Textbook - Pair of Linear Equations in Two Variables 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.

(A) Given equations are:
$3x + 2y - 5 = 0$
$2x - 3y - 7 = 0$
Comparing these 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:
$a_{1} = 3, b_{1} = 2, c_{1} = -5$
$a_{2} = 2, b_{2} = -3, c_{2} = -7$
Now,calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{3}{2}$
$\frac{b_{1}}{b_{2}} = \frac{2}{-3} = -\frac{2}{3}$
Since $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$,the lines intersect at a single point.
Therefore,the pair of linear equations has a unique solution and is consistent.

(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.

(A) Given equations are:
$1) \frac{4}{3} x + 2 y - 8 = 0$
$2) 2 x + 3 y - 12 = 0$
Comparing with the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:
$a_{1} = \frac{4}{3}, b_{1} = 2, c_{1} = -8$
$a_{2} = 2, b_{2} = 3, c_{2} = -12$
Calculating the ratios:
$\frac{a_{1}}{a_{2}} = \frac{4/3}{2} = \frac{4}{6} = \frac{2}{3}$
$\frac{b_{1}}{b_{2}} = \frac{2}{3}$
$\frac{c_{1}}{c_{2}} = \frac{-8}{-12} = \frac{2}{3}$
Since $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$,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 $(4,0)$.

(N/A) Step $1:$ We pick either of the equations and write one variable in terms of the other. Let us consider Equation $(2)$.
$x + 2y = 3$
We can write it as $x = 3 - 2y$ $...(3)$
Step $2:$ Substitute the value of $x$ from Equation $(3)$ into Equation $(1)$. We get:
$7(3 - 2y) - 15y = 2$
$21 - 14y - 15y = 2$
$-29y = 2 - 21$
$-29y = -19$
Therefore,$y = \frac{19}{29}$
Step $3:$ Substituting this value of $y$ in Equation $(3)$,we get:
$x = 3 - 2\left(\frac{19}{29}\right)$
$x = 3 - \frac{38}{29}$
$x = \frac{87 - 38}{29} = \frac{49}{29}$
Therefore,the solution is $x = \frac{49}{29}, y = \frac{19}{29}$.

(N/A) Let $s$ and $t$ be the current ages (in years) of Aftab and his daughter,respectively.
According to the first condition,seven years ago:
$(s - 7) = 7(t - 7)$
$s - 7 = 7t - 49$
$s - 7t = -42$ $...(1)$
According to the second condition,three years from now:
$(s + 3) = 3(t + 3)$
$s + 3 = 3t + 9$
$s - 3t = 6$ $...(2)$
From Equation $(2)$,we get $s = 3t + 6$.
Substituting this value of $s$ into Equation $(1)$:
$(3t + 6) - 7t = -42$
$-4t + 6 = -42$
$-4t = -48$
$t = 12$
Now,substitute $t = 12$ into $s = 3t + 6$:
$s = 3(12) + 6 = 36 + 6 = 42$
Thus,the current age of Aftab is $42$ years and his daughter is $12$ years old.

(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.

(A) $x+y=14$ $....(1)$
$x-y=4$ $....(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$x=14-y$ $....(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$(14-y)-y=4$
$14-2y=4$
$14-4=2y$
$10=2y$
$y=5$ $....(4)$
Now,substituting the value of $y=5$ into equation $(3)$:
$x=14-5$
$x=9$
Therefore,the solution is $x=9$ and $y=5$.

(S=9, T=6) $s-t=3$ $...(1)$
$\frac{s}{3}+\frac{t}{2}=6$ $...(2)$
From equation $(1)$,we express $s$ in terms of $t$:
$s = t + 3$ $...(3)$
Substituting the value of $s$ from equation $(3)$ into equation $(2)$:
$\frac{t+3}{3} + \frac{t}{2} = 6$
Multiply the entire equation by $6$ (the $LCM$ of $3$ and $2$) to clear the denominators:
$2(t+3) + 3t = 36$
$2t + 6 + 3t = 36$
$5t + 6 = 36$
$5t = 30$
$t = 6$ $...(4)$
Now,substitute the value of $t = 6$ into equation $(3)$:
$s = 6 + 3$
$s = 9$
Therefore,the solution is $s = 9$ and $t = 6$.

(D) Given equations are:
$3x - y = 3$ $...(1)$
$9x - 3y = 9$ $...(2)$
From equation $(1)$,we can express $y$ in terms of $x$:
$y = 3x - 3$ $...(3)$
Substituting the value of $y$ from equation $(3)$ into equation $(2)$:
$9x - 3(3x - 3) = 9$
$9x - 9x + 9 = 9$
$9 = 9$
Since $9 = 9$ is a true statement for all values of $x$,the two equations are coincident lines.
Therefore,the given pair of linear equations has infinitely many solutions.
The relationship between the variables is given by $y = 3x - 3$.

(N/A) Given equations are:
$0.2 x + 0.3 y = 1.3$ $...(1)$
$0.4 x + 0.5 y = 2.3$ $...(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$0.2 x = 1.3 - 0.3 y$
$x = \frac{1.3 - 0.3 y}{0.2}$ $...(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$0.4 \left( \frac{1.3 - 0.3 y}{0.2} \right) + 0.5 y = 2.3$
$2(1.3 - 0.3 y) + 0.5 y = 2.3$
$2.6 - 0.6 y + 0.5 y = 2.3$
$2.6 - 0.1 y = 2.3$
$2.6 - 2.3 = 0.1 y$
$0.3 = 0.1 y$
$y = \frac{0.3}{0.1} = 3$
Now,substitute $y = 3$ into equation $(3)$:
$x = \frac{1.3 - 0.3(3)}{0.2}$
$x = \frac{1.3 - 0.9}{0.2}$
$x = \frac{0.4}{0.2} = 2$
Therefore,the solution is $x = 2$ and $y = 3$.

(A) Given equations are:
$\sqrt{2} x + \sqrt{3} y = 0$ $...(1)$
$\sqrt{3} x - \sqrt{8} y = 0$ $...(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$x = -\frac{\sqrt{3} y}{\sqrt{2}}$ $...(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$\sqrt{3} \left( -\frac{\sqrt{3} y}{\sqrt{2}} \right) - \sqrt{8} y = 0$
$-\frac{3 y}{\sqrt{2}} - 2\sqrt{2} y = 0$
Factoring out $y$:
$y \left( -\frac{3}{\sqrt{2}} - 2\sqrt{2} \right) = 0$
Since the term in the bracket is non-zero,we must have $y = 0$.
Substituting $y = 0$ into equation $(3)$:
$x = -\frac{\sqrt{3} (0)}{\sqrt{2}} = 0$
Thus,the solution is $x = 0$ and $y = 0$.

(A) Given equations are:
$(1) \frac{3x}{2} - \frac{5y}{3} = -2$
$(2) \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Step $1$: Simplify the equations.
Multiply $(1)$ by $6$: $9x - 10y = -12$ --- $(3)$
Multiply $(2)$ by $6$: $2x + 3y = 13$ --- $(4)$
Step $2$: Express $x$ in terms of $y$ from equation $(4)$:
$2x = 13 - 3y$
$x = \frac{13 - 3y}{2}$ --- $(5)$
Step $3$: Substitute $(5)$ into $(3)$:
$9(\frac{13 - 3y}{2}) - 10y = -12$
Multiply by $2$: $9(13 - 3y) - 20y = -24$
$117 - 27y - 20y = -24$
$-47y = -24 - 117$
$-47y = -141$
$y = \frac{141}{47} = 3$
Step $4$: Substitute $y = 3$ into $(5)$:
$x = \frac{13 - 3(3)}{2} = \frac{13 - 9}{2} = \frac{4}{2} = 2$
Thus,the solution is $x = 2, y = 3$.

(M = -1) Given equations are:
$2x + 3y = 11$ $...(1)$
$2x - 4y = -24$ $...(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$2x = 11 - 3y$
$x = \frac{11 - 3y}{2}$ $...(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$2\left(\frac{11 - 3y}{2}\right) - 4y = -24$
$11 - 3y - 4y = -24$
$11 - 7y = -24$
$-7y = -24 - 11$
$-7y = -35$
$y = 5$
Now,substitute $y = 5$ into equation $(3)$ to find $x$:
$x = \frac{11 - 3(5)}{2} = \frac{11 - 15}{2} = \frac{-4}{2} = -2$
So,the solution is $x = -2$ and $y = 5$.
Now,substitute these values into the equation $y = mx + 3$ to find $m$:
$5 = m(-2) + 3$
$5 - 3 = -2m$
$2 = -2m$
$m = -1$

(N/A) Let the first number be $x$ and the other number be $y$ such that $y > x$.
According to the given information:
$y = 3x$ $...(1)$
$y - x = 26$ $...(2)$
On substituting the value of $y$ from equation $(1)$ into equation $(2)$,we obtain:
$3x - x = 26$
$2x = 26$
$x = 13$ $...(3)$
Substituting the value of $x$ in equation $(1)$,we obtain:
$y = 3(13) = 39$
Hence,the two numbers are $13$ and $39$.

(A) Let the larger angle be $x$ and the smaller angle be $y$.
We know that the sum of the measures of angles of a supplementary pair is always $180^{\circ}$.
According to the given information:
$x + y = 180^{\circ}$ $...(1)$
$x - y = 18^{\circ}$ $...(2)$
From equation $(1)$,we obtain:
$x = 180^{\circ} - y$ $...(3)$
Substituting the value of $x$ in equation $(2)$,we get:
$(180^{\circ} - y) - y = 18^{\circ}$
$180^{\circ} - 2y = 18^{\circ}$
$180^{\circ} - 18^{\circ} = 2y$
$162^{\circ} = 2y$
$y = 81^{\circ}$ $...(4)$
Substituting the value of $y$ in equation $(3)$,we get:
$x = 180^{\circ} - 81^{\circ}$
$x = 99^{\circ}$
Hence,the two angles are $99^{\circ}$ and $81^{\circ}$.

(A) Let the cost of a bat be $x$ and the cost of a ball be $y$.
According to the given information:
$7x + 6y = 3800$ $...(1)$
$3x + 5y = 1750$ $...(2)$
From equation $(1)$,we express $y$ in terms of $x$:
$6y = 3800 - 7x$
$y = \frac{3800 - 7x}{6}$ $...(3)$
Substituting this value of $y$ into equation $(2)$:
$3x + 5\left(\frac{3800 - 7x}{6}\right) = 1750$
Multiply the entire equation by $6$ to simplify:
$18x + 5(3800 - 7x) = 10500$
$18x + 19000 - 35x = 10500$
$-17x = 10500 - 19000$
$-17x = -8500$
$x = 500$
Now,substitute $x = 500$ into equation $(3)$:
$y = \frac{3800 - 7(500)}{6}$
$y = \frac{3800 - 3500}{6}$
$y = \frac{300}{6} = 50$
Therefore,the cost of one bat is ₹ $500$ and the cost of one ball is ₹ $50$.

(D) Let the fixed charge be ₹ $x$ and the charge per $km$ be ₹ $y$.
According to the given information:
$x + 10y = 105$ $(1)$
$x + 15y = 155$ $(2)$
From equation $(1)$,we obtain:
$x = 105 - 10y$ $(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$(105 - 10y) + 15y = 155$
$105 + 5y = 155$
$5y = 155 - 105$
$5y = 50$
$y = 10$
Substituting $y = 10$ in equation $(3)$:
$x = 105 - 10(10)$
$x = 105 - 100$
$x = 5$
Thus,the fixed charge is ₹ $5$ and the charge per $km$ is ₹ $10$.
For a distance of $25 \, km$,the total charge is:
$x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255$.

(A) Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.
According to the first condition: $\frac{x+2}{y+2} = \frac{9}{11}$.
Cross-multiplying gives: $11(x+2) = 9(y+2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4$ (Equation $1$).
According to the second condition: $\frac{x+3}{y+3} = \frac{5}{6}$.
Cross-multiplying gives: $6(x+3) = 5(y+3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3$ (Equation $2$).
From Equation $2$,$5y = 6x + 3 \implies y = \frac{6x+3}{5}$.
Substituting this into Equation $1$: $11x - 9(\frac{6x+3}{5}) = -4$.
Multiply by $5$: $55x - 9(6x+3) = -20 \implies 55x - 54x - 27 = -20 \implies x = 7$.
Now,find $y$: $y = \frac{6(7)+3}{5} = \frac{42+3}{5} = \frac{45}{5} = 9$.
The fraction is $\frac{7}{9}$.

(N/A) Let the present age of Jacob be $x$ years and the present age of his son be $y$ years.
Case $1$: Five years hence,Jacob's age will be $(x + 5)$ and his son's age will be $(y + 5)$.
According to the problem,$(x + 5) = 3(y + 5) \implies x + 5 = 3y + 15 \implies x - 3y = 10$ (Equation $1$).
Case $2$: Five years ago,Jacob's age was $(x - 5)$ and his son's age was $(y - 5)$.
According to the problem,$(x - 5) = 7(y - 5) \implies x - 5 = 7y - 35 \implies x - 7y = -30$ (Equation $2$).
From Equation $1$,$x = 3y + 10$.
Substituting this value into Equation $2$: $(3y + 10) - 7y = -30 \implies -4y = -40 \implies y = 10$.
Substituting $y = 10$ into $x = 3y + 10$: $x = 3(10) + 10 = 40$.
Therefore,the present age of Jacob is $40$ years and the present age of his son is $10$ years.

(A) Let the incomes of the two persons be ₹ $9x$ and ₹ $7x$,and their expenditures be ₹ $4y$ and ₹ $3y$ respectively.
Since $\text{Income} - \text{Expenditure} = \text{Savings}$,we have:
$9x - 4y = 2000$ $...(1)$
$7x - 3y = 2000$ $...(2)$
To solve these equations,multiply equation $(1)$ by $3$ and equation $(2)$ by $4$:
$27x - 12y = 6000$ $...(3)$
$28x - 12y = 8000$ $...(4)$
Subtracting equation $(3)$ from equation $(4)$:
$(28x - 27x) - (12y - 12y) = 8000 - 6000$
$x = 2000$
Substituting $x = 2000$ into equation $(1)$:
$9(2000) - 4y = 2000$
$18000 - 4y = 2000$
$4y = 16000$
$y = 4000$
The monthly incomes are $9x = 9 \times 2000 = ₹ 18,000$ and $7x = 7 \times 2000 = ₹ 14,000$.

(D) Step $1$: Multiply Equation $(1)$ by $2$ and Equation $(2)$ by $1$ to make the coefficients of $x$ equal. Then we get the equations as:
$4x + 6y = 16$ $...(3)$
$4x + 6y = 7$ $...(4)$
Step $2$: Subtracting Equation $(4)$ from Equation $(3)$:
$(4x - 4x) + (6y - 6y) = 16 - 7$
i.e.,$0 = 9$,which is a false statement.
Therefore,the pair of linear equations has no solution.

(A) Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$.
When the digits are reversed,the new number is $10y + x$.
According to the problem,$(10x + y) + (10y + x) = 66$.
$11x + 11y = 66$,which simplifies to $x + y = 6$ $...(1)$.
Given that the digits differ by $2$,we have two cases:
Case $1$: $x - y = 2$ $...(2)$.
Adding $(1)$ and $(2)$: $2x = 8 \implies x = 4$. Substituting $x=4$ in $(1)$,$y = 2$. The number is $42$.
Case $2$: $y - x = 2$ $...(3)$.
Adding $(1)$ and $(3)$: $2y = 8 \implies y = 4$. Substituting $y=4$ in $(1)$,$x = 2$. The number is $24$.
Thus,there are two such numbers: $42$ and $24$.

(N/A) By elimination method:
$x+y=5$ $...(1)$
$2x-3y=4$ $...(2)$
Multiplying equation $(1)$ by $2$,we obtain:
$2x+2y=10$ $...(3)$
Subtracting equation $(2)$ from equation $(3)$,we obtain:
$(2x+2y) - (2x-3y) = 10 - 4$
$5y = 6$
$y = \frac{6}{5}$ $...(4)$
Substituting the value of $y$ in equation $(1)$,we obtain:
$x + \frac{6}{5} = 5$
$x = 5 - \frac{6}{5} = \frac{25-6}{5} = \frac{19}{5}$
Therefore,$x = \frac{19}{5}, y = \frac{6}{5}$.
By substitution method:
From equation $(1)$,we obtain:
$x = 5 - y$ $...(5)$
Substituting this value in equation $(2)$,we obtain:
$2(5-y) - 3y = 4$
$10 - 2y - 3y = 4$
$10 - 5y = 4$
$-5y = 4 - 10$
$-5y = -6$
$y = \frac{6}{5}$
Substituting the value of $y$ in equation $(5)$,we obtain:
$x = 5 - \frac{6}{5} = \frac{19}{5}$
Therefore,$x = \frac{19}{5}, y = \frac{6}{5}$.

(N/A) By elimination method:
$3x + 4y = 10$ $...(1)$
$2x - 2y = 2$ $...(2)$
Multiplying equation $(2)$ by $2$,we obtain:
$4x - 4y = 4$ $...(3)$
Adding equation $(1)$ and $(3)$,we obtain:
$7x = 14$
$x = 2$ $...(4)$
Substituting $x = 2$ in equation $(1)$,we obtain:
$3(2) + 4y = 10$
$6 + 4y = 10$
$4y = 4$
$y = 1$
Hence,$x = 2, y = 1$.
By substitution method:
From equation $(2)$,we obtain:
$2x = 2 + 2y$
$x = 1 + y$ $...(5)$
Putting this value in equation $(1)$,we obtain:
$3(1 + y) + 4y = 10$
$3 + 3y + 4y = 10$
$7y = 7$
$y = 1$
Substituting the value $y = 1$ in equation $(5)$,we obtain:
$x = 1 + 1 = 2$
Therefore,$x = 2, y = 1$.