Graphical Method of Solution Questions in English

(N/A) The pair of equations formed is:
$y = \frac{1}{2}x$
i.e.,$x - 2y = 0$ ......$(1)$
$3x + 4y = 20$ ......$(2)$
Let us represent these equations graphically. For this,we need at least two solutions for each equation. We give these solutions in the tables below:

$x$	$0$	$2$
$y = \frac{x}{2}$	$0$	$1$

$x$	$0$	$\frac{20}{3}$	$4$
$y = \frac{20 - 3x}{4}$	$5$	$0$	$2$

Recall from Class $IX$ that there are infinitely many solutions for each linear equation. So,each of you can choose any two values,which may not be the ones we have chosen. When one of the variables is zero,the equation reduces to a linear equation in one variable,which can be solved easily. For instance,putting $x = 0$ in Equation $(2)$,we get $4y = 20$,i.e.,$y = 5$. Similarly,putting $y = 0$ in Equation $(2)$,we get $3x = 20$,i.e.,$x = \frac{20}{3}$. But as $\frac{20}{3}$ is not an integer,it will not be easy to plot exactly on the graph paper. So,we choose $y = 2$,which gives $x = 4$,an integral value.
Plot the points $A(0, 0), B(2, 1)$ and $P(0, 5), Q(4, 2)$ corresponding to the solutions in the tables. Now draw the lines $AB$ and $PQ$,representing the equations $x - 2y = 0$ and $3x + 4y = 20$.

(N/A) Let the cost of $1$ pencil be ₹ $x$ and the cost of $1$ eraser be ₹ $y$. The algebraic representation is given by the following equations:
$2x + 3y = 9$ $.......(1)$
$4x + 6y = 18$ $.......(2)$
To represent these graphically,we find points for each line:
For equation $(1)$,$y = \frac{9 - 2x}{3}$:

$x$	$0$	$4.5$
$y$	$3$	$0$

For equation $(2)$,$y = \frac{18 - 4x}{6}$:

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

When we plot these points on a graph,we observe that both lines coincide. This is because the two equations are equivalent,as equation $(2)$ is simply $2$ times equation $(1)$.

(N/A) To represent the equations geometrically,we find two solutions for each equation.
For the equation $x+2y-4=0$:

$x$	$0$	$4$
$y = \frac{4-x}{2}$	$2$	$0$

We plot the points $R(0, 2)$ and $S(4, 0)$ to get the line representing $x+2y-4=0$.
For the equation $2x+4y-12=0$:

$x$	$0$	$6$
$y = \frac{12-2x}{4}$	$3$	$0$

We plot the points $P(0, 3)$ and $Q(6, 0)$ to get the line representing $2x+4y-12=0$.
Observing the graph,we see that the two lines do not intersect anywhere,which means they are parallel to each other.

(N/A) Let the present age of Aftab be $x$ and the present age of his daughter be $y$.
Seven years ago:
Age of Aftab $= x - 7$
Age of his daughter $= y - 7$
According to the problem,$(x - 7) = 7(y - 7) \implies x - 7 = 7y - 49 \implies x - 7y = -42$ $...(1)$
Three years from now:
Age of Aftab $= x + 3$
Age of his daughter $= y + 3$
According to the problem,$(x + 3) = 3(y + 3) \implies x + 3 = 3y + 9 \implies x - 3y = 6$ $...(2)$
Algebraic representation:
$x - 7y = -42$
$x - 3y = 6$
For $x - 7y = -42$,we have $x = 7y - 42$:

$x$	$-7$	$0$	$7$
$y$	$5$	$6$	$7$

For $x - 3y = 6$,we have $x = 3y + 6$:

$x$	$6$	$3$	$0$
$y$	$0$	$-1$	$-2$

(N/A) Let the cost of a bat be ₹ $x$ and the cost of a ball be ₹ $y$.
According to the problem,the algebraic representation is:
$3x + 6y = 3900$
$x + 3y = 1300$
For the first equation,$3x + 6y = 3900$,we can simplify it to $x + 2y = 1300$,so $x = 1300 - 2y$.
| $x$ | $300$ | $100$ | $-100$ |
| $y$ | $500$ | $600$ | $700$ |
For the second equation,$x + 3y = 1300$,we have $x = 1300 - 3y$.
| $x$ | $400$ | $700$ | $1000$ |
| $y$ | $300$ | $200$ | $100$ |
The geometric representation is shown in the provided graph.

(N/A) Let the cost of $1\, kg$ of apples be ₹ $x$ and the cost of $1\, kg$ of grapes be ₹ $y$.
According to the question,the algebraic representation is:
$2x + y = 160$
$4x + 2y = 300$
For the equation $2x + y = 160$,we have $y = 160 - 2x$. The solution table is:

$x$	$50$	$60$	$70$
$y$	$60$	$40$	$20$

For the equation $4x + 2y = 300$,we have $y = \frac{300 - 4x}{2} = 150 - 2x$. The solution table is:

$x$	$70$	$80$	$75$
$y$	$10$	$-10$	$0$

The graphical representation shows two parallel lines,indicating that the system has no solution.

(N/A) Let us draw the graphs of the Equations $(1)$ and $(2).$ For this,we find two solutions for each of the equations,which are given in the tables below:
For Equation $(1): x+3y=6$

$x$	$0$	$6$
$y = \frac{6-x}{3}$	$2$	$0$

For Equation $(2): 2x-3y=12$

$x$	$0$	$3$
$y = \frac{2x-12}{3}$	$-4$	$-2$

Plot the points $A(0, 2), B(6, 0)$ for line $(1)$ and $P(0, -4), Q(3, -2)$ for line $(2)$ on graph paper. Join the points to form the lines $AB$ and $PQ$.
We observe that there is a point $B(6, 0)$ common to both the lines $AB$ and $PQ.$ Since the lines intersect at a point,the pair of linear equations is consistent. The solution is $x=6$ and $y=0.$

(A) Given equations are:
$5x - 8y + 1 = 0$ $...(1)$
$3x - \frac{24}{5}y + \frac{3}{5} = 6$ $...(2)$
Multiply Equation $(2)$ by $\frac{5}{3}$:
$\frac{5}{3} \times (3x - \frac{24}{5}y + \frac{3}{5}) = \frac{5}{3} \times 6$
$5x - 8y + 1 = 10$
Wait,let us re-examine Equation $(2)$ as given: $3x - \frac{24}{5}y + \frac{3}{5} = 6$.
Multiplying by $\frac{5}{3}$ gives $5x - 8y + 1 = 10$,which simplifies to $5x - 8y - 9 = 0$.
Comparing $5x - 8y + 1 = 0$ and $5x - 8y - 9 = 0$,the coefficients of $x$ and $y$ are the same,but the constants are different.
Therefore,the lines are parallel and have no solution.

(A) Let us denote the number of pants by $x$ and the number of skirts by $y$.
According to the problem,the equations formed are:
$y = 2x - 2$ $...(1)$
$y = 4x - 4$ $...(2)$
To solve this pair of linear equations graphically,we find two solutions for each equation:
For equation $(1)$,$y = 2x - 2$:

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

For equation $(2)$,$y = 4x - 4$:

$x$	$0$	$1$
$y$	$-4$	$0$

Plotting these points on a graph and drawing the lines,we find that the two lines intersect at the point $(1, 0)$.
Thus,the solution is $x = 1$ and $y = 0$.
Therefore,Champa purchased $1$ pant and $0$ skirts.

(N/A) Let the number of girls be $x$ and the number of boys be $y$.
According to the question,the algebraic representation is:
$x + y = 10$
$x - y = 4$
For $x + y = 10$,we have $x = 10 - y$:

$x$	$5$	$4$	$6$
$y$	$5$	$6$	$4$

For $x - y = 4$,we have $x = 4 + y$:

$x$	$5$	$4$	$3$
$y$	$1$	$0$	$-1$

By plotting these points on a graph,we observe that the lines intersect at the point $(7, 3)$.
Therefore,the number of girls is $7$ and the number of boys is $3$.

(N/A) Let the cost of $1$ pencil be ₹ $x$ and the cost of $1$ pen be ₹ $y$.
According to the question,the algebraic representation is:
$5x + 7y = 50$
$7x + 5y = 46$
For $5x + 7y = 50$,we have $x = \frac{50 - 7y}{5}$.

$x$	$3$	$10$	$-4$
$y$	$5$	$0$	$10$

For $7x + 5y = 46$,we have $x = \frac{46 - 5y}{7}$.

$x$	$8$	$3$	$-2$
$y$	$-2$	$5$	$12$

By plotting these points on a graph,we observe that the lines intersect at the point $(3, 5)$.
Therefore,the cost of one pencil is ₹ $3$ and the cost of one pen is ₹ $5$.

(A) Given equations are:
$5x - 4y + 8 = 0$
$7x + 6y - 9 = 0$
Comparing these equations with the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,we obtain:
$a_{1} = 5, b_{1} = -4, c_{1} = 8$
$a_{2} = 7, b_{2} = 6, c_{2} = -9$
Now,calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{5}{7}$
$\frac{b_{1}}{b_{2}} = \frac{-4}{6} = -\frac{2}{3}$
Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$,the lines representing the given pair of linear equations intersect at exactly one point.

(C) Given equations are:
$9x + 3y + 12 = 0$
$18x + 6y + 24 = 0$
Comparing these equations with the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,we get:
$a_{1} = 9, b_{1} = 3, c_{1} = 12$
$a_{2} = 18, b_{2} = 6, c_{2} = 24$
Now,calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}$
$\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}$
$\frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}$
Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$,the lines representing the given pair of equations are coincident,and there are infinitely many solutions.

(B) Given equations are:
$6x - 3y + 10 = 0$
$2x - y + 9 = 0$
Comparing these equations with the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,we get:
$a_{1} = 6, b_{1} = -3, c_{1} = 10$
$a_{2} = 2, b_{2} = -1, c_{2} = 9$
Now,calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{6}{2} = \frac{3}{1} = 3$
$\frac{b_{1}}{b_{2}} = \frac{-3}{-1} = \frac{3}{1} = 3$
$\frac{c_{1}}{c_{2}} = \frac{10}{9}$
Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$,the lines representing the given pair of equations are parallel to each other. Therefore,these lines will never intersect at any point,and there is no solution for the given pair of equations.

(B) Given equations are:
$2x - 3y = 8$ (or $2x - 3y - 8 = 0$)
$4x - 6y = 9$ (or $4x - 6y - 9 = 0$)
Comparing with $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:
$a_{1} = 2, b_{1} = -3, c_{1} = -8$
$a_{2} = 4, b_{2} = -6, c_{2} = -9$
Calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_{1}}{b_{2}} = \frac{-3}{-6} = \frac{1}{2}$
$\frac{c_{1}}{c_{2}} = \frac{-8}{-9} = \frac{8}{9}$
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,the pair of linear equations has no solution and is inconsistent.

(A) Given equations are:
$1) \frac{3}{2} x + \frac{5}{3} y = 7 \implies \frac{3}{2} x + \frac{5}{3} y - 7 = 0$
$2) 9 x - 10 y = 14 \implies 9 x - 10 y - 14 = 0$
Comparing with $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$:
$a_{1} = \frac{3}{2}, b_{1} = \frac{5}{3}, c_{1} = -7$
$a_{2} = 9, b_{2} = -10, c_{2} = -14$
Now,calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{3/2}{9} = \frac{3}{18} = \frac{1}{6}$
$\frac{b_{1}}{b_{2}} = \frac{5/3}{-10} = \frac{5}{-30} = -\frac{1}{6}$
Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$,the lines intersect at a unique point.
Therefore,the pair of linear equations is consistent.

(A) Given equations are:
$5x - 3y = 11$ (Equation $1$)
$-10x + 6y = -22$ (Equation $2$)
Comparing with the standard form $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$:
$a_{1} = 5, b_{1} = -3, c_{1} = 11$
$a_{2} = -10, b_{2} = 6, c_{2} = -22$
Calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{5}{-10} = -\frac{1}{2}$
$\frac{b_{1}}{b_{2}} = \frac{-3}{6} = -\frac{1}{2}$
$\frac{c_{1}}{c_{2}} = \frac{11}{-22} = -\frac{1}{2}$
Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = -\frac{1}{2}$,the lines are coincident.
Because the lines are coincident,they have infinitely many solutions.
Therefore,the pair of linear equations is consistent.

(N/A) Given equations:
$x+y=5$
$2x+2y=10$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-10} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,these linear equations represent coincident lines and thus have infinitely many solutions. Hence,the pair of linear equations is consistent.
For $x+y=5$,we have $x = 5-y$:

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

For $2x+2y=10$,we have $x = \frac{10-2y}{2} = 5-y$:

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

From the graph,it can be observed that these lines overlap each other. Therefore,infinitely many solutions are possible for the given pair of equations.

(N/A) Given equations are $x-y=8$ and $3x-3y=16$.
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 1, b_1 = -1, c_1 = -8$
$a_2 = 3, b_2 = -3, c_2 = -16$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}$
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 to each other.
Therefore,there is no common point between the lines,which means the pair of linear equations is inconsistent.

(A) Given equations are $2x + y - 6 = 0$ and $4x - 2y - 4 = 0$.
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = 1, c_1 = -6$
$a_2 = 4, b_2 = -2, c_2 = -4$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines intersect at a unique point,meaning the system is consistent.
For $2x + y - 6 = 0 \implies y = 6 - 2x$:
If $x=0, y=6$; if $x=1, y=4$; if $x=2, y=2$.
For $4x - 2y - 4 = 0 \implies y = 2x - 2$:
If $x=0, y=-2$; if $x=1, y=0$; if $x=2, y=2$.
The lines intersect at $(2, 2)$.

(B) Given equations are:
$2x - 2y - 2 = 0$ (Equation $1$)
$4x - 4y - 5 = 0$ (Equation $2$)
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 2, b_1 = -2, c_1 = -2$
$a_2 = 4, b_2 = -4, c_2 = -5$
Now,find the ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other.
Therefore,the pair of linear equations is inconsistent and has no solution.

(N/A) Let the width of the garden be $x$ and the length be $y$.
According to the question:
$y - x = 4$ $...(1)$
$y + x = 36$ $...(2)$
From equation $(1)$,$y = x + 4$. The table of values is:

$x$	$0$	$8$	$12$
$y$	$4$	$12$	$16$

From equation $(2)$,$y = 36 - x$. The table of values is:

$x$	$0$	$36$	$16$
$y$	$36$	$0$	$20$

By plotting these lines on a graph,we observe that they intersect at the point $(16, 20)$.
Therefore,the width of the garden is $16\, m$ and the length is $20\, m$.

(N/A) For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to represent intersecting lines,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Given the equation $2x + 3y - 8 = 0$,we have $a_1 = 2$ and $b_1 = 3$.
We need to choose $a_2$ and $b_2$ such that $\frac{2}{a_2} \neq \frac{3}{b_2}$.
If we choose $a_2 = 3$ and $b_2 = 2$,then $\frac{2}{3} \neq \frac{3}{2}$,which satisfies the condition.
Thus,one such equation is $3x + 2y - 5 = 0$ (or any other equation satisfying the condition).

(A) For two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be parallel,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given the equation $2x + 3y - 8 = 0$,we have $a_1 = 2, b_1 = 3, c_1 = -8$.
To satisfy the condition,we can choose $a_2 = 4$ and $b_2 = 6$ (which makes $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{2}$) and choose $c_2$ such that $\frac{c_1}{c_2} \neq \frac{1}{2}$.
Let $c_2 = -10$. Then the equation is $4x + 6y - 10 = 0$.
Checking the condition: $\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-10}$,which simplifies to $\frac{1}{2} = \frac{1}{2} \neq \frac{4}{5}$.
Thus,$4x + 6y - 10 = 0$ is a valid answer.

(A) For two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to represent coincident lines,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given the equation $2x + 3y - 8 = 0$,we have $a_1 = 2, b_1 = 3, c_1 = -8$.
To find another equation,we can multiply the given equation by any non-zero constant $k$. Let $k = 2$.
Multiplying the equation by $2$,we get $2(2x + 3y - 8) = 2(0)$,which simplifies to $4x + 6y - 16 = 0$.
Checking the condition: $\frac{2}{4} = \frac{3}{6} = \frac{-8}{-16} = \frac{1}{2}$.
Since the ratios are equal,the lines are coincident. Thus,$4x + 6y - 16 = 0$ is a valid answer.

(N/A) For the equation $x-y+1=0$,we can write $x=y-1$. The table of values is:

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

For the equation $3x+2y-12=0$,we can write $x=\frac{12-2y}{3}$. The table of values is:

$x$	$4$	$2$	$0$
$y$	$0$	$3$	$6$

By plotting these points on a graph,we obtain two intersecting lines. The lines intersect each other at the point $(2,3)$ and intersect the $x$-axis at $(-1,0)$ and $(4,0)$.
Thus,the vertices of the triangle formed by these lines and the $x$-axis are $(2,3)$,$(-1,0)$,and $(4,0)$.

(D) Let the cost of one pencil be ₹ $x$ and the cost of one eraser be ₹ $y$.
According to the problem,the pair of linear equations formed are:
$2x + 3y = 9$ $...(1)$
$4x + 6y = 18$ $...(2)$
We observe that equation $(2)$ is exactly $2$ times equation $(1)$.
Dividing equation $(2)$ by $2$,we get $2x + 3y = 9$,which is identical to equation $(1)$.
Since both equations represent the same line,they are coincident.
Therefore,there are infinitely many solutions for $x$ and $y$ that satisfy these equations. We cannot determine a unique cost for each pencil and eraser.

(B) The given pair of linear equations is:
$x + 2y - 4 = 0$ $...(1)$
$2x + 4y - 12 = 0$ $...(2)$
Comparing these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 2, c_1 = -4$
$a_2 = 2, b_2 = 4, c_2 = -12$
Now,we find the ratios of the coefficients:
$\frac{a_1}{a_2} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$
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,the rails will not cross each other.

(C) The given equations are:
$6x - 3y + 10 = 0$ ... $(1)$
$2x - y + 9 = 0$ ... $(2)$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have:
$a_1 = 6, b_1 = -3, c_1 = 10$
$a_2 = 2, b_2 = -1, c_2 = 9$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{6}{2} = 3$
$\frac{b_1}{b_2} = \frac{-3}{-1} = 3$
$\frac{c_1}{c_2} = \frac{10}{9}$
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,the equations are $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$.
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 2, c_1 = 5$
$a_2 = -3, b_2 = -6, c_2 = 1$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}$
$\frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}$
$\frac{c_1}{c_2} = \frac{5}{1} = 5$
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,the pair of equations has no solution.

(C) The equation $x=a$ represents a vertical line parallel to the $y$-axis,passing through the point $(a, 0)$ on the $x$-axis.
The equation $y=b$ represents a horizontal line parallel to the $x$-axis,passing through the point $(0, b)$ on the $y$-axis.
When these two lines are plotted on the same Cartesian plane,the vertical line $x=a$ and the horizontal line $y=b$ intersect at exactly one point.
The $x$-coordinate of the intersection point is $a$ and the $y$-coordinate is $b$.
Therefore,the lines intersect at the point $(a, b)$.

(D) The condition for two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be coincident is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $3x - y + 8 = 0$ and $6x - ky + 16 = 0$.
Comparing with the standard form,we have $a_1 = 3, b_1 = -1, c_1 = 8$ and $a_2 = 6, b_2 = -k, c_2 = 16$.
Substituting these values into the condition:
$\frac{3}{6} = \frac{-1}{-k} = \frac{8}{16}$.
Simplifying the ratios:
$\frac{1}{2} = \frac{1}{k} = \frac{1}{2}$.
From $\frac{1}{k} = \frac{1}{2}$,we get $k = 2$.

(B) Given equations are:
$4x + 3y - 1 = 5 \implies 4x + 3y = 6$
$12x + 9y = 15$
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
$a_1 = 4, b_1 = 3, c_1 = 6$
$a_2 = 12, b_2 = 9, c_2 = 15$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{4}{12} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{6}{15} = \frac{2}{5}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel and not coincident. Therefore,the answer is No.

(A) Given equations are $x+2y-3=0$ and $3x+6y-9=0$.
Comparing these with the standard form $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$,we get:
$a_1=1, b_1=2, c_1=-3$
$a_2=3, b_2=6, c_2=-9$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the lines are coincident.
Therefore,the pair of equations has infinitely many solutions and is consistent.

(B) The condition for a pair of linear equations to have no solution is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are:
$2x + 4y - 3 = 0$
$x - 2y = 0$
Here,$a_1 = 2, b_1 = 4, c_1 = -3$ and $a_2 = 1, b_2 = -2, c_2 = 0$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{1} = 2$
$\frac{b_1}{b_2} = \frac{4}{-2} = -2$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $(2 \neq -2)$,the lines intersect at a unique point.
Therefore,the given pair of linear equations has a unique solution,not 'no solution'.

(B) The condition for two lines to be coincident is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given the pair of linear equations:
$3x + \frac{1}{7}y - 3 = 0$
$7x + 3y - 7 = 0$
Here,the coefficients are:
$a_1 = 3, b_1 = \frac{1}{7}, c_1 = -3$
$a_2 = 7, b_2 = 3, c_2 = -7$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{3}{7}$
$\frac{b_1}{b_2} = \frac{1/7}{3} = \frac{1}{21}$
$\frac{c_1}{c_2} = \frac{-3}{-7} = \frac{3}{7}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines are not coincident. They intersect at a unique point.

(N/A) The condition for two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be coincident is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are:
$1) \frac{1}{2}x + 1y + \frac{2}{5} = 0$
$2) 4x + 8y + \frac{5}{16} = 0$
Here,$a_1 = \frac{1}{2}, b_1 = 1, c_1 = \frac{2}{5}$ and $a_2 = 4, b_2 = 8, c_2 = \frac{5}{16}$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{1/2}{4} = \frac{1}{8}$
$\frac{b_1}{b_2} = \frac{1}{8}$
$\frac{c_1}{c_2} = \frac{2/5}{5/16} = \frac{2}{5} \times \frac{16}{5} = \frac{32}{25}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel and not coincident. Therefore,the given equations do not represent a pair of coincident lines.

(N/A) The statement is false.
By observing the Cartesian plane,the equation $x=7$ represents a vertical line that passes through the point $(7, 0)$ on the $x$-axis.
Since the line is vertical,it is parallel to the $y$-axis and perpendicular to the $x$-axis.
Therefore,the given statement is incorrect.

(B) The given pair of linear equations is:
$3x - y - 5 = 0$ ... $(i)$
$6x - 2y - p = 0$ ... $(ii)$
Comparing these with the general form $ax + by + c = 0$,we get:
$a_1 = 3, b_1 = -1, c_1 = -5$
$a_2 = 6, b_2 = -2, c_2 = -p$
For two lines to be parallel,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Substituting the values,we get:
$\frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{-p}$
$\frac{1}{2} = \frac{1}{2} \neq \frac{5}{p}$
For the lines to be parallel,we must have $\frac{1}{2} \neq \frac{5}{p}$.
Cross-multiplying,we get $p \neq 10$.
Thus,the lines are parallel for all real values of $p$ except $10$.

(B) આપેલ સમીકરણોને પ્રમાણિત સ્વરૂપમાં લખતા:
$3x + y - 1 = 0$
$(2k - 1)x + (k - 1)y - (2k + 1) = 0$
અહીં,$a_1 = 3, b_1 = 1, c_1 = -1$ અને $a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1)$ છે.
સુરેખ સમીકરણોની જોડને કોઈ ઉકેલ ન હોય તેની શરત છે:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
તેથી,$\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)}$
$\frac{3}{2k - 1} = \frac{1}{k - 1}$ લેતા:
$3(k - 1) = 1(2k - 1)$
$3k - 3 = 2k - 1$
$3k - 2k = -1 + 3$
$k = 2$
આમ,$k = 2$ માટે આપેલ સમીકરણોની જોડને કોઈ ઉકેલ નથી.

(N/A) For the equation $5x - y = 5$,we can write $y = 5x - 5$. The solution table is:

$x$	$0$	$1$	$2$
$y$	$-5$	$0$	$5$

For the equation $3x - y = 3$,we can write $y = 3x - 3$. The solution table is:

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

By plotting these points on a graph,we obtain two lines that intersect at the point $(1, 0)$.
The triangle is formed by the intersection of these two lines and the $y$-axis (where $x = 0$).
The vertices of the triangle are the intersection point of the two lines $(1, 0)$ and the points where each line intersects the $y$-axis,which are $(0, -3)$ and $(0, -5)$.
Thus,the coordinates of the vertices are $(1, 0)$,$(0, -3)$,and $(0, -5)$.

(B) The given linear equations are:
$x-3y-2=0$ ..... $(i)$
$-2x+6y-5=0$ ..... $(ii)$
Comparing these equations with the general form $ax+by+c=0$,we get:
For equation $(i)$: $a_1=1, b_1=-3, c_1=-2$
For equation $(ii)$: $a_2=-2, b_2=6, c_2=-5$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{-2} = -\frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel.
Therefore,the two straight paths represented by the given equations never cross each other.

(A) The given pair of equations is:
$3x + y + 4 = 0 .....(i)$
$6x - 2y + 4 = 0 .....(ii)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 3, b_1 = 1, c_1 = 4$
$a_2 = 6, b_2 = -2, c_2 = 4$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines intersect at a unique point. Therefore,the system is consistent.
For equation $(i)$,$y = -3x - 4$:

$x$	$0$	$-1$	$-2$
$y$	$-4$	$-1$	$2$

For equation $(ii)$,$2y = 6x + 4 \Rightarrow y = 3x + 2$:

$x$	$-1$	$0$	$1$
$y$	$-1$	$2$	$5$

Plotting these lines on a graph,they intersect at the point $(-1, -1)$. Thus,the solution is $x = -1, y = -1$.

(D) The given pair of equations is:
$x - 2y = 6$ ..... $(i)$
$3x - 6y = 0$ ..... $(ii)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = -2, c_1 = -6$
$a_2 = 3, b_2 = -6, c_2 = 0$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-2}{-6} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-6}{0}$ (which is undefined)
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.
Because the lines are parallel,they do not intersect at any point.
Therefore,the given pair of linear equations is inconsistent and has no solution.

(CONSISTENT) Given pair of equations is:
$x+y=3 .....(i)$
$3x+3y=9 .....(ii)$
On comparing with $ax+by+c=0$,we get:
$a_1=1, b_1=1, c_1=-3$ [from Eq. $(i)$]
$a_2=3, b_2=3, c_2=-9$ [from Eq. $(ii)$]
$\frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the given pair of lines is coincident. Therefore,these lines have infinitely many solutions. Hence,the given pair of linear equations is consistent.
Now,for $x+y=3 \Rightarrow y=3-x$:

$x$	$0$	$3$
$y$	$3$	$0$
Points	$A$	$B$

For $3x+3y=9 \Rightarrow y = \frac{9-3x}{3} = 3-x$:

$x$	$0$	$1$	$3$
$y$	$3$	$2$	$0$
Points	$C$	$D$	$E$

Plotting the points $A(0,3)$ and $B(3,0)$,we get the line $AB$. Similarly,plotting the points $C(0,3), D(1,2)$,and $E(3,0)$,we get the same line. We observe that the lines represented by Eqs. $(i)$ and $(ii)$ are coincident.

(N/A) We know that the graph of $x=-2$ is a line parallel to the $y$-axis at a distance of $2$ units to the left of it.
The graph of $y=3$ is a line parallel to the $x$-axis at a distance of $3$ units above it.
The figure enclosed by the lines $x=-2$,$y=3$,the $x$-axis,and the $y$-axis is $OABC$,which is a rectangle.
$A$ is a point on the $y$-axis at a distance of $3$ units above the $x$-axis. So,the coordinates of $A$ are $(0, 3)$.
$C$ is a point on the $x$-axis at a distance of $2$ units to the left of the $y$-axis. So,the coordinates of $C$ are $(-2, 0)$.
$B$ is the intersection point of the lines $x=-2$ and $y=3$. So,the coordinates of $B$ are $(-2, 3)$.
The origin $O$ is $(0, 0)$.
Thus,the vertices of the rectangle $OABC$ are $O(0, 0)$,$A(0, 3)$,$B(-2, 3)$,and $C(-2, 0)$.
The length and breadth of this rectangle are $2$ units and $3$ units,respectively.
Since the area of a rectangle $=$ length $\times$ breadth,
the area of rectangle $OABC = 2 \times 3 = 6 \text{ sq. units}$.

(O(0,0), Q(4,4), D(6,2)) The given linear equations are:
$y=x \quad (i)$
$3y=x \quad (ii)$
$x+y=8 \quad (iii)$
To find the vertices,we determine the intersection points of these lines taken in pairs.
$1$. Intersection of $(i)$ and $(ii)$:
Substituting $y=x$ into $3y=x$,we get $3x=x$,which implies $2x=0$,so $x=0$. Thus,$y=0$. The intersection point is $O(0,0)$.
$2$. Intersection of $(i)$ and $(iii)$:
Substituting $y=x$ into $x+y=8$,we get $x+x=8$,so $2x=8$,which means $x=4$. Thus,$y=4$. The intersection point is $Q(4,4)$.
$3$. Intersection of $(ii)$ and $(iii)$:
From $(ii)$,$x=3y$. Substituting this into $(iii)$,we get $3y+y=8$,so $4y=8$,which means $y=2$. Then $x=3(2)=6$. The intersection point is $D(6,2)$.
Thus,the vertices of the triangle formed by the lines are $O(0,0)$,$Q(4,4)$,and $D(6,2)$.

(8) Given equations of lines are $2x-y-4=0$,$x=3$,and $x=5$.
For the line $2x-y-4=0$,we find two points:

$x$	$0$	$2$
$y=2x-4$	$-4$	$0$

Plot the points $P(0, -4)$ and $Q(2, 0)$ and draw the line passing through them. Also,draw the vertical lines $x=3$ and $x=5$.
The quadrilateral formed by the lines $x=3$,$x=5$,the $x$-axis,and the line $2x-y-4=0$ is a trapezium with parallel sides at $x=3$ and $x=5$.
At $x=3$,$y = 2(3)-4 = 2$. So,point $D$ is $(3, 2)$.
At $x=5$,$y = 2(5)-4 = 6$. So,point $C$ is $(5, 6)$.
The parallel sides of the trapezium are $AD = 2$ units and $BC = 6$ units. The distance between them is $AB = 5-3 = 2$ units.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
$= \frac{1}{2} \times (AD + BC) \times AB$
$= \frac{1}{2} \times (2 + 6) \times 2$
$= 8 \text{ sq units}$.
Thus,the area of the quadrilateral is $8 \text{ sq units}$.

(A) $x+y=7$
$\therefore y=7-x$
For $x=0, y=7-0=7$
For $x=7, y=7-7=0$

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

$\therefore$ Plot the ordered pairs $(0, 7)$ and $(7, 0)$ of the solution set of $x+y=7$ on the graph paper and draw the line by joining them.
$5x+2y=20$
$\therefore 2y=20-5x \quad \therefore y=\frac{20-5x}{2}$
For $x=0, y=\frac{20-0}{2}=10$
For $x=4, y=\frac{20-20}{2}=0$

$x$	$0$	$4$
$y$	$10$	$0$

$\therefore$ Plot the ordered pairs $(0, 10)$ and $(4, 0)$ of the solution set of $5x+2y=20$ on the graph paper and draw the line by joining them.
The intersection point (common point) of these two lines is $(2, 5)$,which satisfies both equations.
Thus,the solution of the pair of linear equations is $(2, 5)$.

(N/A) For the equation $3x + 6y = 3900$:
Dividing by $3$,we get $x + 2y = 1300$,which implies $y = \frac{1300 - x}{2}$.
If $x = 0$,$y = 650$. If $x = 1300$,$y = 0$.

$x$	$0$	$1300$
$y$	$650$	$0$

Plot the points $(0, 650)$ and $(1300, 0)$ and join them to draw the line.
For the equation $x + 3y = 1300$:
This implies $y = \frac{1300 - x}{3}$.
If $x = 1300$,$y = 0$. If $x = 100$,$y = 400$.

$x$	$1300$	$100$
$y$	$0$	$400$

Plot the points $(1300, 0)$ and $(100, 400)$ and join them to draw the line.
The intersection point of these two lines is $(1300, 0)$,which satisfies both equations.
Thus,the solution is $x = 1300, y = 0$.