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

(A) Let the speed of the car starting from $A$ be $x\, km/hr$ and the speed of the car starting from $B$ be $y\, km/hr$.
When moving in the same direction,the relative speed is $(x - y)\, km/hr$. Since they meet in $9$ hours,the distance covered is $9(x - y) = 90$,which simplifies to $x - y = 10$ (Equation $1$).
When moving in opposite directions,the relative speed is $(x + y)\, km/hr$. Since they meet in $\frac{9}{7}$ hours,the distance covered is $\frac{9}{7}(x + y) = 90$,which simplifies to $x + y = 70$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(x - y) + (x + y) = 10 + 70$,so $2x = 80$,which gives $x = 40\, km/hr$.
Substituting $x = 40$ into Equation $1$: $40 - y = 10$,so $y = 30\, km/hr$.
Thus,the speeds of the cars are $40\, km/hr$ and $30\, km/hr$.

(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 $x-2y = 5$.

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

(A) Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
The equations become:
$11v - 7u = 1$ --- $(1)$
$9v - 4u = 6$ --- $(2)$
To eliminate $u$,multiply $(1)$ by $4$ and $(2)$ by $7$:
$44v - 28u = 4$ --- $(3)$
$63v - 28u = 42$ --- $(4)$
Subtract $(3)$ from $(4)$:
$(63v - 44v) = (42 - 4)$
$19v = 38 \implies v = 2$
Since $v = \frac{1}{y} = 2$,we get $y = \frac{1}{2}$.
Substitute $v = 2$ into $(1)$:
$11(2) - 7u = 1$
$22 - 7u = 1$
$7u = 21 \implies u = 3$
Since $u = \frac{1}{x} = 3$,we get $x = \frac{1}{3}$.
Thus,the solution is $(x, y) = (\frac{1}{3}, \frac{1}{2})$.

(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 $(x, y) = (3, 2)$.

(D) Let the two numbers be $x$ and $y$.
According to the first condition: $\frac{x+1}{y+1} = \frac{1}{2} \implies 2x + 2 = y + 1 \implies y = 2x + 1$ (Equation $1$).
According to the second condition: $\frac{x-5}{y-5} = \frac{5}{11} \implies 11x - 55 = 5y - 25 \implies 11x - 5y = 30$ (Equation $2$).
Substitute $y = 2x + 1$ into Equation $2$: $11x - 5(2x + 1) = 30$.
$11x - 10x - 5 = 30 \implies x = 35$.
Now,find $y$: $y = 2(35) + 1 = 70 + 1 = 71$.
Thus,the numbers are $35$ and $71$.

(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 $= xy = 12 \times 5 = 60$. Wait,checking the options provided,there seems to be a discrepancy. Let us re-evaluate the calculation.
If $x=12, y=5$,total $= 60$. If $1$ extra student per row ($6$ per row),rows $= 60/6 = 10$ (which is $12-2=10$). Correct.
If $1$ less student per row ($4$ per row),rows $= 60/4 = 15$ (which is $12+3=15$). Correct.
The calculated answer is $60$. Since $60$ is not in the options,let us re-read the question. If the question implies $x$ students per row and $y$ rows,the logic holds. Given the options,perhaps the question intended different values. However,based on the provided logic,the result is $60$.

(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 $k = 6$.

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

(A) Given equations are:
$(1)$ $x + 2y = 5$
$(2)$ $2x + y = 4$
From equation $(1)$,we get $x = 5 - 2y$.
Substitute this value of $x$ into equation $(2)$:
$2(5 - 2y) + y = 4$
$10 - 4y + y = 4$
$10 - 3y = 4$
$-3y = 4 - 10$
$-3y = -6$
$y = 2$
Now,substitute $y = 2$ into $x = 5 - 2y$:
$x = 5 - 2(2)$
$x = 5 - 4$
$x = 1$
Therefore,the solution is $(x, y) = (1, 2)$.

(B) Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
The given equations become:
$3u - 2v = 5$ --- $(1)$
$4u - 5v = 2$ --- $(2)$
To solve for $u$ and $v$,multiply $(1)$ by $5$ and $(2)$ by $2$:
$15u - 10v = 25$ --- $(3)$
$8u - 10v = 4$ --- $(4)$
Subtract $(4)$ from $(3)$:
$(15u - 8u) = 25 - 4$
$7u = 21 \implies u = 3$
Substitute $u = 3$ into $(1)$:
$3(3) - 2v = 5$
$9 - 2v = 5$
$-2v = -4 \implies v = 2$
Thus,$\frac{1}{x} = 3$ and $\frac{1}{y} = 2$.
The value of $\frac{1}{x} - \frac{1}{y} = 3 - 2 = 1$.

(C) Given equations are:
$1) 3x - 2y = 1$
$2) 5x + y = 6$
To eliminate $y$,the coefficients of $y$ in both equations must be equal in magnitude but opposite in sign.
The coefficient of $y$ in the first equation is $-2$.
The coefficient of $y$ in the second equation is $1$.
To make the coefficient of $y$ in the second equation equal to $2$ (so that $-2y + 2y = 0$),we must multiply the entire second equation by $2$.

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

(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.

(C) Given the equations:
$2x + y = 5$ --- $(1)$
$y - 1 = 0$ --- $(2)$
From equation $(2)$,we get $y = 1$.
Substitute the value of $y = 1$ into equation $(1)$:
$2x + 1 = 5$
$2x = 5 - 1$
$2x = 4$
$x = 4 / 2$
$x = 2$
Therefore,the value of $x$ is $2$.

(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 $10(5) + 4 = 54$.

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

(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.

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the sum of the digits is equal to the product of the digits:
$x + y = xy$
Rearranging the equation:
$xy - x - y = 0$
Adding $1$ to both sides to factorize:
$xy - x - y + 1 = 1$
$(x - 1)(y - 1) = 1$
Since $x$ and $y$ are digits ($1$ to $9$ for $x$,$0$ to $9$ for $y$),the only integer solution for $(x - 1)(y - 1) = 1$ is:
$x - 1 = 1 \implies x = 2$
$y - 1 = 1 \implies y = 2$
Thus,the number is $10(2) + 2 = 22$.

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

(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 $k = 3 \times 2 = 6$.

(C) Let the two numbers be $x$ and $y,$ where $x > y.$
According to the problem:
$x + y = 20$ (Equation $1$)
$x - y = 8$ (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(x + y) + (x - y) = 20 + 8$
$2x = 28$
$x = 14$
Substituting $x = 14$ in Equation $1$:
$14 + y = 20$
$y = 20 - 14 = 6$
The two numbers are $14$ and $6.$ The larger number is $14.$

(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 $\left(\frac{22}{31}, \frac{11}{23}\right)$.

(A) Let $x = \frac{1}{a}$ and $y = \frac{1}{b}$. The equations become:
$8x - 9y = 1$ ---$(1)$
$10x + 6y = 7$ ---$(2)$
Multiply equation $(1)$ by $2$ and equation $(2)$ by $3$ to eliminate $y$:
$16x - 18y = 2$ ---$(3)$
$30x + 18y = 21$ ---$(4)$
Adding $(3)$ and $(4)$:
$46x = 23 \implies x = \frac{23}{46} = \frac{1}{2}$.
Since $x = \frac{1}{a}$,we have $a = 2$.
Substitute $x = \frac{1}{2}$ into $(1)$:
$8(\frac{1}{2}) - 9y = 1 \implies 4 - 9y = 1 \implies 9y = 3 \implies y = \frac{1}{3}$.
Since $y = \frac{1}{b}$,we have $b = 3$.
Thus,$(a, b) = (2, 3)$.

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

(D) linear equation in two variables is an equation of the form $ax + by + c = 0$,where $a, b,$ and $c$ are real numbers and $a$ and $b$ are not both zero. 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 $1$.

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

(A) 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 $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.

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