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

(A) Let the speed of the train be $x \, km/hr$ and the speed of the car be $y \, km/hr$.
Case $1$: Time taken = $\frac{120}{x} + \frac{480}{y} = 8$ hours. Dividing by $8$,we get $\frac{15}{x} + \frac{60}{y} = 1$.
Case $2$: Time taken = $8$ hours $20$ minutes = $8 + \frac{20}{60} = 8 + \frac{1}{3} = \frac{25}{3}$ hours. The equation is $\frac{200}{x} + \frac{400}{y} = \frac{25}{3}$. Dividing by $25$,we get $\frac{8}{x} + \frac{16}{y} = \frac{1}{3}$.
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become $15u + 60v = 1$ and $8u + 16v = \frac{1}{3}$.
Solving these,we get $u = \frac{1}{60}$ and $v = \frac{1}{80}$.
Thus,$x = 60 \, km/hr$ and $y = 80 \, km/hr$.

(A) Let the time taken by one man to finish the work be $x$ days and by one woman be $y$ days.
Work done by one man in $1$ day = $1/x$.
Work done by one woman in $1$ day = $1/y$.
According to the problem:
$10(8/x + 12/y) = 1 \implies 8/x + 12/y = 1/10$ (Equation $1$)
$14(6/x + 8/y) = 1 \implies 6/x + 8/y = 1/14$ (Equation $2$)
Let $u = 1/x$ and $v = 1/y$.
$8u + 12v = 1/10 \implies 80u + 120v = 1$ (Equation $3$)
$6u + 8v = 1/14 \implies 84u + 112v = 1$ (Equation $4$)
Solving the system: From Equation $3$,$v = (1 - 80u)/120$. Substituting into Equation $4$:
$84u + 112((1 - 80u)/120) = 1$
$84u + (14/15)(1 - 80u) = 1$
$1260u + 14 - 1120u = 15$
$140u = 1 \implies u = 1/140$.
So,$x = 140$ days.
Substituting $u = 1/140$ into $80(1/140) + 120v = 1$:
$4/7 + 120v = 1 \implies 120v = 3/7 \implies v = 1/280$.
So,$y = 280$ days.
Therefore,one man takes $140$ days and one woman takes $280$ days.

(D) The given equations are:
$x - ky - 2 = 0$ ...... $(1)$
$3x + 2y + 5 = 0$ ...... $(2)$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = -k, c_1 = -2$
$a_2 = 3, b_2 = 2, c_2 = 5$
For a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Substituting the values:
$\frac{1}{3} \neq \frac{-k}{2}$
Multiplying both sides by $6$:
$2 \neq -3k$
$k \neq -\frac{2}{3}$
Thus,the system has a unique solution for all real values of $k$ except $k = -\frac{2}{3}$.

(A) The given equations are:
$3x - 4y + 7 = 0$ ... $(1)$
$kx + 3y - 5 = 0$ ... $(2)$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 3, b_1 = -4, c_1 = 7$
$a_2 = k, b_2 = 3, c_2 = -5$
For a pair of linear equations to have no solution,the condition is:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Substituting the values:
$\frac{3}{k} = \frac{-4}{3} \neq \frac{7}{-5}$
Since $\frac{-4}{3} \neq \frac{7}{-5}$,we only need to solve for:
$\frac{3}{k} = \frac{-4}{3}$
Cross-multiplying gives:
$-4k = 9$
$k = -\frac{9}{4}$
Thus,the value of $k$ is $-\frac{9}{4}$.

(B) The given equations are:
$5x + 2y - k = 0$ .......... $(1)$
$10x + 4y - 3 = 0$ .......... $(2)$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 5, b_1 = 2, c_1 = -k$
$a_2 = 10, b_2 = 4, c_2 = -3$
For a pair of linear equations to have infinite solutions,the condition is:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Substituting the values:
$\frac{5}{10} = \frac{2}{4} = \frac{-k}{-3}$
$\frac{1}{2} = \frac{1}{2} = \frac{k}{3}$
Taking the equality $\frac{1}{2} = \frac{k}{3}$,we get:
$k = \frac{3}{2}$
Thus,for $k = \frac{3}{2}$,the system has infinite solutions.

(N/A) Let the present age of Abhishek be $x$ years and the present age of Sweta be $y$ years.
Case $1$: Three years ago,Abhishek's age was $(x-3)$ and Sweta's age was $(y-3)$.
According to the problem: $(x-3) = 4(y-3)$
$x-3 = 4y-12$
$x-4y = -9$ --- (Equation $1$)
Case $2$: After five years,Abhishek's age will be $(x+5)$ and Sweta's age will be $(y+5)$.
According to the problem: $(x+5) = 2(y+5)$
$x+5 = 2y+10$
$x-2y = 5$ --- (Equation $2$)
Thus,the required pair of linear equations is $x-4y = -9$ and $x-2y = 5$.

(N/A) The given equations are:
$ax + by - 1 = 0$ --- $(1)$
$bx + ay - \frac{2ab}{a^2 + b^2} = 0$ --- $(2)$
Using the cross-multiplication method:
$\frac{x}{b(-\frac{2ab}{a^2+b^2}) - a(-1)} = \frac{-y}{a(-\frac{2ab}{a^2+b^2}) - b(-1)} = \frac{1}{a^2 - b^2}$
$\frac{x}{\frac{-2ab^2 + a(a^2+b^2)}{a^2+b^2}} = \frac{-y}{\frac{-2a^2b + b(a^2+b^2)}{a^2+b^2}} = \frac{1}{a^2 - b^2}$
$\frac{x}{\frac{a^3 - ab^2}{a^2+b^2}} = \frac{-y}{\frac{b^3 - a^2b}{a^2+b^2}} = \frac{1}{a^2 - b^2}$
$\frac{x}{\frac{a(a^2-b^2)}{a^2+b^2}} = \frac{y}{\frac{b(a^2-b^2)}{a^2+b^2}} = \frac{1}{a^2 - b^2}$
$x = \frac{a(a^2-b^2)}{(a^2+b^2)(a^2-b^2)} = \frac{a}{a^2+b^2}$
$y = \frac{b(a^2-b^2)}{(a^2+b^2)(a^2-b^2)} = \frac{b}{a^2+b^2}$
Thus,the solution is $(x, y) = (\frac{a}{a^2+b^2}, \frac{b}{a^2+b^2})$.

(C) Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
The equations become:
$25u - 7v = -2$ --- $(1)$
$15u - 7v = -4$ --- $(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(25u - 7v) - (15u - 7v) = -2 - (-4)$
$10u = 2 \implies u = \frac{2}{10} = \frac{1}{5}$.
Substituting $u = \frac{1}{5}$ in equation $(1)$:
$25(\frac{1}{5}) - 7v = -2$
$5 - 7v = -2$
$-7v = -7 \implies v = 1$.
Now,$\frac{1}{x+y} = \frac{1}{5} \implies x+y = 5$ --- $(3)$
And $\frac{1}{x-y} = 1 \implies x-y = 1$ --- $(4)$
Adding $(3)$ and $(4)$:
$2x = 6 \implies x = 3$.
Substituting $x = 3$ in $(3)$:
$3 + y = 5 \implies y = 2$.
Thus,the solution is $(x, y) = (3, 2)$.

(A) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the first condition: $10x + y = 8(x + y) + 1$.
$10x + y = 8x + 8y + 1 \implies 2x - 7y = 1$ --- (Equation $1$).
According to the second condition: $10x + y = 13(x - y) + 2$.
$10x + y = 13x - 13y + 2 \implies -3x + 14y = 2$ --- (Equation $2$).
Multiply Equation $1$ by $2$: $4x - 14y = 2$ --- (Equation $3$).
Add Equation $2$ and Equation $3$: $(-3x + 14y) + (4x - 14y) = 2 + 2$.
$x = 4$.
Substitute $x = 4$ into Equation $1$: $2(4) - 7y = 1 \implies 8 - 7y = 1 \implies 7y = 7 \implies y = 1$.
The number is $10(4) + 1 = 41$.

(B) Let the numerator be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.
According to the first condition,the numerator is $4$ less than the denominator: $x = y - 4$,which implies $y = x + 4$.
According to the second condition,if the numerator is decreased by $2$ and the denominator is increased by $1$,the denominator becomes eight times the numerator: $(y + 1) = 8(x - 2)$.
Substitute $y = x + 4$ into the second equation: $(x + 4 + 1) = 8(x - 2)$.
$x + 5 = 8x - 16$.
$5 + 16 = 8x - x$.
$21 = 7x$,so $x = 3$.
Now,find $y$: $y = x + 4 = 3 + 4 = 7$.
The fraction is $\frac{3}{7}$.

(D) Let the number of rows be $x$ and the number of students in each row be $y$. The total number of students is $xy$.
According to the first condition: $(x - 2)(y + 1) = xy$
$xy + x - 2y - 2 = xy$
$x - 2y = 2$ --- $(1)$
According to the second condition: $(x + 3)(y - 1) = xy$
$xy - x + 3y - 3 = xy$
$-x + 3y = 3$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$(x - 2y) + (-x + 3y) = 2 + 3$
$y = 5$
Substituting $y = 5$ into equation $(1)$:
$x - 2(5) = 2$
$x - 10 = 2$
$x = 12$
Total number of students $= xy = 12 \times 5 = 60$.

(D) Let the length of the rectangle be $x$ and the breadth be $y$. The area is $A = xy$.
According to the first condition: $(x + 2)(y - 2) = xy - 28$.
Expanding this: $xy - 2x + 2y - 4 = xy - 28$.
Simplifying: $-2x + 2y = -24$,which gives $x - y = 12$ (Equation $1$).
According to the second condition: $(x - 1)(y + 2) = xy + 33$.
Expanding this: $xy + 2x - y - 2 = xy + 33$.
Simplifying: $2x - y = 35$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(2x - y) - (x - y) = 35 - 12$,so $x = 23$.
Substituting $x = 23$ into Equation $1$: $23 - y = 12$,so $y = 11$.
The area of the rectangle is $A = x \times y = 23 \times 11 = 253$ square units.

(B) For a system of two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Given equations are $2x - 3y = 1$ and $kx + 5y = 7$.
Here,$a_1 = 2, b_1 = -3$ and $a_2 = k, b_2 = 5$.
Substituting these values into the condition: $\frac{2}{k} \neq \frac{-3}{5}$.
Cross-multiplying gives: $2 \times 5 \neq -3 \times k$,which simplifies to $10 \neq -3k$.
Therefore,$k \neq -\frac{10}{3}$.
Thus,the system has a unique solution for all real values of $k$ except $k = -\frac{10}{3}$.

(B) For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Given equations are $2x - ky + 3 = 0$ and $3x + 2y - 1 = 0$.
Here,$a_1 = 2, b_1 = -k, c_1 = 3$ and $a_2 = 3, b_2 = 2, c_2 = -1$.
Applying the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$:
$\frac{2}{3} = \frac{-k}{2}$
Multiplying both sides by $2$,we get $4 = -3k$.
Therefore,$k = -\frac{4}{3}$.
Checking the condition $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$:
$\frac{-(-4/3)}{2} = \frac{4/3}{2} = \frac{2}{3}$.
Since $\frac{c_1}{c_2} = \frac{3}{-1} = -3$,and $\frac{2}{3} \neq -3$,the condition holds true for $k = -\frac{4}{3}$.

(A) 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 $kx + 3y = (k - 3)$ and $12x + ky = k$.
Here,$a_1 = k, b_1 = 3, c_1 = k - 3$ and $a_2 = 12, b_2 = k, c_2 = k$.
Applying the condition: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$.
Taking the first two parts: $\frac{k}{12} = \frac{3}{k} \implies k^2 = 36 \implies k = \pm 6$.
Taking the last two parts: $\frac{3}{k} = \frac{k - 3}{k} \implies 3k = k(k - 3) \implies 3k = k^2 - 3k \implies k^2 - 6k = 0 \implies k(k - 6) = 0$.
This gives $k = 0$ or $k = 6$.
For the condition to hold for all ratios,we must satisfy both equations. The common value is $k = 6$.

(A) Given equations are:
$x+y-1=0 \implies x+y=1$ (Equation $1$)
$3x+3y-2=0 \implies 3(x+y)=2 \implies x+y=\frac{2}{3}$ (Equation $2$)
Comparing the coefficients of the system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{a_1}{a_2} = \frac{1}{3}$,$\frac{b_1}{b_2} = \frac{1}{3}$,and $\frac{c_1}{c_2} = \frac{-1}{-2} = \frac{1}{2}$.
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,there is no solution to this system of equations.
The solution set is $\phi$ (the empty set).

(B) linear equation in two variables is an equation of the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
In this equation,the variables $x$ and $y$ must have a degree of $1$.
Let us analyze the options:
$A) x = 3y - 1$ can be written as $x - 3y + 1 = 0$,which is a linear equation in two variables.
$B) x^2 - 5 = 0$ is a quadratic equation in one variable because the degree of $x$ is $2$ and the variable $y$ is absent.
$C) y = 2x + 3$ can be written as $2x - y + 3 = 0$,which is a linear equation in two variables.
$D) x - y = 0$ is a linear equation in two variables.
Therefore,$x^2 - 5 = 0$ is not a linear equation in two variables.

(D) Given equations are:
$2x + 3y = 7$ --- $(1)$
$3x + 2y = 3$ --- $(2)$
Adding equation $(1)$ and $(2)$:
$(2x + 3x) + (3y + 2y) = 7 + 3$
$5x + 5y = 10$
Dividing the entire equation by $5$:
$x + y = 2$
Thus,the value of $x + y$ is $2$.

(A) Given the equations: $\frac{x}{2} = 2$ and $\frac{6}{y} = 2$.
From the first equation: $x = 2 \times 2 = 4$.
From the second equation: $6 = 2y$,which implies $y = \frac{6}{2} = 3$.
Now,calculate the value of $x - y$: $x - y = 4 - 3 = 1$.
Therefore,the correct value is $1$.

(B) Let the tens digit be $x$ and the units digit be $y$.
The number is $10x + y$.
According to the problem:
$x + y = 9$ (Equation $1$)
$x - y = 1$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (x - y) = 9 + 1$
$2x = 10$
$x = 5$
Substituting $x = 5$ in Equation $1$:
$5 + y = 9$
$y = 4$
Therefore,the number is $10(5) + 4 = 54$.

(D) Given the equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$.
The conditions provided are:
$a_{1}b_{2} - a_{2}b_{1} = 0 \implies \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$
$b_{1}c_{2} - b_{2}c_{1} = 0 \implies \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$c_{1}a_{2} - c_{2}a_{1} = 0 \implies \frac{c_{1}}{c_{2}} = \frac{a_{1}}{a_{2}}$
Combining these,we get $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = k$ (where $k$ is a constant).
This implies that the two equations are dependent (coincident lines).
For coincident lines,there are infinitely many solutions.
Therefore,the solution set is an infinite set.

(A) Let the current ages of the four persons be $a, b, c,$ and $d$.
Four years ago,the sum of their ages was $(a-4) + (b-4) + (c-4) + (d-4) = 40$.
This simplifies to $a + b + c + d - 16 = 40$,which means $a + b + c + d = 56$.
This is the sum of their current ages.
After two years,the sum of their ages will be $(a+2) + (b+2) + (c+2) + (d+2)$.
This equals $(a + b + c + d) + 8$.
Substituting the current sum,we get $56 + 8 = 64$ years.

(B) Given equations are:
$1) \, x + y - 1 = 0 \implies x + y = 1$
$2) \, 2x + 2y - 5 = 0 \implies 2(x + y) = 5 \implies x + y = 2.5$
Comparing the two equations,we see that $x + y$ cannot be both $1$ and $2.5$ simultaneously.
Alternatively,checking the ratio of coefficients:
$\frac{a_1}{a_2} = \frac{1}{2}$,$\frac{b_1}{b_2} = \frac{1}{2}$,and $\frac{c_1}{c_2} = \frac{-1}{-5} = \frac{1}{5}$.
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 system of equations has no solution.

(C) Given equations are:
$x + y - 1 = 0$ --- $(1)$
$2x + 2y - 2 = 0$ --- $(2)$
Divide equation $(2)$ by $2$:
$x + y - 1 = 0$
Since both equations are identical,they represent the same line.
For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the system has infinitely many solutions.
Here,$\frac{1}{2} = \frac{1}{2} = \frac{-1}{-2} = \frac{1}{2}$.
Therefore,the system has infinitely many solutions,which corresponds to an infinite set.

(B) For a system of two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the coefficients must satisfy the condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $2x + 3y = 5$ and $4x + ky = 10$.
Here,$a_1 = 2, b_1 = 3, c_1 = 5$ and $a_2 = 4, b_2 = k, c_2 = 10$.
Substituting these values into the condition: $\frac{2}{4} = \frac{3}{k} = \frac{5}{10}$.
Simplifying the ratios: $\frac{1}{2} = \frac{3}{k} = \frac{1}{2}$.
From $\frac{1}{2} = \frac{3}{k}$,we get $k = 3 \times 2 = 6$.

(D) Given equations are:
$8y - 3x = 5xy$ ---$(1)$
$6y - 5x = -2xy$ ---$(2)$
Case $1$: If $x = 0$,then from $(1)$,$8y = 0 \implies y = 0$. Checking in $(2)$,$6(0) - 5(0) = -2(0)(0) \implies 0 = 0$. Thus,$(0, 0)$ is a solution.
Case $2$: If $x \neq 0$ and $y \neq 0$,divide both equations by $xy$:
$(1)$ $\implies \frac{8}{x} - \frac{3}{y} = 5$
$(2)$ $\implies \frac{6}{x} - \frac{5}{y} = -2$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
$8u - 3v = 5$ ---$(3)$
$6u - 5v = -2$ ---$(4)$
Multiply $(3)$ by $5$ and $(4)$ by $3$:
$40u - 15v = 25$
$18u - 15v = -6$
Subtracting: $22u = 31 \implies u = \frac{31}{22} \implies x = \frac{22}{31}$.
Substitute $u$ in $(3)$: $8(\frac{31}{22}) - 3v = 5 \implies \frac{124}{11} - 3v = 5 \implies 3v = \frac{124}{11} - 5 = \frac{69}{11} \implies v = \frac{23}{11} \implies y = \frac{11}{23}$.
Thus,the solutions are $(0, 0)$ and $\left(\frac{22}{31}, \frac{11}{23}\right)$.

(C) linear equation in two variables $x$ and $y$ is an equation that can be written in the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
This condition is mathematically expressed as $a^{2} + b^{2} \neq 0$,which ensures that at least one of the coefficients of the variables is non-zero.
Therefore,the correct general form is $ax + by + c = 0$.

(D) linear equation in two variables is defined as an equation of the form $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero simultaneously.
If $a = 0$ and $b = 0$,the equation becomes $0x + 0y + c = 0$,which simplifies to $c = 0$. This does not represent a linear equation in two variables because the variables $x$ and $y$ are eliminated.
Therefore,the condition $a = 0$ and $b = 0$ is not true for a linear equation in two variables.

(D) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.
Given that the product of the digits is $xy = 10$.
When $27$ is added to the number,the digits are interchanged,so the new number is $10y + x$.
According to the problem: $(10x + y) + 27 = 10y + x$.
Rearranging the terms: $9x - 9y = -27$,which simplifies to $x - y = -3$ or $y - x = 3$.
Substitute $y = x + 3$ into the product equation: $x(x + 3) = 10$.
$x^2 + 3x - 10 = 0$.
$(x + 5)(x - 2) = 0$.
Since $x$ must be a positive digit,$x = 2$.
Then $y = 2 + 3 = 5$.
The number is $10(2) + 5 = 25$.

(D) linear equation in two variables is an equation of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero. The degree of the variables must be $1$.
In the equation $x^{2} - 6x + 7 = 0$,the highest power (degree) of the variable $x$ is $2$. Therefore,it is a quadratic equation,not a linear equation.
The other options,$6x - 5 = y$,$x - 6y = 0$,and $6x + y = 7$,are all linear equations in two variables because the degree of the variables is $1$.

(A) For a pair of linear equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,the condition for a unique solution is that the lines must intersect at exactly one point.
This occurs when the ratios of the coefficients of $x$ and $y$ are not equal,i.e.,$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.
By cross-multiplying this inequality,we get $a_{1}b_{2} \neq a_{2}b_{1}$.
Therefore,the correct option is $A$.

(C) For a pair of linear equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:
$1$. If $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$,the lines intersect at a single point,resulting in a unique solution.
$2$. If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$,the lines are parallel,resulting in no solution.
$3$. If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$,the lines are coincident,meaning they overlap at every point,resulting in infinitely many solutions.
Therefore,the condition for infinitely many solutions is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.

(A) For a pair of linear equations $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$,the condition for having no solution is that the lines represented by the equations are parallel.
This occurs when the ratios of the coefficients of $x$ and $y$ are equal,but they are not equal to the ratio of the constant terms.
The mathematical condition for this is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.

(D) Given the pair of linear equations:
$x+y-2=0$ $...(1)$
$2x+2y-4=0$ $...(2)$
Divide equation $(2)$ by $2$:
$\frac{2x}{2} + \frac{2y}{2} - \frac{4}{2} = \frac{0}{2}$
$x+y-2=0$ $...(3)$
Comparing equation $(1)$ and equation $(3)$,we observe that both equations are identical.
Since the equations are coincident,they represent the same line.
Therefore,the pair of equations has infinitely many solutions,and the solution set is an infinite set.

(D) Given the equation $\frac{x}{2} = \frac{6}{y} = 3$.
First,solve for $x$:
$\frac{x}{2} = 3 \implies x = 3 \times 2 = 6$.
Next,solve for $y$:
$\frac{6}{y} = 3 \implies y = \frac{6}{3} = 2$.
Finally,calculate $x + y$:
$x + y = 6 + 2 = 8$.

(D) Given that the cost of one pant is ₹ $x$ and the cost of one shirt is ₹ $y$.
According to the problem,the cost of $5$ pants is $5x$ and the cost of $8$ shirts is $8y$.
The total cost of $5$ pants and $8$ shirts is given as ₹ $3100$.
Therefore,the linear equation representing this situation is $5x + 8y = 3100$.

(C) Let the length of the rectangle be $l$ and the breadth be $b$.
According to the problem,the length is five less than thrice the breadth.
Thrice the breadth is $3b$.
Five less than thrice the breadth is $3b - 5$.
Therefore,the relationship is $l = 3b - 5$.

(B) Given equations are:
$5x - y = 9$ ---$(1)$
$10x - 2y = 18$ ---$(2)$
Divide equation $(2)$ by $2$:
$5x - y = 9$ ---$(3)$
Comparing equation $(1)$ and equation $(3)$,we see that both equations are identical.
For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the equations are coincident and have infinitely many solutions.
Since the equations are identical,they represent the same line,meaning they are consistent and have infinitely many solutions.

(B) Given equation: $3x - 2y + 5 = 0$
To express $y$ in terms of $x$,we isolate the $y$ term:
$-2y = -3x - 5$
Multiply the entire equation by $-1$:
$2y = 3x + 5$
Divide by $2$:
$y = \frac{3x + 5}{2}$

(C) The standard form of a linear equation in two variables is given by $ax + by + c = 0$,where $a$,$b$,and $c$ are real numbers and $a$ and $b$ are not both zero.
Given the equation $2x = 5y$.
To express this in the form $ax + by + c = 0$,we subtract $5y$ from both sides:
$2x - 5y = 0$.
This can be written as $2x - 5y + 0 = 0$,where $a = 2$,$b = -5$,and $c = 0$.
Comparing this with the given options,$2x - 5y + 0 = 0$ is the standard representation.

(C) Given equation is $\frac{2}{3} x + \frac{3}{2} y = 5$.
To simplify,multiply the entire equation by the least common multiple of the denominators $3$ and $2$,which is $6$.
$\frac{2}{3} x \times 6 + \frac{3}{2} y \times 6 = 5 \times 6$
$4 x + 9 y = 30$.
Thus,the equation $4 x + 9 y = 30$ is equivalent to the given equation.

(B) To find the correct equation,we substitute $x = -1$ and $y = -2$ into each given option:
For option $A$: $2(-1) - (-2) = -2 + 2 = 0 \neq -4$.
For option $B$: $3(-1) + (-2) = -3 - 2 = -5$. This matches the right-hand side.
For option $C$: $3(-1) - 2(-2) = -3 + 4 = 1 \neq 7$.
For option $D$: $(-1) - 2(-2) = -1 + 4 = 3 \neq 0$.
Thus,the equation satisfied by the given values is $3x + y = -5$.

(D) Given the equation: $5x + 3y = 10xy$.
Divide both sides of the equation by $xy$:
$\frac{5x}{xy} + \frac{3y}{xy} = \frac{10xy}{xy}$
$\frac{5}{y} + \frac{3}{x} = 10$ --- (Equation $1$)
We need to find the value of $\frac{6}{x} + \frac{10}{y}$.
Multiply Equation $1$ by $2$:
$2 \times (\frac{3}{x} + \frac{5}{y}) = 2 \times 10$
$\frac{6}{x} + \frac{10}{y} = 20$.

(D) Given equations are:
$(1)$ $\frac{2}{x} + \frac{3}{y} = 3$
$(2)$ $\frac{3}{x} + \frac{2}{y} = 7$
Adding equation $(1)$ and equation $(2)$:
$(\frac{2}{x} + \frac{3}{y}) + (\frac{3}{x} + \frac{2}{y}) = 3 + 7$
$\frac{5}{x} + \frac{5}{y} = 10$
Dividing the entire equation by $5$:
$\frac{1}{x} + \frac{1}{y} = 2$
To find the value of $\frac{3}{x} + \frac{3}{y}$,multiply the equation $\frac{1}{x} + \frac{1}{y} = 2$ by $3$:
$3 \times (\frac{1}{x} + \frac{1}{y}) = 3 \times 2$
$\frac{3}{x} + \frac{3}{y} = 6$

(C) Given equations are:
$1) x + y - 1 = 0 \implies x + y = 1$
$2) 2x + 2y - 3 = 0 \implies 2x + 2y = 3$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 1, b_1 = 1, c_1 = -1$
$a_2 = 2, b_2 = 2, c_2 = -3$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-1}{-3} = \frac{1}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other.
Therefore,the system of equations has no solution,which corresponds to an empty set.

(C) Given the pair of linear equations:
$2x + 3y = 1$
$3x - 2y = 8$
Comparing these with the standard form $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$,we get:
$a_{1} = 2, b_{1} = 3$
$a_{2} = 3, b_{2} = -2$
Now,calculating the value of $a_{1}b_{2} - a_{2}b_{1}$:
$a_{1}b_{2} - a_{2}b_{1} = (2)(-2) - (3)(3)$
$= -4 - 9$
$= -13$

(C) The given equations are $5x - 4y - 1 = 0$ and $10x - 8y - 2 = 0$.
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:
$a_{1} = 5, b_{1} = -4$
$a_{2} = 10, b_{2} = -8$
Now,calculating the value of $a_{1}b_{2} - a_{2}b_{1}$:
$a_{1}b_{2} - a_{2}b_{1} = (5)(-8) - (10)(-4)$
$= -40 - (-40)$
$= -40 + 40$
$= 0$

(C) Given equations are:
$(1)$ $7x - 13y = 1$
$(2)$ $13x - 7y = 19$
Subtract equation $(1)$ from equation $(2)$:
$(13x - 7y) - (7x - 13y) = 19 - 1$
$13x - 7y - 7x + 13y = 18$
$6x + 6y = 18$
Divide by $6$:
$x + y = 3$ $(3)$
Now,add equation $(1)$ and equation $(2)$:
$(7x - 13y) + (13x - 7y) = 1 + 19$
$20x - 20y = 20$
Divide by $20$:
$x - y = 1$ $(4)$
Thus,the value of $x - y$ is $1$.

(B) Given equations are:
$13x - 7y = 19$ ---$(i)$
$7x - 13y = 31$ ---(ii)
Adding equation $(i)$ and (ii):
$(13x - 7y) + (7x - 13y) = 19 + 31$
$20x - 20y = 50$
Dividing by $10$:
$2x - 2y = 5$ ---(iii)
Subtracting equation (ii) from $(i)$:
$(13x - 7y) - (7x - 13y) = 19 - 31$
$6x + 6y = -12$
Dividing by $6$:
$x + y = -2$
Thus,the value of $x + y$ is $-2$.

(B) 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 $x + y + 1 = 0$ and $3x + 3y + k = 0$.
Here,$a_1 = 1, b_1 = 1, c_1 = 1$ and $a_2 = 3, b_2 = 3, c_2 = k$.
Applying the condition: $\frac{1}{3} = \frac{1}{3} = \frac{1}{k}$.
Therefore,$\frac{1}{3} = \frac{1}{k}$,which implies $k = 3$.