Mix Examples - Pair of Linear Equations in Two Variables Questions in English

(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.

(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.

(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 $m$:
$5 = m(-2) + 3$
$5 - 3 = -2m$
$2 = -2m$
$m = -1$

(A) Given equations are:
$5x - 15y = 8$ --- $(1)$
$3x - 9y = \frac{24}{5}$ --- $(2)$
Comparing these with the standard form $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:
$a_1 = 5, b_1 = -15, c_1 = 8$
$a_2 = 3, b_2 = -9, c_2 = \frac{24}{5}$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{5}{3}$
$\frac{b_1}{b_2} = \frac{-15}{-9} = \frac{5}{3}$
$\frac{c_1}{c_2} = \frac{8}{24/5} = \frac{8 \times 5}{24} = \frac{40}{24} = \frac{5}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{5}{3}$,the lines are coincident.
Therefore,the pair of equations has infinitely many solutions.

(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.
According to the problem,the sum of the digits is $x + y = 9$ (Equation $1$).
When $27$ is added to the number,the digits are reversed,so the new number is $10y + x$.
Thus,$(10x + y) + 27 = 10y + x$.
Rearranging the terms: $10x - x + y - 10y = -27$,which simplifies to $9x - 9y = -27$.
Dividing by $9$,we get $x - y = -3$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(x + y) + (x - y) = 9 + (-3)$,which gives $2x = 6$,so $x = 3$.
Substituting $x = 3$ into Equation $1$: $3 + y = 9$,so $y = 6$.
The number is $10(3) + 6 = 36$.

(A) pair of linear equations is said to be consistent if it has at least one solution.
$1$. If the lines intersect at a single point,the system has a unique solution,and the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$2$. If the lines are coincident,the system has infinitely many solutions,and the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Since both cases provide at least one solution,the lines are either intersecting or coincident.

(B) The given pair of equations are $y=0$ and $y=-7$.
$y=0$ represents the $x$-axis.
$y=-7$ represents a line parallel to the $x$-axis at a distance of $7$ units below it.
Since both lines are parallel to each other,they will never intersect.
Therefore,the pair of equations has no solution.

(A) The condition for two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to be parallel is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given the equations:
$3x + 2ky - 2 = 0$
$2x + 5y + 1 = 0$
Here,$a_1 = 3, b_1 = 2k, c_1 = -2$ and $a_2 = 2, b_2 = 5, c_2 = 1$.
For the lines to be parallel,we must have $\frac{a_1}{a_2} = \frac{b_1}{b_2}$.
Substituting the values,we get $\frac{3}{2} = \frac{2k}{5}$.
Cross-multiplying,we get $3 \times 5 = 2 \times 2k$,which implies $15 = 4k$.
Therefore,$k = \frac{15}{4}$.

(B) The condition for a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
The given equations are $cx - y - 2 = 0$ and $6x - 2y - 3 = 0$.
Here,$a_1 = c, b_1 = -1, c_1 = -2$ and $a_2 = 6, b_2 = -2, c_2 = -3$.
Applying the condition: $\frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3}$.
This simplifies to $\frac{c}{6} = \frac{1}{2} = \frac{2}{3}$.
Since $\frac{1}{2} \neq \frac{2}{3}$,there is no value of $c$ that satisfies both equalities simultaneously.
Therefore,the pair of equations will have no value of $c$ for which they have infinitely many solutions.

(C) For a pair of linear equations to be dependent,they must represent the same line,meaning their coefficients must be proportional: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k$.
Given equation: $-5x + 7y - 2 = 0$.
Comparing with $a_1x + b_1y + c_1 = 0$,we have $a_1 = -5, b_1 = 7, c_1 = -2$.
For the second equation,we multiply the given equation by a constant $k$. Let $k = 2$:
$2(-5x + 7y - 2) = 2(0)$
$-10x + 14y - 4 = 0$
Rearranging this,we get $10x - 14y + 4 = 0$.
Comparing this with option $(C)$: $10x - 14y = -4$ is equivalent to $10x - 14y + 4 = 0$.

(C) For a pair of linear equations to have a unique solution $(x=2, y=-3)$,these values must satisfy both equations in the pair.
Let us check the options:
Option $(A): x+y=1 \implies 2+(-3) = -1 \neq 1$. (Incorrect)
Option $(B): 2x-y=1 \implies 2(2)-(-3) = 4+3 = 7 \neq 1$. (Incorrect)
Option $(C): x-4y-14=0 \implies 2-4(-3)-14 = 2+12-14 = 0$. (Satisfied)
$5x-y-13=0 \implies 5(2)-(-3)-13 = 10+3-13 = 0$. (Satisfied)
Since both equations in option $(C)$ are satisfied by $(x=2, y=-3)$ and the lines are not coincident (slopes are $1/4$ and $5$),the solution is unique.
Note: Option $(D)$ represents coincident lines ($4x+10y=-22$ is $2$ times $2x+5y=-11$),which have infinitely many solutions,not a unique one.

(A) Yes,it is true.
Given equations are:
$1) -x + 2y + 2 = 0$
$2) \frac{1}{2}x - \frac{1}{4}y - 1 = 0$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = -1, b_1 = 2, c_1 = 2$
$a_2 = \frac{1}{2}, b_2 = -\frac{1}{4}, c_2 = -1$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{-1}{1/2} = -2$
$\frac{b_1}{b_2} = \frac{2}{-1/4} = -8$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (i.e.,$-2 \neq -8$),the pair of linear equations has a unique solution.

(B) The condition for a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are:
$6x + y - 6 = 0$ --- $(i)$
$2x - y = 0$ --- $(ii)$
Here,$a_1 = 6, b_1 = 1, c_1 = -6$ and $a_2 = 2, b_2 = -1, c_2 = 0$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{6}{2} = 3$
$\frac{b_1}{b_2} = \frac{1}{-1} = -1$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $(3 \neq -1)$,the lines intersect at a single point.
Therefore,the given pair of linear equations has a unique solution,not 'no solution'.

(B) No,the given pair of equations is:
$3x + y - 3 = 0$ and $2x + \frac{2}{3}y - 2 = 0$
Here,the coefficients are:
$a_1 = 3, b_1 = 1, c_1 = -3$
$a_2 = 2, b_2 = \frac{2}{3}, c_2 = -2$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{3}{2}$
$\frac{b_1}{b_2} = \frac{1}{2/3} = \frac{3}{2}$
$\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{3}{2}$,the lines are coincident.
Therefore,the given pair of linear equations has infinitely many solutions,not no solution.

(A) The condition for two lines to be coincident is given by the ratio of their coefficients:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
First,rewrite the given equations in the standard form $ax + by + c = 0$:
$1) -2x - 3y - 1 = 0$
$2) 4x + 6y + 2 = 0$
Here,the coefficients are:
$a_1 = -2, b_1 = -3, c_1 = -1$
$a_2 = 4, b_2 = 6, c_2 = 2$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{-2}{4} = -\frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-1}{2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$,the given pair of linear equations represents coincident lines.

(B) pair of linear equations is consistent if it has at least one solution. This occurs when:
$1$. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)
$2$. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Infinitely many solutions)
Given equations:
$-3x - 4y - 12 = 0$
$3x + 4y - 12 = 0$
Here,$a_1 = -3, b_1 = -4, c_1 = -12$ and $a_2 = 3, b_2 = 4, c_2 = -12$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{-3}{3} = -1$
$\frac{b_1}{b_2} = \frac{-4}{4} = -1$
$\frac{c_1}{c_2} = \frac{-12}{-12} = 1$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel and do not intersect.
Therefore,the pair of linear equations is inconsistent.

(A) pair of linear equations is consistent if it has at least one solution. The conditions are:
$1$. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)
$2$. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Infinitely many solutions)
Given equations:
$Eq_1: \frac{3}{5} x - y - \frac{1}{2} = 0$
$Eq_2: \frac{1}{5} x - 3 y - \frac{1}{6} = 0$
Comparing with $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$:
$a_1 = \frac{3}{5}, b_1 = -1, c_1 = -\frac{1}{2}$
$a_2 = \frac{1}{5}, b_2 = -3, c_2 = -\frac{1}{6}$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{3/5}{1/5} = 3$
$\frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $(3 \neq \frac{1}{3})$,the system has a unique solution.
Therefore,the given pair of linear equations is consistent.

(A) pair of linear equations is consistent if it has at least one solution. The conditions are:
$1$. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)
$2$. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Infinitely many solutions)
Given equations:
$2ax + by - a = 0$
$4ax + 2by - 2a = 0$
Here,$a_1 = 2a, b_1 = b, c_1 = -a$ and $a_2 = 4a, b_2 = 2b, c_2 = -2a$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2a}{4a} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{b}{2b} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-a}{-2a} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$,the system has infinitely many solutions. Therefore,the pair of linear equations is consistent.

(B) pair of linear equations is consistent if it has at least one solution. This occurs when:
$1$. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)
$2$. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Infinitely many solutions)
Given equations:
$x + 3y = 11$ (Equation $1$)
$2(2x + 6y) = 22 \implies 4x + 12y = 22 \implies 2x + 6y = 11$ (Equation $2$)
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
For Equation $1$: $a_1 = 1, b_1 = 3, c_1 = -11$
For Equation $2$: $a_2 = 2, b_2 = 6, c_2 = -11$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-11}{-11} = 1$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel and have no solution.
Therefore,the given pair of linear equations is inconsistent.

(B) The given pair of linear equations is $\lambda x + 3y + 7 = 0$ and $2x + 6y - 14 = 0$.
For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Here,$a_1 = \lambda, b_1 = 3, c_1 = 7$ and $a_2 = 2, b_2 = 6, c_2 = -14$.
Substituting these values into the condition,we get $\frac{\lambda}{2} = \frac{3}{6} = \frac{7}{-14}$.
Simplifying the ratios,we get $\frac{\lambda}{2} = \frac{1}{2} = -\frac{1}{2}$.
Since $\frac{1}{2} \neq -\frac{1}{2}$,the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ can never be satisfied for any value of $\lambda$.
Therefore,the statement is false.

(B) The given pair of linear equations is:
$x - 2y = 8$
$5x - 10y = c$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = -2, c_1 = -8$
$a_2 = 5, b_2 = -10, c_2 = -c$
For a system of linear equations to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Here,$\frac{a_1}{a_2} = \frac{1}{5}$ and $\frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2}$,the system can never have a unique solution regardless of the value of $c$.
If $c = 40$,then $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{5}$,which means the system has infinitely many solutions.
If $c \neq 40$,the system has no solution.
Therefore,the statement is False.

(D) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $4x + 5y = 2$ and $(2p + 7q)x + (p + 8q)y = 2q - p + 1$.
Here,$\frac{a_1}{a_2} = \frac{4}{2p + 7q}$,$\frac{b_1}{b_2} = \frac{5}{p + 8q}$,and $\frac{c_1}{c_2} = \frac{2}{2q - p + 1}$.
Equating $\frac{a_1}{a_2} = \frac{b_1}{b_2}$:
$\frac{4}{2p + 7q} = \frac{5}{p + 8q} \implies 4(p + 8q) = 5(2p + 7q) \implies 4p + 32q = 10p + 35q \implies 6p + 3q = 0 \implies q = -2p$ (Equation $1$).
Equating $\frac{a_1}{a_2} = \frac{c_1}{c_2}$:
$\frac{4}{2p + 7q} = \frac{2}{2q - p + 1} \implies 4(2q - p + 1) = 2(2p + 7q) \implies 8q - 4p + 4 = 4p + 14q \implies 8p + 6q = 4 \implies 4p + 3q = 2$ (Equation $2$).
Substituting $q = -2p$ into Equation $2$:
$4p + 3(-2p) = 2 \implies 4p - 6p = 2 \implies -2p = 2 \implies p = -1$.
Now,$q = -2(-1) = 2$.
Thus,for $p = -1$ and $q = 2$,the equations have infinitely many solutions.

(A) Given equations are:
$21x + 47y = 110$ --- $(1)$
$47x + 21y = 162$ --- $(2)$
Adding $(1)$ and $(2)$:
$(21x + 47x) + (47y + 21y) = 110 + 162$
$68x + 68y = 272$
Dividing by $68$:
$x + y = 4$ --- $(3)$
Subtracting $(1)$ from $(2)$:
$(47x - 21x) + (21y - 47y) = 162 - 110$
$26x - 26y = 52$
Dividing by $26$:
$x - y = 2$ --- $(4)$
Adding $(3)$ and $(4)$:
$(x + y) + (x - y) = 4 + 2$
$2x = 6$
$x = 3$
Substituting $x = 3$ in $(3)$:
$3 + y = 4$
$y = 1$
Thus,the solution is $x = 3, y = 1$.

(B) To draw the graphs of the given equations,we find two solutions for each equation:
For $x-y+2=0$:

$x$	$0$	$-2$
$y=x+2$	$2$	$0$

For $4x-y-4=0$:

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

Plot the points $A(0, 2)$,$B(-2, 0)$,$P(0, -4)$,and $Q(1, 0)$ on a graph and draw the lines $AB$ and $PQ$. The lines intersect at point $R(2, 4)$.
The triangle formed by these lines and the $x$-axis is $\triangle BQR$,with vertices $B(-2, 0)$,$Q(1, 0)$,and $R(2, 4)$.
Area of triangle $= \frac{1}{2} \times \text{Base} \times \text{Altitude}$.
Here,Base $= BQ = |1 - (-2)| = 3$ units.
Altitude $= \text{Ordinate of } R = 4$ units.
Area $= \frac{1}{2} \times 3 \times 4 = 6$ sq. units.

(A) The given pair of linear equations is $\lambda x + y = \lambda^2$ and $x + \lambda y = 1$.
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have $a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2$ and $a_2 = 1, b_2 = \lambda, c_2 = -1$.
$(i)$ For no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
$\frac{\lambda}{1} = \frac{1}{\lambda} \neq \frac{-\lambda^2}{-1} \Rightarrow \lambda^2 = 1$ and $\lambda^2 \neq 1$. This is impossible for $\lambda = 1$. For $\lambda = -1$,$\frac{-1}{1} = \frac{1}{-1} \neq \frac{-(-1)^2}{-1} \Rightarrow -1 = -1 \neq 1$. Thus,for $\lambda = -1$,there is no solution.
$(ii)$ For infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{\lambda}{1} = \frac{1}{\lambda} = \frac{-\lambda^2}{-1} \Rightarrow \lambda^2 = 1$ and $\lambda = \lambda^3$. This holds for $\lambda = 1$.
$(iii)$ For a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
$\frac{\lambda}{1} \neq \frac{1}{\lambda} \Rightarrow \lambda^2 \neq 1 \Rightarrow \lambda \neq \pm 1$. Thus,all real values of $\lambda$ except $\pm 1$ yield a unique solution.

(D) The given pair of linear equations is:
$kx + 3y = (k - 3)$
$12x + ky = k$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = k, b_1 = 3, c_1 = -(k - 3)$
$a_2 = 12, b_2 = k, c_2 = -k$
For the pair of linear equations to have no solution,the condition is:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
$\frac{k}{12} = \frac{3}{k} \neq \frac{-(k - 3)}{-k}$
Taking the first two parts:
$\frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$
Now,checking the condition $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:
If $k = 6$,then $\frac{3}{6} = \frac{1}{2}$ and $\frac{k-3}{k} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}$. Since $\frac{1}{2} = \frac{1}{2}$,$k = 6$ gives infinitely many solutions.
If $k = -6$,then $\frac{3}{-6} = -\frac{1}{2}$ and $\frac{k-3}{k} = \frac{-6-3}{-6} = \frac{-9}{-6} = \frac{3}{2}$. Since $-\frac{1}{2} \neq \frac{3}{2}$,the condition holds.
Thus,the value of $k$ for which there is no solution is $-6$.

(A) The given pair of linear equations is:
$x + 2y - 1 = 0$ .....$(i)$
$(a - b)x + (a + b)y - (a + b - 2) = 0$ .....$(ii)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 2, c_1 = -1$
$a_2 = (a - b), b_2 = (a + b), c_2 = -(a + b - 2)$
For infinitely many solutions,the condition is:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
$\frac{1}{a - b} = \frac{2}{a + b} = \frac{-1}{-(a + b - 2)}$
Taking the first two parts:
$\frac{1}{a - b} = \frac{2}{a + b} \Rightarrow a + b = 2a - 2b \Rightarrow a = 3b$ .....$(iii)$
Taking the last two parts:
$\frac{2}{a + b} = \frac{1}{a + b - 2} \Rightarrow 2a + 2b - 4 = a + b \Rightarrow a + b = 4$ .....$(iv)$
Substituting the value of $a$ from $(iii)$ into $(iv)$:
$3b + b = 4 \Rightarrow 4b = 4 \Rightarrow b = 1$
Substituting $b = 1$ into $(iii)$:
$a = 3(1) = 3$
Thus,for $a = 3$ and $b = 1$,the given pair of equations has infinitely many solutions.

(C) The given pair of linear equations is:
$-x + py - 1 = 0 \quad ...(i)$
$px - y - 1 = 0 \quad ...(ii)$
Comparing these with the standard form $ax + by + c = 0$,we get:
$a_1 = -1, b_1 = p, c_1 = -1$ (from Eq. $i$)
$a_2 = p, b_2 = -1, c_2 = -1$ (from Eq. $ii$)
For a pair of linear equations to have no solution,the lines must be parallel,which implies:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Substituting the values:
$\frac{-1}{p} = \frac{p}{-1} \neq \frac{-1}{-1}$
Taking the first two parts:
$\frac{-1}{p} = \frac{p}{-1} \Rightarrow p^2 = 1 \Rightarrow p = \pm 1$
Taking the condition $\frac{p}{-1} \neq \frac{-1}{-1}$:
$\frac{p}{-1} \neq 1 \Rightarrow p \neq -1$
Since $p = \pm 1$ and $p \neq -1$,the only valid value is $p = 1$.

(D) Given,the pair of linear equations is:
$-3x + 5y - 7 = 0 \dots (i)$
$2px - 3y - 1 = 0 \dots (ii)$
Comparing these with the standard form $ax + by + c = 0$,we get:
$a_1 = -3, b_1 = 5, c_1 = -7$
$a_2 = 2p, b_2 = -3, c_2 = -1$
For the lines to intersect at a unique point,the condition for a unique solution is:
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
Substituting the values:
$\frac{-3}{2p} \neq \frac{5}{-3}$
Cross-multiplying,we get:
$(-3) \times (-3) \neq 5 \times (2p)$
$9 \neq 10p$
$p \neq \frac{9}{10}$
Thus,the lines intersect at a unique point for all real values of $p$ except $\frac{9}{10}$.

(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 $7x = 7 \times 2000 = ₹ 14,000$.

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

(N/A) Given equations are:
$1) 3x - 5y = 4$
$2) 9x - 2y = 7$
Elimination Method:
Multiply equation $(1)$ by $3$ to make the coefficients of $x$ equal:
$9x - 15y = 12$ $(3)$
Subtract equation $(2)$ from $(3)$:
$(9x - 15y) - (9x - 2y) = 12 - 7$
$-13y = 5$
$y = -5/13$
Substitute $y = -5/13$ into equation $(1)$:
$3x - 5(-5/13) = 4$
$3x + 25/13 = 4$
$3x = 4 - 25/13 = (52 - 25)/13 = 27/13$
$x = 9/13$
Substitution Method:
From equation $(1)$,$3x = 5y + 4$,so $x = (5y + 4)/3$.
Substitute this into equation $(2)$:
$9((5y + 4)/3) - 2y = 7$
$3(5y + 4) - 2y = 7$
$15y + 12 - 2y = 7$
$13y = -5$
$y = -5/13$
Then $x = (5(-5/13) + 4)/3 = (-25/13 + 52/13)/3 = (27/13)/3 = 9/13$.
Thus,the solution is $x = 9/13$ and $y = -5/13$.

(X=2, Y=-3) Given equations:
$(1)$ $\frac{x}{2} + \frac{2y}{3} = -1$
$(2)$ $x - \frac{y}{3} = 3$
Simplify equation $(1)$ by multiplying by $6$: $3x + 4y = -6$ $(3)$
Simplify equation $(2)$ by multiplying by $3$: $3x - y = 9$ $(4)$
Elimination Method:
Subtract $(4)$ from $(3)$: $(3x + 4y) - (3x - y) = -6 - 9$
$5y = -15 \implies y = -3$
Substitute $y = -3$ into $(4)$: $3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2$
Substitution Method:
From $(4)$,$y = 3x - 9$
Substitute into $(3)$: $3x + 4(3x - 9) = -6$
$3x + 12x - 36 = -6 \implies 15x = 30 \implies x = 2$
Substitute $x = 2$ into $y = 3x - 9$: $y = 3(2) - 9 = 6 - 9 = -3$
Final solution: $x = 2, y = -3$.

(N/A) Let ₹ $x$ be the fare from the bus stand in Bangalore to Malleswaram,and ₹ $y$ be the fare to Yeshwanthpur. From the given information,we have:
$2x + 3y = 46$,i.e.,$2x + 3y - 46 = 0$ $...(1)$
$3x + 5y = 74$,i.e.,$3x + 5y - 74 = 0$ $...(2)$
To solve the equations by the cross-multiplication method,we use the coefficients as follows:
$\frac{x}{(3)(-74) - (5)(-46)} = \frac{y}{(-46)(3) - (-74)(2)} = \frac{1}{(2)(5) - (3)(3)}$
i.e.,$\frac{x}{-222 + 230} = \frac{y}{-138 + 148} = \frac{1}{10 - 9}$
i.e.,$\frac{x}{8} = \frac{y}{10} = \frac{1}{1}$
Thus,$\frac{x}{8} = 1$ and $\frac{y}{10} = 1$,which gives $x = 8$ and $y = 10$.
Hence,the fare from the bus stand in Bangalore to Malleswaram is ₹ $8$ and the fare to Yeshwanthpur is ₹ $10$.

(B) The given equations are $4x + py + 8 = 0$ and $2x + 2y + 2 = 0$.
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:
$a_{1} = 4, b_{1} = p, c_{1} = 8$
$a_{2} = 2, b_{2} = 2, c_{2} = 2$
For a pair of linear equations to have a unique solution,the condition is $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.
Substituting the values,we get $\frac{4}{2} \neq \frac{p}{2}$.
This simplifies to $2 \neq \frac{p}{2}$.
Multiplying both sides by $2$,we get $p \neq 4$.
Thus,for all real values of $p$ except $4$,the pair of equations has a unique solution.

(A) For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$.
Here,$a_1 = k, b_1 = 3, c_1 = -(k - 3)$ and $a_2 = 12, b_2 = k, c_2 = -k$.
Applying the condition: $\frac{k}{12} = \frac{3}{k} = \frac{-(k - 3)}{-k}$.
From $\frac{k}{12} = \frac{3}{k}$,we get $k^2 = 36$,so $k = \pm 6$.
From $\frac{3}{k} = \frac{k - 3}{k}$,we get $3k = k(k - 3) \implies 3k = k^2 - 3k \implies k^2 - 6k = 0 \implies k(k - 6) = 0$. Thus,$k = 0$ or $k = 6$.
The common value satisfying both conditions is $k = 6$.

(B) Given equations are:
$x - 3y - 3 = 0$ ... $(1)$
$3x - 9y - 2 = 0$ ... $(2)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = -3, c_1 = -3$
$a_2 = 3, b_2 = -9, c_2 = -2$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$
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,the system of equations has no solution.

(A) Given equations are:
$2x + y - 5 = 0$
$3x + 2y - 8 = 0$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = 1, c_1 = -5$
$a_2 = 3, b_2 = 2, c_2 = -8$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-8} = \frac{5}{8}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the pair of equations has a unique solution.
Using the cross-multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
$\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$
$\frac{x}{-8 + 10} = \frac{y}{-15 + 16} = \frac{1}{4 - 3}$
$\frac{x}{2} = \frac{y}{1} = \frac{1}{1}$
Thus,$x = 2$ and $y = 1$.

(D) Given equations are:
$3x - 5y - 20 = 0$ --- $(1)$
$6x - 10y - 40 = 0$ --- $(2)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 3, b_1 = -5, c_1 = -20$
$a_2 = 6, b_2 = -10, c_2 = -40$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the lines are coincident.
Therefore,the system of linear equations has infinitely many solutions.

(A) Given equations are:
$x - 3y - 7 = 0$ ... $(1)$
$3x - 3y - 15 = 0$ ... $(2)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 1, b_1 = -3, c_1 = -7$
$a_2 = 3, b_2 = -3, c_2 = -15$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-3} = 1, \quad \frac{c_1}{c_2} = \frac{-7}{-15} = \frac{7}{15}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (i.e.,$\frac{1}{3} \neq 1$),the system has a unique solution.
Using the cross-multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
$\frac{x}{(-3)(-15) - (-3)(-7)} = \frac{y}{(-7)(3) - (-15)(1)} = \frac{1}{(1)(-3) - (3)(-3)}$
$\frac{x}{45 - 21} = \frac{y}{-21 + 15} = \frac{1}{-3 + 9}$
$\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$
For $x$: $\frac{x}{24} = \frac{1}{6} \implies x = \frac{24}{6} = 4$
For $y$: $\frac{y}{-6} = \frac{1}{6} \implies y = \frac{-6}{6} = -1$
Thus,the unique solution is $x = 4, y = -1$.

(A) The given equations are:
$2x + 3y - 7 = 0$
$(a-b)x + (a+b)y - (3a + b - 2) = 0$
For a system of linear equations $\frac{a_1}{a_2}x + \frac{b_1}{b_2}y + \frac{c_1}{c_2} = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Here,$\frac{a_1}{a_2} = \frac{2}{a-b}$,$\frac{b_1}{b_2} = \frac{3}{a+b}$,and $\frac{c_1}{c_2} = \frac{-7}{-(3a+b-2)} = \frac{7}{3a+b-2}$.
Equating the ratios:
$1) \frac{2}{a-b} = \frac{3}{a+b} \implies 2a + 2b = 3a - 3b \implies a - 5b = 0 \implies a = 5b$ (Equation $1$)
$2) \frac{2}{a-b} = \frac{7}{3a+b-2} \implies 6a + 2b - 4 = 7a - 7b \implies a - 9b = -4$ (Equation $2$)
Substituting $a = 5b$ into Equation $2$:
$5b - 9b = -4 \implies -4b = -4 \implies b = 1$.
Now,substituting $b = 1$ into $a = 5b$:
$a = 5(1) = 5$.
Thus,the values are $a = 5$ and $b = 1$.

(A) Given equations are:
$8x + 5y = 9$ $...(i)$
$3x + 2y = 4$ $...(ii)$
Substitution Method:
From equation $(ii)$,we get:
$3x = 4 - 2y \implies x = \frac{4 - 2y}{3}$ $...(iii)$
Substituting $(iii)$ in $(i)$:
$8(\frac{4 - 2y}{3}) + 5y = 9$
$32 - 16y + 15y = 27$
$-y = 27 - 32$
$-y = -5 \implies y = 5$
Substituting $y = 5$ in $(iii)$:
$x = \frac{4 - 2(5)}{3} = \frac{4 - 10}{3} = \frac{-6}{3} = -2$
Thus,$x = -2, y = 5$.
Cross-multiplication Method:
$8x + 5y - 9 = 0$
$3x + 2y - 4 = 0$
Using the formula $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$:
$\frac{x}{(5)(-4) - (2)(-9)} = \frac{y}{(-9)(3) - (-4)(8)} = \frac{1}{(8)(2) - (3)(5)}$
$\frac{x}{-20 + 18} = \frac{y}{-27 + 32} = \frac{1}{16 - 15}$
$\frac{x}{-2} = \frac{y}{5} = \frac{1}{1}$
$x = -2, y = 5$.

(N/A) Let $x$ be the fixed monthly hostel charge and $y$ be the cost of food per day.
According to the given information:
For student $A$: $x + 20y = 1000$ $...(1)$
For student $B$: $x + 26y = 1180$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$,we get:
$(x + 26y) - (x + 20y) = 1180 - 1000$
$6y = 180$
$y = 30$
Substituting $y = 30$ in equation $(1)$:
$x + 20(30) = 1000$
$x + 600 = 1000$
$x = 400$
Thus,the fixed monthly charge is ₹ $400$ and the cost of food per day is ₹ $30$.

(A) Let the fraction be $\frac{x}{y}$.
According to the given information:
$\frac{x-1}{y} = \frac{1}{3} \implies 3x - y = 3$ $...(1)$
$\frac{x}{y+8} = \frac{1}{4} \implies 4x - y = 8$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$,we obtain:
$(4x - y) - (3x - y) = 8 - 3$
$x = 5$
Substituting the value of $x$ in equation $(1)$:
$3(5) - y = 3$
$15 - y = 3$
$y = 12$
Hence,the fraction is $\frac{5}{12}$.

(20) Let the number of right answers be $x$ and the number of wrong answers be $y$.
According to the given information:
For the first case: $3x - y = 40$ $...(1)$
For the second case: $4x - 2y = 50$
Dividing the second equation by $2$,we get: $2x - y = 25$ $...(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(3x - y) - (2x - y) = 40 - 25$
$x = 15$
Substituting $x = 15$ in equation $(2)$:
$2(15) - y = 25$
$30 - y = 25$
$y = 5$
Therefore,the number of right answers is $15$ and the number of wrong answers is $5$.
Total number of questions $= x + y = 15 + 5 = 20$.

(A) Let the speed of the $1^{st}$ car be $u \ km/h$ and the $2^{nd}$ car be $v \ km/h$.
When the cars travel in the same direction,their relative speed is $(u - v) \ km/h$. Since they meet in $5 \ hours$ covering a distance of $100 \ km$,we have:
$5(u - v) = 100 \Rightarrow u - v = 20 \quad \dots(1)$
When the cars travel towards each other,their relative speed is $(u + v) \ km/h$. Since they meet in $1 \ hour$ covering a distance of $100 \ km$,we have:
$1(u + v) = 100 \Rightarrow u + v = 100 \quad \dots(2)$
Adding equations $(1)$ and $(2)$:
$(u - v) + (u + v) = 20 + 100$
$2u = 120 \Rightarrow u = 60 \ km/h$
Substituting $u = 60$ in equation $(2)$:
$60 + v = 100 \Rightarrow v = 40 \ km/h$
Thus,the speed of the first car is $60 \ km/h$ and the speed of the second car is $40 \ km/h$.

(A) Let the length and breadth of the rectangle be $x$ units and $y$ units respectively.
Area $= x y$
According to the first condition:
$(x - 5)(y + 3) = xy - 9$
$xy + 3x - 5y - 15 = xy - 9$
$3x - 5y = 6$ ...$(1)$
According to the second condition:
$(x + 3)(y + 2) = xy + 67$
$xy + 2x + 3y + 6 = xy + 67$
$2x + 3y = 61$ ...$(2)$
Solving equations $(1)$ and $(2)$ by the cross-multiplication method:
$\frac{x}{(-5)(-61) - (3)(-6)} = \frac{y}{(-6)(2) - (-61)(3)} = \frac{1}{(3)(3) - (-5)(2)}$
$\frac{x}{305 + 18} = \frac{y}{-12 + 183} = \frac{1}{9 + 10}$
$\frac{x}{323} = \frac{y}{171} = \frac{1}{19}$
$x = \frac{323}{19} = 17$
$y = \frac{171}{19} = 9$
Thus,the length is $17$ units and the breadth is $9$ units.

(N/A) Let the speed of the boat in still water be $x\, km/h$ and the speed of the stream be $y\, km/h.$
Then the speed of the boat downstream $= (x+y)\, km/h$ and the speed of the boat upstream $= (x-y)\, km/h.$
Using the formula,$\text{time} = \frac{\text{distance}}{\text{speed}}.$
In the first case,the total time taken is $10\, hours$ for $30\, km$ upstream and $44\, km$ downstream:
$\frac{30}{x-y} + \frac{44}{x+y} = 10 \quad ...(1)$
In the second case,the total time taken is $13\, hours$ for $40\, km$ upstream and $55\, km$ downstream:
$\frac{40}{x-y} + \frac{55}{x+y} = 13 \quad ...(2)$
Let $\frac{1}{x-y} = u$ and $\frac{1}{x+y} = v \quad ...(3)$
Substituting these into equations $(1)$ and $(2)$:
$30u + 44v = 10 \quad ...(4)$
$40u + 55v = 13 \quad ...(5)$
Solving these using the elimination method:
Multiply $(4)$ by $4$ and $(5)$ by $3$:
$120u + 176v = 40$
$120u + 165v = 39$
Subtracting the equations: $11v = 1 \implies v = \frac{1}{11}.$
Substituting $v = \frac{1}{11}$ in $(4)$:
$30u + 44(\frac{1}{11}) = 10 \implies 30u + 4 = 10 \implies 30u = 6 \implies u = \frac{1}{5}.$
Now,$\frac{1}{x-y} = \frac{1}{5} \implies x-y = 5$ and $\frac{1}{x+y} = \frac{1}{11} \implies x+y = 11.$
Adding these: $2x = 16 \implies x = 8.$
Subtracting these: $2y = 6 \implies y = 3.$
Thus,the speed of the boat in still water is $8\, km/h$ and the speed of the stream is $3\, km/h.$