $(3, 1)$.
Since both lines pass through the point $(1, 2)$,the solution to the system of equations is $(1, 2)$.

(D) Given the pair of linear equations:
$2x + 3y = 4$ --- $(1)$
$6x + 9y = 17$ --- $(2)$
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$,we have:
$a_1 = 2, b_1 = 3, c_1 = 4$
$a_2 = 6, b_2 = 9, c_2 = 17$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{4}{17}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines represented by these equations are parallel.
Therefore,there is no common point of intersection,which means the system has no solution.
The solution set is the empty set,denoted by $\phi$.

(D) Given equations are:
$x+y-3=0$ --- $(1)$
$3x+3y-9=0$ --- $(2)$
Dividing equation $(2)$ by $3$,we get:
$x+y-3=0$
Since both equations are identical,they represent the same line.
Therefore,every point on the line $x+y=3$ is a solution to the system.
Thus,the system has infinitely many solutions,and the solution set is an infinite set.

(C) pair of linear equations in two variables $x$ and $y$ is given by $x = 6$ and $y = 4$.
These equations represent lines parallel to the $y$-axis and $x$-axis,respectively.
The point of intersection of these two lines provides the unique solution $(x, y)$.
Substituting the given values,we get the point $(6, 4)$.
Therefore,the solution set is $\{(6, 4)\}$.

(A) For a system of two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have an empty set (no solution),the lines must be parallel.
This condition is given by $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are $x + 2y - 3 = 0$ and $5x + ky + 7 = 0$.
Here,$a_1 = 1, b_1 = 2, c_1 = -3$ and $a_2 = 5, b_2 = k, c_2 = 7$.
Applying the condition for parallel lines: $\frac{1}{5} = \frac{2}{k}$.
Cross-multiplying gives $k = 5 \times 2 = 10$.
Checking the inequality condition: $\frac{c_1}{c_2} = \frac{-3}{7}$. Since $\frac{1}{5} \neq \frac{-3}{7}$,the condition holds.
Therefore,the value of $k$ is $10$.

(D) linear equation in two variables is an equation of the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
In this form,the variables $x$ and $y$ must have an exponent of $1$.
Checking the options:
$A$: $x^{2}+x-3=0$ is a quadratic equation in one variable.
$B$: $x^{2}+y=0$ is not linear because the variable $x$ has an exponent of $2$.
$C$: $x-3=y^{2}$ is not linear because the variable $y$ has an exponent of $2$.
$D$: $x+3=y$ can be rewritten as $x - y + 3 = 0$,which fits the standard form $ax + by + c = 0$ with $a=1, b=-1, c=3$.
Therefore,$x+3=y$ is a linear equation in two variables.

(D) linear equation in two variables is an equation of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero.
$1$. In option $A$,the term $\frac{3}{x}$ makes it a non-linear equation.
$2$. In option $B$,the term $\frac{1}{y}$ makes it a non-linear equation.
$3$. In option $C$,the variables have a degree of $2$,making it a quadratic equation.
$4$. In option $D$,$x = y$ can be written as $1x - 1y + 0 = 0$,which fits the standard form of a linear equation in two variables with degree $1$.

(A) The given equations are $x = 1$ and $y = 2$.
These equations represent two lines in a Cartesian plane.
The line $x = 1$ is a vertical line passing through the point $(1, 0)$ on the $x$-axis.
The line $y = 2$ is a horizontal line passing through the point $(0, 2)$ on the $y$-axis.
The intersection point of these two lines is the point where the $x$-coordinate is $1$ and the $y$-coordinate is $2$.
Therefore,the solution set is the set containing the ordered pair $(1, 2)$,which is written as $\{(1, 2)\}$.

(B) Given the pair of linear equations:
$1$) $x + 2 = 0$
$2$) $y - 1 = 0$
From equation $(1)$,we get:
$x = -2$
From equation $(2)$,we get:
$y = 1$
Therefore,the solution $(x, y)$ is $(-2, 1)$.

(C) Given equations are $5x - 5y = -5$ and $\frac{3x}{2} - \frac{3y}{2} + \frac{3}{2} = 0$.
Step $1$: Simplify the first equation: $5x - 5y = -5$. Dividing by $5$,we get $x - y = -1$.
Step $2$: Simplify the second equation: $\frac{3x}{2} - \frac{3y}{2} = -\frac{3}{2}$. Multiplying by $\frac{2}{3}$,we get $x - y = -1$.
Step $3$: Since both equations simplify to the same linear equation $x - y = -1$,they represent the same line.
Step $4$: $A$ pair of coincident lines has infinitely many solutions. Therefore,the solution set is an infinite set.

(B) Given equations are $2x + 6y = 10$ and $3x + 9y - 15 = 0$.
Divide the first equation by $2$: $x + 3y = 5$.
Divide the second equation by $3$: $x + 3y - 5 = 0$,which implies $x + 3y = 5$.
Since both equations represent the same line,the ratio of coefficients is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{1} = \frac{3}{3} = \frac{5}{5}$.
Therefore,the pair of equations has infinitely many solutions.

(D) Given equations are $x+y+1=0$ and $3x+3y+2=0$.
Comparing these with $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$,we have $a_1=1, b_1=1, c_1=1$ and $a_2=3, b_2=3, c_2=2$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{1}{3}$,$\frac{b_1}{b_2} = \frac{1}{3}$,and $\frac{c_1}{c_2} = \frac{1}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines represented by these equations are parallel.
Parallel lines do not intersect at any point,which means the system has no solution.
Therefore,the solution set is an empty set.

(C) For a pair of linear equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,the condition for a unique solution is $a_{1}/a_{2} \neq b_{1}/b_{2}$,which is equivalent to $a_{1}b_{2} - a_{2}b_{1} \neq 0$.
Here,$a_{1} = 2, b_{1} = 1, c_{1} = -4$ and $a_{2} = 1, b_{2} = 2, c_{2} = -5$.
Calculating the determinant: $a_{1}b_{2} - a_{2}b_{1} = (2)(2) - (1)(1) = 4 - 1 = 3$.
Since $3 \neq 0$,the pair of equations has a unique solution.

(C) Given equations are $2x + y = 6$ $(1)$ and $x + \frac{1}{2}y = 4$ $(2)$.
Multiply equation $(2)$ by $2$,we get $2x + y = 8$.
Comparing the two equations:
$2x + y = 6$
$2x + y = 8$
Here,the coefficients of $x$ and $y$ are the same,but the constants are different $(6 \neq 8)$.
This represents two parallel lines.
Therefore,the system has no solution.

(B) Given equations are:
$1) 2x + y = 7$
$2) 5x - 2y = 4$
To determine the nature of the lines,we compare the coefficients of the equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$.
Here,$a_1 = 2, b_1 = 1, c_1 = 7$ and $a_2 = 5, b_2 = -2, c_2 = 4$.
We calculate the ratio of the coefficients:
$\frac{a_1}{a_2} = \frac{2}{5}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines intersect at exactly one point.
Therefore,the pair of equations represents intersecting lines.

(C) Given equations are:
$3x + 4y = 10$ --- $(1)$
$3x + 4y = 15$ --- $(2)$
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
$a_1 = 3, b_1 = 4, c_1 = 10$
$a_2 = 3, b_2 = 4, c_2 = 15$
Now,find the ratios:
$\frac{a_1}{a_2} = \frac{3}{3} = 1$
$\frac{b_1}{b_2} = \frac{4}{4} = 1$
$\frac{c_1}{c_2} = \frac{10}{15} = \frac{2}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other and do not intersect.

(D) Given equations are:
$x + y = 7$ ---$(1)$
$3x + 3y = 21$ ---$(2)$
Divide equation $(2)$ by $3$:
$x + y = 7$ ---$(3)$
Since equation $(1)$ and equation $(3)$ are identical,they represent the same line.
Therefore,the pair of equations represents coincident lines,which is a single line.

(B) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$,the system is inconsistent if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are $2x + 3y = 5$ and $2x + 3y = 8$.
Here,$a_1 = 2, b_1 = 3, c_1 = 5$ and $a_2 = 2, b_2 = 3, c_2 = 8$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{2} = 1$
$\frac{b_1}{b_2} = \frac{3}{3} = 1$
$\frac{c_1}{c_2} = \frac{5}{8}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $(1 = 1 \neq \frac{5}{8})$,the lines are parallel and have no common solution.
Therefore,the pair of equations is inconsistent.

(A) Given the pair of linear equations:
$2x + y = 3$ (Equation $1$)
$x + 2y = 4$ (Equation $2$)
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
$a_1 = 2, b_1 = 1, c_1 = 3$
$a_2 = 1, b_2 = 2, c_2 = 4$
Now,find the ratios of the coefficients:
$\frac{a_1}{a_2} = \frac{2}{1} = 2$
$\frac{b_1}{b_2} = \frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines intersect at a unique point.
Therefore,the system of equations has a unique solution and is consistent.

(C) The standard form of a linear equation in two variables is given by $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation: $3x = 2y - 1$.
To bring this into the standard form $ax + by + c = 0$,we move all terms to the left-hand side of the equation.
Subtract $2y$ from both sides: $3x - 2y = -1$.
Add $1$ to both sides: $3x - 2y + 1 = 0$.
Comparing this with $ax + by + c = 0$,we have $a = 3$,$b = -2$,and $c = 1$.
Thus,the standard form is $3x - 2y + 1 = 0$.

(A) The given equation is $y = -3$,which can be rewritten as $0x + 1y + 3 = 0$.
Comparing this with the standard form $ax + by + c = 0$:
Here,the coefficient of $x$ is $a = 0$.
The coefficient of $y$ is $b = 1$.
The constant term is $c = 3$.
Therefore,the value of $a$ is $0$.

(B) The given equation is $x=4$.
This can be written in the standard form $ax+by+c=0$ as $1x + 0y - 4 = 0$.
Comparing this with $ax+by+c=0$,we identify the coefficients as $a=1$,$b=0$,and $c=-4$.
Therefore,the value of $b$ is $0$.

(C) Given equations are $3x - 2y = 2$ and $6x - 4y = 4$.
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
For the first equation: $a_1 = 3, b_1 = -2, c_1 = 2$.
For the second equation: $a_2 = 6, b_2 = -4, c_2 = 4$.
Now,find the ratios:
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{2}{4} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$,the lines are coincident lines.

(C) Two lines are parallel if their slopes are equal.
The given line is $x+3y=5$,which can be written in the slope-intercept form $y = mx + c$ as $3y = -x + 5$,or $y = -\frac{1}{3}x + \frac{5}{3}$.
The slope of this line is $m = -\frac{1}{3}$.
$A$ line parallel to this will have the same slope $m = -\frac{1}{3}$ and will be of the form $x+3y=k$ for some constant $k$.
Comparing the given options,the equation $x+3y=10$ has the same coefficients for $x$ and $y$ as the original line,meaning it has the same slope.
Therefore,$x+3y=10$ is parallel to $x+3y=5$.

(A) To find the point where the line intersects the $X$-axis,we set the $y$-coordinate to $0$.
Substituting $y = 0$ into the equation $2x - y = 6$:
$2x - 0 = 6$
$2x = 6$
$x = 3$
Therefore,the point of intersection is $(3, 0)$.

(C) To find the point where the line intersects the $Y$-axis,we set the $x$-coordinate to $0$.
Substituting $x=0$ into the equation $x-2y=2$:
$0-2y=2$
$-2y=2$
$y=-1$
Therefore,the line intersects the $Y$-axis at the point $(0,-1)$.

(A) system of linear equations in two variables $x$ and $y$ is represented as a coordinate pair $(x, y)$.
Given the equations $x = 4$ and $y = 5$,the value of $x$ is fixed at $4$ and the value of $y$ is fixed at $5$.
Therefore,the unique solution point where these two lines intersect is $(4, 5)$.
Thus,the solution set is ${(4, 5)}$.

(D) Given equations are $x+y-1=0$ and $2x+2y-2=0$.
Comparing with $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$,we have $a_1=1, b_1=1, c_1=-1$ and $a_2=2, b_2=2, c_2=-2$.
Calculating the ratios: $\frac{a_1}{a_2} = \frac{1}{2}$,$\frac{b_1}{b_2} = \frac{1}{2}$,and $\frac{c_1}{c_2} = \frac{-1}{-2} = \frac{1}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the lines are coincident.
Coincident lines have infinitely many solutions,meaning the solution set is an infinite set.

(C) The equation $x=0$ represents the $y$-axis.
The equation $y=0$ represents the $x$-axis.
The intersection point of the $x$-axis and the $y$-axis is the origin.
The coordinates of the origin are $(0,0)$.
Therefore,the intersection point of the lines $x=0$ and $y=0$ is $(0,0)$.

(A) The given equations are $x = -2$ and $y = 3$.
These equations represent two lines in the Cartesian plane.
The line $x = -2$ is a vertical line passing through the point $(-2, 0)$ on the $x$-axis.
The line $y = 3$ is a horizontal line passing through the point $(0, 3)$ on the $y$-axis.
The intersection point of these two lines is the unique solution to the system.
By substituting the values,the point of intersection is $(-2, 3)$.
Since a solution set is represented as a set containing the coordinate pair,the correct representation is $\{(-2, 3)\}$.

(A) To determine if a point $(x, y)$ lies on the graph of the equation $2x - y = 1$,we substitute the $x$ and $y$ coordinates of the point into the equation.
If the left-hand side equals the right-hand side,the point lies on the graph.
$A) (0, 0): 2(0) - 0 = 0 \neq 1$. Thus,the graph does not pass through $(0, 0)$.
$B) (1, 1): 2(1) - 1 = 2 - 1 = 1$. The graph passes through $(1, 1)$.
$C) (3, 5): 2(3) - 5 = 6 - 5 = 1$. The graph passes through $(3, 5)$.
$D) (2, 3): 2(2) - 3 = 4 - 3 = 1$. The graph passes through $(2, 3)$.
Therefore,the graph does not pass through $(0, 0)$.

(C) line perpendicular to the $Y-axis$ is a horizontal line.
Any horizontal line has the general form $y = k$,where $k$ is a constant.
Let us analyze the given options:
$A) 3x + 2y = 5$ represents a slanted line.
$B) x = 3$ represents a vertical line (perpendicular to the $X-axis$).
$C) 2y - 3 = 0$ can be rewritten as $2y = 3$ or $y = 1.5$. This is a horizontal line,which is perpendicular to the $Y-axis$.
$D) x - y = 0$ represents a slanted line passing through the origin.
Therefore,the correct option is $C$.

(C) line perpendicular to the $X$-axis is a vertical line,which is represented by the equation of the form $x = k$,where $k$ is a constant.
In the given options:
$A) x - y = 0 \implies y = x$ (a line passing through the origin with slope $1$)
$B) x + y = 0 \implies y = -x$ (a line passing through the origin with slope $-1$)
$C) 3x + 2 = 0 \implies 3x = -2 \implies x = -2/3$ (this is a vertical line,perpendicular to the $X$-axis)
$D) 2y - 3 = 0 \implies 2y = 3 \implies y = 3/2$ (this is a horizontal line,parallel to the $X$-axis)
Therefore,the correct option is $C$.

(B) line is parallel to the $Y$-axis if its equation is of the form $x = k$,where $k$ is a constant.
Checking the options:
$A$) $3y - 1 = 0 \implies y = 1/3$ (This is a line parallel to the $X$-axis).
$B$) $2x + 5 = 0 \implies 2x = -5 \implies x = -2.5$ (This is a line parallel to the $Y$-axis).
$C$) $2x + 3y = 1$ (This is a slanted line intersecting both axes).
$D$) $2x - 3y = 1$ (This is a slanted line intersecting both axes).
Therefore,the correct option is $B$.