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

(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 $10(3) + 6 = 36$.

(C) The given equations are:
$6x - 3y + 10 = 0$ ... $(1)$
$2x - y + 9 = 0$ ... $(2)$
Comparing these with the general form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have:
$a_1 = 6, b_1 = -3, c_1 = 10$
$a_2 = 2, b_2 = -1, c_2 = 9$
Now,calculate the ratios:
$\frac{a_1}{a_2} = \frac{6}{2} = 3$
$\frac{b_1}{b_2} = \frac{-3}{-1} = 3$
$\frac{c_1}{c_2} = \frac{10}{9}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other and do not intersect.

(D) Given,the equations are $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$.
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 2, c_1 = 5$
$a_2 = -3, b_2 = -6, c_2 = 1$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{-3} = -\frac{1}{3}$
$\frac{b_1}{b_2} = \frac{2}{-6} = -\frac{1}{3}$
$\frac{c_1}{c_2} = \frac{5}{1} = 5$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines represented by these equations are parallel.
Therefore,the pair of equations has no solution.

(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.

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

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

(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) Given that $(x=a, y=b)$ is the solution of the system of linear equations:
$x-y=2$ $\cdots$ $(i)$
$x+y=4$ $\cdots$ $(ii)$
Since $(a, b)$ satisfies both equations,we substitute $x=a$ and $y=b$ into the equations:
$a-b=2$ $\cdots$ $(iii)$
$a+b=4$ $\cdots$ $(iv)$
Adding equations $(iii)$ and $(iv)$:
$(a-b) + (a+b) = 2 + 4$
$2a = 6$
$a = 3$
Substituting $a=3$ into equation $(iv)$:
$3 + b = 4$
$b = 4 - 3$
$b = 1$
Thus,the values of $a$ and $b$ are $3$ and $1$ respectively.

(B) Let the number of $Rs. 1$ coins be $x$ and the number of $Rs. 2$ coins be $y$.
According to the problem,the total number of coins is $50$,so $x + y = 50$ (Equation $i$).
The total value of the money is $Rs. 75$,so $1x + 2y = 75$ (Equation $ii$).
Subtracting Equation $i$ from Equation $ii$:
$(x + 2y) - (x + y) = 75 - 50$
$y = 25$.
Substituting $y = 25$ into Equation $i$:
$x + 25 = 50$
$x = 25$.
Therefore,the number of $Rs. 1$ coins is $25$ and the number of $Rs. 2$ coins is $25$.

(C) Let $x$ be the present age of the father and $y$ be the present age of the son in years.
According to the first condition,the father's age is six times the son's age:
$x = 6y$ --- $(i)$
According to the second condition,four years hence,the father's age will be four times the son's age:
$(x + 4) = 4(y + 4)$
$x + 4 = 4y + 16$
$x - 4y = 12$ --- $(ii)$
Substituting the value of $x$ from $(i)$ into $(ii)$:
$6y - 4y = 12$
$2y = 12$
$y = 6$
Now,substituting $y = 6$ into $(i)$:
$x = 6(6) = 36$
Therefore,the present age of the son is $6$ years and the present age of the father is $36$ years.

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

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

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

(B) The condition for 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.

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

(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.

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

(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.

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

(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.

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

(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 $\frac{9}{10}$.

(B) The given pair of linear equations is:
$2x + 3y - 5 = 0$ ... $(i)$
$px - 6y - 8 = 0$ ... $(ii)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 2, b_1 = 3, c_1 = -5$
$a_2 = p, b_2 = -6, c_2 = -8$
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:
$\frac{2}{p} \neq \frac{3}{-6}$
$\frac{2}{p} \neq -\frac{1}{2}$
By cross-multiplying:
$p \neq -4$
Thus,the system has a unique solution for all real values of $p$ except $-4$,i.e.,$p \in \mathbb{R} - \{-4\}$.

(B) The given pair of linear equations is:
$2x + 3y = 7$ ... $(i)$
$2px + py = 28 - qy$ ... $(ii)$
Rearranging equation $(ii)$,we get:
$2px + (p + q)y = 28$
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 $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Comparing the given equations with the standard form:
$a_1 = 2, b_1 = 3, c_1 = -7$
$a_2 = 2p, b_2 = (p + q), c_2 = -28$
Applying the condition:
$\frac{2}{2p} = \frac{3}{p + q} = \frac{-7}{-28}$
$\frac{1}{p} = \frac{3}{p + q} = \frac{1}{4}$
Taking the first and third parts:
$\frac{1}{p} = \frac{1}{4} \Rightarrow p = 4$
Taking the second and third parts:
$\frac{3}{p + q} = \frac{1}{4} \Rightarrow p + q = 12$
Substituting $p = 4$ into $p + q = 12$:
$4 + q = 12 \Rightarrow q = 8$
Thus,the values are $p = 4$ and $q = 8$.

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

(N/A) pair of linear equations in two variables is given by:
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
For the system to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Given the solution $x = -1$ and $y = 3$,these values must satisfy both equations:
$1$) $a_1(-1) + b_1(3) + c_1 = 0 \Rightarrow -a_1 + 3b_1 + c_1 = 0$
$2$) $a_2(-1) + b_2(3) + c_2 = 0 \Rightarrow -a_2 + 3b_2 + c_2 = 0$
We can choose arbitrary values for the coefficients that satisfy these conditions. For example,let $a_1 = 1, b_1 = 1$. Then $-1 + 3(1) + c_1 = 0 \Rightarrow c_1 = -2$. So,$x + y - 2 = 0$.
Let $a_2 = 1, b_2 = 2$. Then $-1 + 3(2) + c_2 = 0 \Rightarrow -1 + 6 + c_2 = 0 \Rightarrow c_2 = -5$. So,$x + 2y - 5 = 0$.
Since there are infinitely many combinations of $a_1, b_1, c_1$ and $a_2, b_2, c_2$ that satisfy the given solution and the condition for a unique solution,we can write infinitely many such pairs.

(A) Given equations are:
$2x + y = 23$ $(i)$
$4x - y = 19$ $(ii)$
Adding equation $(i)$ and $(ii)$:
$(2x + y) + (4x - y) = 23 + 19$
$6x = 42$
$x = 7$
Substituting $x = 7$ in equation $(i)$:
$2(7) + y = 23$
$14 + y = 23$
$y = 23 - 14 = 9$
Now,calculating the required values:
$5y - 2x = 5(9) - 2(7) = 45 - 14 = 31$
$\frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = -\frac{5}{7}$
Thus,the values are $31$ and $-\frac{5}{7}$.

(B) By the property of a rectangle,opposite sides are equal.
For lengths: $x + 3y = 13$ --- $(i)$
For breadths: $3x + y = 7$ --- $(ii)$
To solve this pair of linear equations,multiply equation $(ii)$ by $3$:
$9x + 3y = 21$ --- $(iii)$
Now,subtract equation $(i)$ from equation $(iii)$:
$(9x + 3y) - (x + 3y) = 21 - 13$
$8x = 8$
$x = 1$
Substitute $x = 1$ into equation $(i)$:
$1 + 3y = 13$
$3y = 12$
$y = 4$
Thus,the required values are $x = 1$ and $y = 4$.

(C) Given equations are:
$(i) \ x + y = 3.3$
$(ii) \ \frac{0.6}{3x - 2y} = -1$
From equation $(ii)$:
$0.6 = -1(3x - 2y)$
$0.6 = -3x + 2y$
$3x - 2y = -0.6 \quad (iii)$
To solve the system,multiply equation $(i)$ by $2$:
$2x + 2y = 6.6 \quad (iv)$
Add equation $(iii)$ and $(iv)$:
$(3x - 2y) + (2x + 2y) = -0.6 + 6.6$
$5x = 6$
$x = \frac{6}{5} = 1.2$
Substitute $x = 1.2$ into equation $(i)$:
$1.2 + y = 3.3$
$y = 3.3 - 1.2$
$y = 2.1$
Thus,the solution is $x = 1.2$ and $y = 2.1$.

(D) Given,the pair of linear equations is:
$\frac{x}{3} + \frac{y}{4} = 4$ $(i)$
Multiplying both sides by the $LCM(3, 4) = 12$,we get:
$4x + 3y = 48$ $(iii)$
And the second equation is:
$\frac{5x}{6} - \frac{y}{8} = 4$ $(ii)$
Multiplying both sides by the $LCM(6, 8) = 24$,we get:
$20x - 3y = 96$ $(iv)$
Now,adding equations $(iii)$ and $(iv)$,we get:
$(4x + 3y) + (20x - 3y) = 48 + 96$
$24x = 144$
$x = \frac{144}{24} = 6$
Now,substitute the value of $x = 6$ in equation $(iii)$:
$4(6) + 3y = 48$
$24 + 3y = 48$
$3y = 48 - 24$
$3y = 24$
$y = 8$
Thus,the solution is $x = 6$ and $y = 8$.

(A) Given pair of linear equations are:
$4x + \frac{6}{y} = 15$ ..... $(i)$
$6x - \frac{8}{y} = 14, y \neq 0$ ..... $(ii)$
Let $u = \frac{1}{y}$,then the equations become:
$4x + 6u = 15$ ..... $(iii)$
$6x - 8u = 14$ ..... $(iv)$
To eliminate $u$,multiply equation $(iii)$ by $4$ and equation $(iv)$ by $3$:
$16x + 24u = 60$ ..... $(v)$
$18x - 24u = 42$ ..... $(vi)$
Adding equation $(v)$ and $(vi)$:
$(16x + 18x) + (24u - 24u) = 60 + 42$
$34x = 102$
$x = \frac{102}{34} = 3$
Substitute $x = 3$ in equation $(iii)$:
$4(3) + 6u = 15$
$12 + 6u = 15$
$6u = 15 - 12 = 3$
$u = \frac{3}{6} = \frac{1}{2}$
Since $u = \frac{1}{y}$,we have $\frac{1}{y} = \frac{1}{2}$,which implies $y = 2$.
Thus,the solution is $x = 3$ and $y = 2$.

(B) Given pair of linear equations is:
$\frac{1}{2x} - \frac{1}{y} = -1$ ........ $(i)$
$\frac{1}{x} + \frac{1}{2y} = 8$ ........ $(ii)$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. Substituting these in the equations:
$\frac{u}{2} - v = -1 \Rightarrow u - 2v = -2$ ........ $(iii)$
$u + \frac{v}{2} = 8 \Rightarrow 2u + v = 16$ ........ $(iv)$
Multiply equation $(iv)$ by $2$ to eliminate $v$:
$4u + 2v = 32$ ........ $(v)$
Adding $(iii)$ and $(v)$:
$(u - 2v) + (4u + 2v) = -2 + 32$
$5u = 30 \Rightarrow u = 6$
Substitute $u = 6$ in equation $(iv)$:
$2(6) + v = 16 \Rightarrow 12 + v = 16 \Rightarrow v = 4$
Since $u = \frac{1}{x} = 6$,we get $x = \frac{1}{6}$.
Since $v = \frac{1}{y} = 4$,we get $y = \frac{1}{4}$.
Thus,the solution is $x = \frac{1}{6}$ and $y = \frac{1}{4}$.