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

(C) Given equations are:
$(1) \quad 4x - 3y = 8$
$(2) \quad 6x - y = \frac{29}{3}$
To eliminate $y$,multiply equation $(2)$ by $3$:
$18x - 3y = 29 \quad (3)$
Subtract equation $(1)$ from equation $(3)$:
$(18x - 3y) - (4x - 3y) = 29 - 8$
$14x = 21$
$x = \frac{21}{14} = \frac{3}{2}$
Substitute $x = \frac{3}{2}$ into equation $(2)$:
$6(\frac{3}{2}) - y = \frac{29}{3}$
$9 - y = \frac{29}{3}$
$y = 9 - \frac{29}{3} = \frac{27 - 29}{3} = -\frac{2}{3}$
Thus,the solution is $(x, y) = (\frac{3}{2}, -\frac{2}{3})$.

(D) Given equations are:
$(1) \ 5x - 3y = 1$
$(2) \ 2x + 5y = 19$
To eliminate $y$,multiply equation $(1)$ by $5$ and equation $(2)$ by $3$:
$25x - 15y = 5 \quad (3)$
$6x + 15y = 57 \quad (4)$
Adding equations $(3)$ and $(4)$:
$(25x + 6x) + (-15y + 15y) = 5 + 57$
$31x = 62$
$x = 2$
Substitute $x = 2$ into equation $(1)$:
$5(2) - 3y = 1$
$10 - 3y = 1$
$-3y = -9$
$y = 3$
Thus,the solution is $(x, y) = (2, 3)$.

(A) Given equations are:
$(1)$ $x + 2y = \frac{3}{2}$
$(2)$ $2x + y = \frac{3}{2}$
To eliminate $y$,multiply equation $(1)$ by $1$ and equation $(2)$ by $2$:
$x + 2y = \frac{3}{2}$
$4x + 2y = 3$
Subtracting the first equation from the second:
$(4x - x) + (2y - 2y) = 3 - \frac{3}{2}$
$3x = \frac{6 - 3}{2} = \frac{3}{2}$
$x = \frac{3}{2 \times 3} = \frac{1}{2}$
Substitute $x = \frac{1}{2}$ into equation $(1)$:
$\frac{1}{2} + 2y = \frac{3}{2}$
$2y = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1$
$y = \frac{1}{2}$
Thus,the solution is $\left(\frac{1}{2}, \frac{1}{2}\right)$.

(B) Given equations are:
$(1) \ 2x + y = 35$
$(2) \ 3x + 4y = 65$
From equation $(1)$,we can express $y$ in terms of $x$:
$y = 35 - 2x$
Substitute this value of $y$ into equation $(2)$:
$3x + 4(35 - 2x) = 65$
$3x + 140 - 8x = 65$
$-5x = 65 - 140$
$-5x = -75$
$x = 15$
Now,substitute $x = 15$ back into the expression for $y$:
$y = 35 - 2(15)$
$y = 35 - 30$
$y = 5$
Finally,calculate the ratio $\frac{x}{y}$:
$\frac{x}{y} = \frac{15}{5} = 3$
Therefore,the correct option is $B$.

(A) Let the number of $25$ paise coins be $x$ and the number of $50$ paise coins be $y$.
According to the problem,the total number of coins is $100$,so $x + y = 100$ (Equation $1$).
The value of $x$ coins of $25$ paise is $0.25x$ rupees,and the value of $y$ coins of $50$ paise is $0.50y$ rupees.
The total amount is Rs. $42.50$,so $0.25x + 0.50y = 42.50$ (Equation $2$).
Multiply Equation $1$ by $0.25$: $0.25x + 0.25y = 25$ (Equation $3$).
Subtract Equation $3$ from Equation $2$: $(0.25x + 0.50y) - (0.25x + 0.25y) = 42.50 - 25$.
This simplifies to $0.25y = 17.50$,so $y = 17.50 / 0.25 = 70$.
Substitute $y = 70$ into Equation $1$: $x + 70 = 100$,so $x = 30$.
Therefore,there are $30$ coins of $25$ paise and $70$ coins of $50$ paise.

(D) Let the digit at the unit's place be $x$ and the digit at the ten's place be $y$.
According to the problem,the digit at the ten's place is double the digit at the unit's place,so $y = 2x$.
The original number can be expressed as $10y + x$.
Substituting $y = 2x$,the original number is $10(2x) + x = 20x + x = 21x$.
When the digits are interchanged,the new number becomes $10x + y$.
Substituting $y = 2x$,the new number is $10x + 2x = 12x$.
According to the problem,the new number is $27$ less than the original number,so $21x - 12x = 27$.
$9x = 27$,which gives $x = 3$.
Since $y = 2x$,we have $y = 2(3) = 6$.
The original number is $10(6) + 3 = 63$.

(A) Let the two numbers be $x$ and $y$.
According to the problem,the sum of the numbers is $9$,so $x + y = 9$ (Equation $1$).
The difference of five times the first number and thrice the other number is $5$,so $5x - 3y = 5$ (Equation $2$).
From Equation $1$,we get $y = 9 - x$.
Substitute this into Equation $2$: $5x - 3(9 - x) = 5$.
$5x - 27 + 3x = 5$.
$8x = 32$.
$x = 4$.
Now,substitute $x = 4$ into $y = 9 - x$: $y = 9 - 4 = 5$.
Thus,the numbers are $4$ and $5$.

(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 $x + y = 13$ (Equation $1$).
The number obtained by interchanging the digits is $10y + x$.
According to the problem,$(10y + x) = (10x + y) + 9$.
Simplifying this,$9y - 9x = 9$,which gives $y - x = 1$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $(x + y) + (y - x) = 13 + 1$,so $2y = 14$,which means $y = 7$.
Substituting $y = 7$ into Equation $1$: $x + 7 = 13$,so $x = 6$.
The original number is $10x + y = 10(6) + 7 = 67$.

(A) Let the breadth of the rectangle be $x \,cm$.
According to the problem, the length is $1.5$ times the breadth, so length $= 1.5x \,cm$.
The perimeter of a rectangle is given by the formula $P = 2 \times (\text{length} + \text{breadth})$.
Given $P = 100 \,cm$, we have $100 = 2 \times (1.5x + x)$.
$100 = 2 \times (2.5x)$.
$100 = 5x$.
$x = 20 \,cm$.
Therefore, the breadth is $20 \,cm$ and the length is $1.5 \times 20 = 30 \,cm$.

(A) Let the present ages of Rajan and Jay be $3x$ and $2x$ respectively.
After five years,their ages will be $(3x + 5)$ and $(2x + 5)$.
According to the problem,the ratio of their ages after five years is $4:3$,so:
$\frac{3x + 5}{2x + 5} = \frac{4}{3}$
Cross-multiplying gives: $3(3x + 5) = 4(2x + 5)$
$9x + 15 = 8x + 20$
$9x - 8x = 20 - 15$
$x = 5$
Therefore,the present age of Rajan is $3 \times 5 = 15$ years and the present age of Jay is $2 \times 5 = 10$ years.

(A) Let the cost of one pen be Rs. $x$ and the cost of one pencil be Rs. $y$.
According to the problem,we have the following system of linear equations:
$3x + 4y = 23$ --- $(1)$
$2x + 3y = 16$ --- $(2)$
To solve this,multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$6x + 8y = 46$ --- $(3)$
$6x + 9y = 48$ --- $(4)$
Subtracting equation $(3)$ from equation $(4)$:
$(6x + 9y) - (6x + 8y) = 48 - 46$
$y = 2$
Substitute $y = 2$ into equation $(1)$:
$3x + 4(2) = 23$
$3x + 8 = 23$
$3x = 15$
$x = 5$
Therefore,the cost of a pen is Rs. $5$ and the cost of a pencil is Rs. $2$.

(A) Let the number of $25$ paise coins be $x$ and the number of $50$ paise coins be $y$.
According to the problem,the total number of coins is $20$,so $x + y = 20$ (Equation $1$).
The total value of the coins is Rs. $7$,which is $700$ paise. Thus,$25x + 50y = 700$ (Equation $2$).
Dividing Equation $2$ by $25$,we get $x + 2y = 28$ (Equation $3$).
Subtracting Equation $1$ from Equation $3$: $(x + 2y) - (x + y) = 28 - 20$,which gives $y = 8$.
Substituting $y = 8$ into Equation $1$: $x + 8 = 20$,which gives $x = 12$.
Therefore,there are $12$ coins of $25$ paise and $8$ coins of $50$ paise.

(N/A) Given equations are:
$1) 0.3 x + 0.4 y = 1$
$2) 6 x + 8 y = 20$
Multiply equation $(1)$ by $10$ to clear the decimals: $3 x + 4 y = 10$ (Equation $3$).
Now,multiply equation $(3)$ by $2$: $6 x + 8 y = 20$ (Equation $4$).
Comparing equation $(2)$ and equation $(4)$,we see they are identical: $6 x + 8 y = 20$.
Since both equations represent the same line,there are infinitely many solutions.
The solution set is $\{(x, y) \mid 3 x + 4 y = 10, x, y \in R \}$.

(D) Given equations are:
$2x + 5y = 7$ --- $(i)$
$4x + 10y = 12$ --- $(ii)$
To eliminate $x$,multiply equation $(i)$ by $2$:
$4x + 10y = 14$ --- $(iii)$
Now,subtract equation $(ii)$ from equation $(iii)$:
$(4x + 10y) - (4x + 10y) = 14 - 12$
$0 = 2$
Since $0 = 2$ is a false statement,the given pair of linear equations has no solution. Thus,the system is inconsistent.

(A) Let the amount invested at $12 \%$ be Rs. $x$ and the amount invested at $10 \%$ be Rs. $y$.
According to the problem,the total interest is Rs. $130$:
$0.12x + 0.10y = 130$ --- (Equation $1$)
If the amounts are interchanged,the interest increases by Rs. $4$,so the new interest is Rs. $134$:
$0.10x + 0.12y = 134$ --- (Equation $2$)
Adding Equation $1$ and Equation $2$:
$0.22x + 0.22y = 264 \implies x + y = 1200$ --- (Equation $3$)
Subtracting Equation $2$ from Equation $1$:
$0.02x - 0.02y = -4 \implies x - y = -200$ --- (Equation $4$)
Adding Equation $3$ and Equation $4$:
$2x = 1000 \implies x = 500$
Substituting $x = 500$ in Equation $3$:
$500 + y = 1200 \implies y = 700$
Therefore,Rajvansh invested Rs. $500$ at $12 \%$ and Rs. $700$ at $10 \%$.

(B) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 2, c_1 = 1$
$a_2 = 2, b_2 = -3, c_2 = -12$
Using the cross-multiplication formula:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substituting the values:
$\frac{x}{(2)(-12) - (-3)(1)} = \frac{y}{(1)(2) - (-12)(1)} = \frac{1}{(1)(-3) - (2)(2)}$
$\frac{x}{-24 + 3} = \frac{y}{2 + 12} = \frac{1}{-3 - 4}$
$\frac{x}{-21} = \frac{y}{14} = \frac{1}{-7}$
Now,solving for $x$:
$\frac{x}{-21} = \frac{1}{-7} \implies x = \frac{-21}{-7} = 3$
Solving for $y$:
$\frac{y}{14} = \frac{1}{-7} \implies y = \frac{14}{-7} = -2$
Thus,the solution is $x = 3, y = -2$.

(C) Converting into the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$2x + y - 35 = 0$ ........ $(1)$
$3x + 4y - 65 = 0$ ........ $(2)$
Comparing with the standard form,we get:
$a_1 = 2, b_1 = 1, c_1 = -35$
$a_2 = 3, b_2 = 4, c_2 = -65$
Using the cross-multiplication formula:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substituting the values:
$\frac{x}{(1)(-65) - (4)(-35)} = \frac{y}{(-35)(3) - (-65)(2)} = \frac{1}{(2)(4) - (3)(1)}$
$\frac{x}{-65 + 140} = \frac{y}{-105 + 130} = \frac{1}{8 - 3}$
$\frac{x}{75} = \frac{y}{25} = \frac{1}{5}$
Now,solving for $x$:
$\frac{x}{75} = \frac{1}{5} \implies x = \frac{75}{5} = 15$
Solving for $y$:
$\frac{y}{25} = \frac{1}{5} \implies y = \frac{25}{5} = 5$
Thus,the solution is $(x, y) = (15, 5)$.

(D) Converting the equations into the standard form $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$:
From $\frac{x}{a} + \frac{y}{b} = 2$,we get $bx + ay = 2ab$,so $bx + ay - 2ab = 0$ ... $(1)$
From $ax - by = a^{2} - b^{2}$,we get $ax - by - (a^{2} - b^{2}) = 0$ ... $(2)$
Comparing with the standard form:
$a_{1} = b, b_{1} = a, c_{1} = -2ab$
$a_{2} = a, b_{2} = -b, c_{2} = -(a^{2} - b^{2}) = b^{2} - a^{2}$
Using the cross multiplication method:
$\frac{x}{b_{1}c_{2} - b_{2}c_{1}} = \frac{y}{c_{1}a_{2} - c_{2}a_{1}} = \frac{1}{a_{1}b_{2} - a_{2}b_{1}}$
$\frac{x}{a(b^{2} - a^{2}) - (-b)(-2ab)} = \frac{y}{(-2ab)(a) - (b^{2} - a^{2})(b)} = \frac{1}{b(-b) - a(a)}$
$\frac{x}{ab^{2} - a^{3} - 2ab^{2}} = \frac{y}{-2a^{2}b - b^{3} + a^{2}b} = \frac{1}{-b^{2} - a^{2}}$
$\frac{x}{-a^{3} - ab^{2}} = \frac{y}{-a^{2}b - b^{3}} = \frac{1}{-(a^{2} + b^{2})}$
$\frac{x}{-a(a^{2} + b^{2})} = \frac{y}{-b(a^{2} + b^{2})} = \frac{1}{-(a^{2} + b^{2})}$
Equating to the constant term:
$x = \frac{-a(a^{2} + b^{2})}{-(a^{2} + b^{2})} = a$
$y = \frac{-b(a^{2} + b^{2})}{-(a^{2} + b^{2})} = b$
Thus,the solution is $(x, y) = (a, b)$.

(A) Comparing the given equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = 1, c_1 = -(a + b) = -a - b$
$a_2 = a, b_2 = -b, c_2 = -(a^2 - b^2) = -a^2 + b^2$
Using the cross multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substituting the values:
$\frac{x}{(1)(-a^2 + b^2) - (-b)(-a - b)} = \frac{y}{(-a - b)(a) - (-a^2 + b^2)(1)} = \frac{1}{(1)(-b) - (a)(1)}$
Simplifying the denominators:
$\frac{x}{-a^2 + b^2 - (ab + b^2)} = \frac{y}{-a^2 - ab + a^2 - b^2} = \frac{1}{-b - a}$
$\frac{x}{-a^2 - ab} = \frac{y}{-b^2 - ab} = \frac{1}{-(a + b)}$
Now,solving for $x$:
$x = \frac{-a^2 - ab}{-(a + b)} = \frac{-a(a + b)}{-(a + b)} = a$
Solving for $y$:
$y = \frac{-b^2 - ab}{-(a + b)} = \frac{-b(b + a)}{-(a + b)} = b$
Thus,the solution is $(x, y) = (a, b)$.

(B) The given equations are:
$ax + by + (-a + b) = 0$
$bx - ay + (-a - b) = 0$
Comparing with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = a, b_1 = b, c_1 = -a + b$
$a_2 = b, b_2 = -a, c_2 = -a - b$
Using the cross-multiplication method:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substituting the values:
$\frac{x}{b(-a - b) - (-a)(-a + b)} = \frac{y}{(-a + b)(b) - (-a - b)(a)} = \frac{1}{a(-a) - b(b)}$
$\frac{x}{-ab - b^2 - (a^2 - ab)} = \frac{y}{-ab + b^2 - (-a^2 - ab)} = \frac{1}{-a^2 - b^2}$
$\frac{x}{-ab - b^2 - a^2 + ab} = \frac{y}{-ab + b^2 + a^2 + ab} = \frac{1}{-a^2 - b^2}$
$\frac{x}{-(a^2 + b^2)} = \frac{y}{a^2 + b^2} = \frac{1}{-(a^2 + b^2)}$
Now,solving for $x$ and $y$:
$x = \frac{-(a^2 + b^2)}{-(a^2 + b^2)} = 1$
$y = \frac{a^2 + b^2}{-(a^2 + b^2)} = -1$
Thus,the solution is $(x, y) = (1, -1)$.

(C) Let the digit at the ten's place be $x$ and the digit at the unit's place be $y$ in the original two-digit number.
Then,the original number $= 10x + y$.
The sum of the digits is $x + y = 9$ ......... $(1)$
The number obtained by interchanging the digits is $10y + x$.
According to the problem,the new number exceeds the original number by $45$:
$10y + x = (10x + y) + 45$
$10y + x - 10x - y = 45$
$-9x + 9y = 45$
Dividing by $9$,we get $-x + y = 5$ ......... $(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (-x + y) = 9 + 5$
$2y = 14$
$y = 7$
Substituting $y = 7$ into equation $(1)$:
$x + 7 = 9$
$x = 2$
Therefore,the original number $= 10x + y = 10(2) + 7 = 27$.

(D) First,rewrite the equations in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$1$) $x - 3y + 5 = 0$
$2$) $3x - 7y + 13 = 0$
Here,$a_1 = 1, b_1 = -3, c_1 = 5$ and $a_2 = 3, b_2 = -7, c_2 = 13$.
Using the cross-multiplication formula:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substitute the values:
$\frac{x}{(-3)(13) - (-7)(5)} = \frac{y}{(5)(3) - (13)(1)} = \frac{1}{(1)(-7) - (3)(-3)}$
$\frac{x}{-39 + 35} = \frac{y}{15 - 13} = \frac{1}{-7 + 9}$
$\frac{x}{-4} = \frac{y}{2} = \frac{1}{2}$
Solving for $x$: $x = \frac{-4}{2} = -2$
Solving for $y$: $y = \frac{2}{2} = 1$
Thus,the solution is $(-2, 1)$.

(A) Given equations are:
$3x - 4y - 17 = 0$ --- $(1)$
$4x - 5y - 21 = 0$ --- $(2)$
Using the cross-multiplication method:
$\frac{x}{(-4)(-21) - (-5)(-17)} = \frac{y}{(-17)(4) - (-21)(3)} = \frac{1}{(3)(-5) - (4)(-4)}$
$\frac{x}{84 - 85} = \frac{y}{-68 + 63} = \frac{1}{-15 + 16}$
$\frac{x}{-1} = \frac{y}{-5} = \frac{1}{1}$
Therefore,$x = -1$ and $y = -5$.
The solution is $(-1, -5)$.

(B) The given equations are:
$3x + y - 7 = 0$ --- $(i)$
$2x + 3y - 14 = 0$ --- $(ii)$
Using the cross-multiplication method,the formula is:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = 3, b_1 = 1, c_1 = -7$ and $a_2 = 2, b_2 = 3, c_2 = -14$.
Substituting the values:
$\frac{x}{(1)(-14) - (3)(-7)} = \frac{y}{(-7)(2) - (-14)(3)} = \frac{1}{(3)(3) - (2)(1)}$
$\frac{x}{-14 + 21} = \frac{y}{-14 + 42} = \frac{1}{9 - 2}$
$\frac{x}{7} = \frac{y}{28} = \frac{1}{7}$
For $x$: $\frac{x}{7} = \frac{1}{7} \implies x = 1$
For $y$: $\frac{y}{28} = \frac{1}{7} \implies y = 4$
Thus,the solution is $(1, 4)$.

(C) Given equations are:
$3x + y - 5 = 0$ --- $(1)$
$5x + 3y - 3 = 0$ --- $(2)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 3, b_1 = 1, c_1 = -5$
$a_2 = 5, b_2 = 3, c_2 = -3$
Using the cross-multiplication formula:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Substituting the values:
$\frac{x}{(1)(-3) - (3)(-5)} = \frac{y}{(-5)(5) - (-3)(3)} = \frac{1}{(3)(3) - (5)(1)}$
$\frac{x}{-3 + 15} = \frac{y}{-25 + 9} = \frac{1}{9 - 5}$
$\frac{x}{12} = \frac{y}{-16} = \frac{1}{4}$
For $x$: $\frac{x}{12} = \frac{1}{4} \implies x = 3$
For $y$: $\frac{y}{-16} = \frac{1}{4} \implies y = -4$
Thus,the solution is $(3, -4)$.

(D) The given equations are:
$4x + 6y - 11 = 0$ ... $(i)$
$5x - 8y - 6 = 0$ ... $(ii)$
Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = 4, b_1 = 6, c_1 = -11$ and $a_2 = 5, b_2 = -8, c_2 = -6$.
$\frac{x}{(6)(-6) - (-8)(-11)} = \frac{y}{(-11)(5) - (-6)(4)} = \frac{1}{(4)(-8) - (5)(6)}$
$\frac{x}{-36 - 88} = \frac{y}{-55 + 24} = \frac{1}{-32 - 30}$
$\frac{x}{-124} = \frac{y}{-31} = \frac{1}{-62}$
For $x$: $x = \frac{-124}{-62} = 2$
For $y$: $y = \frac{-31}{-62} = \frac{1}{2}$
Thus,the solution is $(x, y) = (2, \frac{1}{2})$.

(A) First,simplify the equations:
$1$) Multiply the first equation by $20$: $4x - 5y = 4.5$ or $8x - 10y = 9$,so $8x - 10y - 9 = 0$.
$2$) Multiply the second equation by $12$: $4x - 6y = 5$,so $4x - 6y - 5 = 0$.
Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = 8, b_1 = -10, c_1 = -9$ and $a_2 = 4, b_2 = -6, c_2 = -5$.
$\frac{x}{(-10)(-5) - (-6)(-9)} = \frac{y}{(-9)(4) - (-5)(8)} = \frac{1}{(8)(-6) - (4)(-10)}$
$\frac{x}{50 - 54} = \frac{y}{-36 + 40} = \frac{1}{-48 + 40}$
$\frac{x}{-4} = \frac{y}{4} = \frac{1}{-8}$
$x = \frac{-4}{-8} = \frac{1}{2}$ and $y = \frac{4}{-8} = -\frac{1}{2}$.
Thus,the solution is $(\frac{1}{2}, -\frac{1}{2})$.

(B) Given equations are:
$5x + 2y - 10 = 0$ ---$(1)$
$4x + 3y - 1 = 0$ ---$(2)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 5, b_1 = 2, c_1 = -10$
$a_2 = 4, b_2 = 3, c_2 = -1$
Using the cross-multiplication formula:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
$\frac{x}{(2)(-1) - (3)(-10)} = \frac{y}{(-10)(4) - (-1)(5)} = \frac{1}{(5)(3) - (4)(2)}$
$\frac{x}{-2 + 30} = \frac{y}{-40 + 5} = \frac{1}{15 - 8}$
$\frac{x}{28} = \frac{y}{-35} = \frac{1}{7}$
For $x$: $\frac{x}{28} = \frac{1}{7} \implies x = \frac{28}{7} = 4$
For $y$: $\frac{y}{-35} = \frac{1}{7} \implies y = \frac{-35}{7} = -5$
Thus,the solution is $(4, -5)$.

(C) Given equations are:
$4x - 19y + 13 = 0$ ... $(i)$
$13x - 23y + 19 = 0$ ... $(ii)$
Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = 4, b_1 = -19, c_1 = 13$ and $a_2 = 13, b_2 = -23, c_2 = 19$.
$\frac{x}{(-19)(19) - (-23)(13)} = \frac{y}{(13)(13) - (19)(4)} = \frac{1}{(4)(-23) - (13)(-19)}$
$\frac{x}{-361 + 299} = \frac{y}{169 - 76} = \frac{1}{-92 + 247}$
$\frac{x}{-62} = \frac{y}{93} = \frac{1}{155}$
$x = -62 / 155 = -2 / 5$
$y = 93 / 155 = 3 / 5$
Thus,the solution is $(-2/5, 3/5)$.

(D) The given equations are:
$1$) $\frac{x}{a} + \frac{y}{b} - (a + b) = 0$
$2$) $\frac{x}{a^2} + \frac{y}{b^2} - 2 = 0$
Using the cross-multiplication method for equations $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$:
$\frac{x}{B_1C_2 - B_2C_1} = \frac{y}{C_1A_2 - C_2A_1} = \frac{1}{A_1B_2 - A_2B_1}$
Here,$A_1 = \frac{1}{a}, B_1 = \frac{1}{b}, C_1 = -(a + b)$
$A_2 = \frac{1}{a^2}, B_2 = \frac{1}{b^2}, C_2 = -2$
$\frac{x}{(\frac{1}{b})(-2) - (\frac{1}{b^2})(-(a + b))} = \frac{y}{(-(a + b))(\frac{1}{a^2}) - (-2)(\frac{1}{a})} = \frac{1}{(\frac{1}{a})(\frac{1}{b^2}) - (\frac{1}{a^2})(\frac{1}{b})}$
Simplifying the denominators:
$\frac{x}{-\frac{2}{b} + \frac{a+b}{b^2}} = \frac{y}{-\frac{a+b}{a^2} + \frac{2}{a}} = \frac{1}{\frac{1}{ab^2} - \frac{1}{a^2b}}$
$\frac{x}{\frac{-2b + a + b}{b^2}} = \frac{y}{\frac{-a - b + 2a}{a^2}} = \frac{1}{\frac{a - b}{a^2b^2}}$
$\frac{x}{\frac{a - b}{b^2}} = \frac{y}{\frac{a - b}{a^2}} = \frac{a^2b^2}{a - b}$
$x = \frac{a^2b^2}{a - b} \cdot \frac{a - b}{b^2} = a^2$
$y = \frac{a^2b^2}{a - b} \cdot \frac{a - b}{a^2} = b^2$
Thus,the solution is $(a^2, b^2)$.

(A) The given equations are:
$ax - by - \frac{a-b}{2} = 0$ --- $(1)$
$x + 3y - 2 = 0$ --- $(2)$
Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = a, b_1 = -b, c_1 = -\frac{a-b}{2}$ and $a_2 = 1, b_2 = 3, c_2 = -2$.
$\frac{x}{(-b)(-2) - (3)(-\frac{a-b}{2})} = \frac{y}{(-\frac{a-b}{2})(1) - (-2)(a)} = \frac{1}{(a)(3) - (1)(-b)}$
$\frac{x}{2b + \frac{3a-3b}{2}} = \frac{y}{-\frac{a}{2} + \frac{b}{2} + 2a} = \frac{1}{3a + b}$
$\frac{x}{\frac{4b + 3a - 3b}{2}} = \frac{y}{\frac{-a + b + 4a}{2}} = \frac{1}{3a + b}$
$\frac{x}{\frac{3a + b}{2}} = \frac{y}{\frac{3a + b}{2}} = \frac{1}{3a + b}$
Equating with the constant term:
$x = \frac{3a+b}{2} \cdot \frac{1}{3a+b} = \frac{1}{2}$
$y = \frac{3a+b}{2} \cdot \frac{1}{3a+b} = \frac{1}{2}$
Thus,the solution is $(\frac{1}{2}, \frac{1}{2})$.

(B) Given equations are:
$1$) $\frac{x}{a} - \frac{y}{b} = 0 \implies bx - ay = 0$
$2$) $ax + by = a^2 + b^2 \implies ax + by - (a^2 + b^2) = 0$
Using the cross-multiplication method for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = b, b_1 = -a, c_1 = 0$ and $a_2 = a, b_2 = b, c_2 = -(a^2 + b^2)$.
$\frac{x}{(-a)(-(a^2 + b^2)) - (b)(0)} = \frac{y}{(0)(a) - (-(a^2 + b^2))(b)} = \frac{1}{(b)(b) - (a)(-a)}$
$\frac{x}{a(a^2 + b^2)} = \frac{y}{b(a^2 + b^2)} = \frac{1}{b^2 + a^2}$
$x = \frac{a(a^2 + b^2)}{a^2 + b^2} = a$
$y = \frac{b(a^2 + b^2)}{a^2 + b^2} = b$
Thus,the solution is $(a, b)$.

(C) Given equations are:
$1$) $\frac{x}{a} + \frac{y}{b} = 0 \implies bx + ay = 0$
$2$) $(a+b)x + (a-b)y = a^2 + b^2$
Rewriting in standard form $a_1x + b_1y + c_1 = 0$:
$1$) $bx + ay + 0 = 0$
$2$) $(a+b)x + (a-b)y - (a^2 + b^2) = 0$
Using cross-multiplication method $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$:
$\frac{x}{a(-(a^2+b^2)) - (a-b)(0)} = \frac{y}{0(a+b) - (- (a^2+b^2))b} = \frac{1}{b(a-b) - a(a+b)}$
$\frac{x}{-a(a^2+b^2)} = \frac{y}{b(a^2+b^2)} = \frac{1}{ab - b^2 - a^2 - ab}$
$\frac{x}{-a(a^2+b^2)} = \frac{y}{b(a^2+b^2)} = \frac{1}{-(a^2+b^2)}$
For $x$: $x = \frac{-a(a^2+b^2)}{-(a^2+b^2)} = a$
For $y$: $y = \frac{b(a^2+b^2)}{-(a^2+b^2)} = -b$
Thus,the solution is $(a, -b)$.

(D) Given equations:
$0.5x + 0.3y - 0.1 = 0$ --- $(1)$
$0.6x + 0.3y - 0.3 = 0$ --- $(2)$
Multiply both equations by $10$ to simplify:
$5x + 3y - 1 = 0$
$6x + 3y - 3 = 0$
Using the cross-multiplication formula for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$
Here,$a_1 = 5, b_1 = 3, c_1 = -1$ and $a_2 = 6, b_2 = 3, c_2 = -3$.
$\frac{x}{(3)(-3) - (3)(-1)} = \frac{y}{(-1)(6) - (-3)(5)} = \frac{1}{(5)(3) - (6)(3)}$
$\frac{x}{-9 + 3} = \frac{y}{-6 + 15} = \frac{1}{15 - 18}$
$\frac{x}{-6} = \frac{y}{9} = \frac{1}{-3}$
$x = \frac{-6}{-3} = 2$
$y = \frac{9}{-3} = -3$
Thus,the solution is $(2, -3)$.

(A) Let the larger number be $x$ and the smaller number be $y$.
According to the first condition,$x = y + 3$ or $x - y = 3$ (Equation $1$).
According to the second condition,$2x + 17 = 3y$ or $2x - 3y = -17$ (Equation $2$).
Substitute $x = y + 3$ into Equation $2$:
$2(y + 3) - 3y = -17$
$2y + 6 - 3y = -17$
$-y + 6 = -17$
$-y = -23$
$y = 23$.
Now,find $x$ using $x = y + 3$:
$x = 23 + 3 = 26$.
Therefore,the numbers are $26$ and $23$.

(B) 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 $y = 5$ in equation $(1)$:
$x - 2(5) = 2$
$x - 10 = 2$
$x = 12$
Total number of students $= xy = 12 \times 5 = 60$.

(C) Let the tens digit be $x$ and the units digit be $y$. The original number is $10x + y$.
The number obtained by interchanging the digits is $10y + x$.
According to the first condition: $(10x + y) + (10y + x) = 110$.
$11x + 11y = 110 \implies x + y = 10$ (Equation $1$).
According to the second condition: $(10x + y - 10) = 5(x + y) + 4$.
Substitute $x + y = 10$ into the equation: $10x + y - 10 = 5(10) + 4$.
$10x + y - 10 = 54 \implies 10x + y = 64$ (Equation $2$).
From $x + y = 10$ and $10x + y = 64$,we subtract the equations: $(10x + y) - (x + y) = 64 - 10$.
$9x = 54 \implies x = 6$.
Substituting $x = 6$ into $x + y = 10$,we get $y = 4$.
Thus,the original number is $10(6) + 4 = 64$.

(A) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$m \angle A + m \angle C = 180^{\circ}$ and $m \angle B + m \angle D = 180^{\circ}$.
For $A$ and $C$: $(2x + 4) + (2y + 10) = 180 \implies 2x + 2y + 14 = 180 \implies 2x + 2y = 166 \implies x + y = 83$ (Equation $1$).
For $B$ and $D$: $(y + 3) + (4x - 5) = 180 \implies 4x + y - 2 = 180 \implies 4x + y = 182$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(4x + y) - (x + y) = 182 - 83 \implies 3x = 99 \implies x = 33$.
Substituting $x = 33$ into Equation $1$: $33 + y = 83 \implies y = 50$.
Calculating the angles:
$m \angle A = 2(33) + 4 = 66 + 4 = 70^{\circ}$.
$m \angle B = 50 + 3 = 53^{\circ}$.
$m \angle C = 2(50) + 10 = 100 + 10 = 110^{\circ}$.
$m \angle D = 4(33) - 5 = 132 - 5 = 127^{\circ}$.

(A) Let the breadth of the rectangle be $x$ and the length be $y$.
According to the problem,the length is $5$ less than twice the breadth: $y = 2x - 5$,which can be written as $2x - y - 5 = 0$.
The perimeter of a rectangle is $2(l + b) = 110$,so $l + b = 55$,which means $x + y = 55$,or $x + y - 55 = 0$.
We have the system of equations:
$2x - y - 5 = 0$
$x + y - 55 = 0$
Using the cross-multiplication method:
$\frac{x}{(-1)(-55) - (1)(-5)} = \frac{-y}{(2)(-55) - (1)(-5)} = \frac{1}{(2)(1) - (1)(-1)}$
$\frac{x}{55 + 5} = \frac{-y}{-110 + 5} = \frac{1}{2 + 1}$
$\frac{x}{60} = \frac{-y}{-105} = \frac{1}{3}$
$x = \frac{60}{3} = 20$ and $y = \frac{105}{3} = 35$.
The area of the rectangle is $l \times b = 35 \times 20 = 700$ square units.

(N/A) Given equations are $2x + y = \frac{7xy}{3}$ and $x + 3y = \frac{11xy}{3}$.
It is clear that $x = 0$ and $y = 0$ satisfy both equations. Thus,$(0, 0)$ is a solution.
Assuming $x \neq 0$ and $y \neq 0$,we divide both equations by $xy$:
For the first equation: $\frac{2x}{xy} + \frac{y}{xy} = \frac{7}{3} \implies \frac{2}{y} + \frac{1}{x} = \frac{7}{3} \implies \frac{6}{y} + \frac{3}{x} = 7$ ... $(1)$
For the second equation: $\frac{x}{xy} + \frac{3y}{xy} = \frac{11}{3} \implies \frac{1}{y} + \frac{3}{x} = \frac{11}{3} \implies \frac{3}{y} + \frac{9}{x} = 11$ ... $(2)$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$3u + 6v = 7$ ... $(3)$
$9u + 3v = 11$ ... $(4)$
Multiply equation $(4)$ by $2$: $18u + 6v = 22$ ... $(5)$
Subtract equation $(3)$ from $(5)$: $(18u - 3u) + (6v - 6v) = 22 - 7 \implies 15u = 15 \implies u = 1$.
Substitute $u = 1$ into equation $(3)$: $3(1) + 6v = 7 \implies 6v = 4 \implies v = \frac{2}{3}$.
Since $u = \frac{1}{x} = 1 \implies x = 1$ and $v = \frac{1}{y} = \frac{2}{3} \implies y = \frac{3}{2}$.
The solutions are $(x, y) = (0, 0)$ and $(x, y) = (1, \frac{3}{2})$.

(C) Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. The equations become:
$2x + 3y = 1$ ....... $(1)$
$x + y = 1$ ........... $(2)$
Multiply equation $(2)$ by $2$ to get $2x + 2y = 2$ ....... $(3)$
Subtract equation $(3)$ from equation $(1)$:
$(2x + 3y) - (2x + 2y) = 1 - 2$
$y = -1$
Substitute $y = -1$ into equation $(2)$:
$x + (-1) = 1$
$x = 2$
Since $x = \frac{1}{a} = 2$,we get $a = \frac{1}{2}$.
Since $y = \frac{1}{b} = -1$,we get $b = -1$.
Thus,the solution is $(a, b) = \left(\frac{1}{2}, -1\right)$.

(D) Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. The equations become:
$5x + \frac{3}{2}y = 1$ and $\frac{1}{2}x - 3y = 1$
Multiplying the first equation by $2$,we get $10x + 3y = 2$ ... $(1)$
Multiplying the second equation by $2$,we get $x - 6y = 2$ ... $(2)$
To eliminate $y$,multiply equation $(1)$ by $2$ and add to equation $(2)$:
$20x + 6y = 4$
$x - 6y = 2$
Adding these gives $21x = 6$,so $x = \frac{6}{21} = \frac{2}{7}$.
Substitute $x = \frac{2}{7}$ into equation $(1)$:
$10(\frac{2}{7}) + 3y = 2$
$\frac{20}{7} + 3y = 2$
$3y = 2 - \frac{20}{7} = \frac{14 - 20}{7} = -\frac{6}{7}$
$y = -\frac{6}{7} \times \frac{1}{3} = -\frac{2}{7}$.
Since $x = \frac{1}{a} = \frac{2}{7}$,we have $a = \frac{7}{2}$.
Since $y = \frac{1}{b} = -\frac{2}{7}$,we have $b = -\frac{7}{2}$.
Thus,the solution is $(a, b) = (\frac{7}{2}, -\frac{7}{2})$.

(A) Given equations are:
$\frac{2}{a} + \frac{3}{b} = -1$ --- $(1)$
$\frac{4}{a} - \frac{9}{b} = -7$ --- $(2)$
Let $\frac{1}{a} = x$ and $\frac{1}{b} = y$. Then the equations become:
$2x + 3y = -1$ --- $(3)$
$4x - 9y = -7$ --- $(4)$
Multiply equation $(3)$ by $3$ to eliminate $y$:
$6x + 9y = -3$ --- $(5)$
Adding equation $(4)$ and $(5)$:
$(4x - 9y) + (6x + 9y) = -7 + (-3)$
$10x = -10$
$x = -1$
Since $x = \frac{1}{a}$,we have $\frac{1}{a} = -1$,so $a = -1$.
Substitute $x = -1$ into equation $(3)$:
$2(-1) + 3y = -1$
$-2 + 3y = -1$
$3y = 1$
$y = \frac{1}{3}$
Since $y = \frac{1}{b}$,we have $\frac{1}{b} = \frac{1}{3}$,so $b = 3$.
Thus,the solution is $(a, b) = (-1, 3)$.

(A) Let the speed of the boat in still water be $x \, km/hr$ and the speed of the stream be $y \, km/hr$. It is necessary that $x > y$.
The speed of the boat downstream is $(x+y) \, km/hr$ and the speed of the boat upstream is $(x-y) \, km/hr$.
Using the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$,we have:
For the first case: $\frac{32}{x-y} + \frac{36}{x+y} = 7$ --- $(1)$
For the second case: $\frac{40}{x-y} + \frac{48}{x+y} = 9$ --- $(2)$
Let $\frac{1}{x-y} = a$ and $\frac{1}{x+y} = b$. The equations become:
$32a + 36b = 7$ --- $(3)$
$40a + 48b = 9$ --- $(4)$
Solving these linear equations by elimination or substitution:
Multiply $(3)$ by $4$ and $(4)$ by $3$:
$128a + 144b = 28$
$120a + 144b = 27$
Subtracting the two equations: $8a = 1 \implies a = \frac{1}{8}$.
Substituting $a = \frac{1}{8}$ in $(3)$:
$32(\frac{1}{8}) + 36b = 7 \implies 4 + 36b = 7 \implies 36b = 3 \implies b = \frac{1}{12}$.
Now,$x-y = 8$ and $x+y = 12$.
Adding these: $2x = 20 \implies x = 10$.
Subtracting these: $2y = 4 \implies y = 2$.
Thus,the speed of the boat in still water is $10 \, km/hr$ and the speed of the stream is $2 \, km/hr$.

(C) Given equations are:
$(1)$ $x + y = 5xy$
$(2)$ $3x + 2y = 13xy$
Dividing both equations by $xy$ (assuming $x \neq 0, y \neq 0$):
From $(1)$: $1/y + 1/x = 5$ => $1/x + 1/y = 5$ --- $(3)$
From $(2)$: $3/y + 2/x = 13$ => $2/x + 3/y = 13$ --- $(4)$
Let $u = 1/x$ and $v = 1/y$. The equations become:
$(3)$ $u + v = 5$
$(4)$ $2u + 3v = 13$
Multiply $(3)$ by $2$: $2u + 2v = 10$ --- $(5)$
Subtract $(5)$ from $(4)$: $(2u + 3v) - (2u + 2v) = 13 - 10$ => $v = 3$
Substitute $v = 3$ in $(3)$: $u + 3 = 5$ => $u = 2$
Since $u = 1/x = 2$,$x = 1/2$.
Since $v = 1/y = 3$,$y = 1/3$.
Also,if $x=0$ and $y=0$,both equations are satisfied $(0=0)$.
Thus,the solutions are $(0, 0)$ and $(1/2, 1/3)$.

(A) Let $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$.
Substituting these into the given equations,we get:
$5u + v = 2$ --- $(1)$
$6u - 3v = 1$ --- $(2)$
Multiply equation $(1)$ by $3$ to eliminate $v$:
$15u + 3v = 6$ --- $(3)$
Adding equation $(2)$ and $(3)$:
$(6u - 3v) + (15u + 3v) = 1 + 6$
$21u = 7$
$u = \frac{7}{21} = \frac{1}{3}$
Substitute $u = \frac{1}{3}$ into equation $(1)$:
$5(\frac{1}{3}) + v = 2$
$v = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$
Now,solve for $x$ and $y$:
$\frac{1}{x-1} = \frac{1}{3} \implies x-1 = 3 \implies x = 4$
$\frac{1}{y-2} = \frac{1}{3} \implies y-2 = 3 \implies y = 5$
Thus,the solution is $(x, y) = (4, 5)$.

(A) Given equations are:
$1) 2(3u - v) = 5uv \implies 6u - 2v = 5uv$
$2) 2(u + 3v) = 5uv \implies 2u + 6v = 5uv$
Divide both equations by $uv$ (assuming $u, v \neq 0$):
$1) \frac{6}{v} - \frac{2}{u} = 5$
$2) \frac{2}{v} + \frac{6}{u} = 5$
Let $x = \frac{1}{u}$ and $y = \frac{1}{v}$. The equations become:
$1) -2x + 6y = 5$
$2) 6x + 2y = 5$
Multiply equation $(2)$ by $3$: $18x + 6y = 15$. Subtract equation $(1)$ from this:
$(18x + 6y) - (-2x + 6y) = 15 - 5 \implies 20x = 10 \implies x = \frac{1}{2}$.
Substitute $x = \frac{1}{2}$ into $6x + 2y = 5$:
$6(\frac{1}{2}) + 2y = 5 \implies 3 + 2y = 5 \implies 2y = 2 \implies y = 1$.
Since $x = \frac{1}{u} = \frac{1}{2} \implies u = 2$ and $y = \frac{1}{v} = 1 \implies v = 1$.
Thus,$(u, v) = (2, 1)$.

(B) Given equations are:
$(1)$ $\frac{2}{x} + \frac{3}{y} = \frac{9}{xy}$
$(2)$ $\frac{4}{x} + \frac{9}{y} = \frac{21}{xy}$
Multiply both equations by $xy$ (since $x \neq 0, y \neq 0$):
$(1)$ $2y + 3x = 9 \implies 3x + 2y = 9$
$(2)$ $4y + 9x = 21 \implies 9x + 4y = 21$
Multiply the first equation by $2$ to eliminate $y$:
$2(3x + 2y) = 2(9) \implies 6x + 4y = 18$
Subtract this from the second equation:
$(9x + 4y) - (6x + 4y) = 21 - 18$
$3x = 3 \implies x = 1$
Substitute $x = 1$ into $3x + 2y = 9$:
$3(1) + 2y = 9$
$3 + 2y = 9 \implies 2y = 6 \implies y = 3$
Thus,the solution is $(x, y) = (1, 3)$.

(C) Let $x = \frac{1}{a}$ and $y = \frac{1}{b}$.
The equations become:
$5x - 6y = 3$ --- $(1)$
$x + 4y = 11$ --- $(2)$
From equation $(2)$,$x = 11 - 4y$.
Substitute $x$ in equation $(1)$:
$5(11 - 4y) - 6y = 3$
$55 - 20y - 6y = 3$
$55 - 26y = 3$
$26y = 52$
$y = 2$.
Since $y = \frac{1}{b} = 2$,then $b = \frac{1}{2}$.
Substitute $y = 2$ in $x = 11 - 4y$:
$x = 11 - 4(2) = 11 - 8 = 3$.
Since $x = \frac{1}{a} = 3$,then $a = \frac{1}{3}$.
Thus,$(a, b) = \left(\frac{1}{3}, \frac{1}{2}\right)$.

(D) Let $x = \frac{1}{a}$ and $y = \frac{1}{b}$.
The equations become:
$9x + 8y = 7$ --- $(1)$
$2x - 3y = -8$ --- $(2)$
Multiply $(1)$ by $3$ and $(2)$ by $8$ to eliminate $y$:
$27x + 24y = 21$ --- $(3)$
$16x - 24y = -64$ --- $(4)$
Adding $(3)$ and $(4)$:
$43x = -43 \implies x = -1$.
Since $x = \frac{1}{a} = -1$,we get $a = -1$.
Substitute $x = -1$ into $(2)$:
$2(-1) - 3y = -8$
$-2 - 3y = -8$
$-3y = -6 \implies y = 2$.
Since $y = \frac{1}{b} = 2$,we get $b = \frac{1}{2}$.
Thus,$(a, b) = \left(-1, \frac{1}{2}\right)$.