Textbook - Pair of Linear Equations in Two Variables Questions in English

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

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

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

(N/A) Let the present age of Nuri be $x$ years and the present age of Sonu be $y$ years.
According to the given information:
Five years ago,Nuri's age was $(x-5)$ and Sonu's age was $(y-5)$.
$(x-5) = 3(y-5)$
$x - 5 = 3y - 15$
$x - 3y = -10$ $...(1)$
Ten years later,Nuri's age will be $(x+10)$ and Sonu's age will be $(y+10)$.
$(x+10) = 2(y+10)$
$x + 10 = 2y + 20$
$x - 2y = 10$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(x - 2y) - (x - 3y) = 10 - (-10)$
$x - 2y - x + 3y = 10 + 10$
$y = 20$
Substituting $y = 20$ in equation $(1)$:
$x - 3(20) = -10$
$x - 60 = -10$
$x = 50$
Thus,the present age of Nuri is $50$ years and the present age of Sonu is $20$ years.

(A) Let the unit digit be $x$ and the tens digit be $y$. Then,the number is $10y + x$.
According to the first condition,the sum of the digits is $9$:
$x + y = 9$ $...(1)$
According to the second condition,nine times the number is twice the number obtained by reversing the digits:
$9(10y + x) = 2(10x + y)$
$90y + 9x = 20x + 2y$
$88y - 11x = 0$
Dividing by $11$,we get:
$8y - x = 0$ or $-x + 8y = 0$ $...(2)$
Adding equation $(1)$ and $(2)$:
$(x + y) + (-x + 8y) = 9 + 0$
$9y = 9$
$y = 1$
Substituting $y = 1$ in equation $(1)$:
$x + 1 = 9$
$x = 8$
Therefore,the number is $10y + x = 10(1) + 8 = 18$.

(A) Let the number of ₹ $50$ notes be $x$ and the number of ₹ $100$ notes be $y$.
According to the given information:
$1$. The total number of notes is $25$,so $x + y = 25$ $...(1)$
$2$. The total value of the notes is ₹ $2000$,so $50x + 100y = 2000$ $...(2)$
To solve by the elimination method,multiply equation $(1)$ by $50$:
$50x + 50y = 1250$ $...(3)$
Subtract equation $(3)$ from equation $(2)$:
$(50x + 100y) - (50x + 50y) = 2000 - 1250$
$50y = 750$
$y = 15$
Substitute $y = 15$ into equation $(1)$:
$x + 15 = 25$
$x = 10$
Therefore,Meena received $10$ notes of ₹ $50$ and $15$ notes of ₹ $100$.

(N/A) Let the fixed charge for the first three days be $Rs$ $x$ and the additional charge for each day thereafter be $Rs$ $y$.
According to the given information:
For Saritha: The book was kept for $7$ days. This includes $3$ fixed days and $4$ extra days. So,$x + 4y = 27$ $...(1)$
For Susy: The book was kept for $5$ days. This includes $3$ fixed days and $2$ extra days. So,$x + 2y = 21$ $...(2)$
Subtracting equation $(2)$ from equation $(1)$:
$(x + 4y) - (x + 2y) = 27 - 21$
$2y = 6$
$y = 3$ $...(3)$
Substituting the value of $y = 3$ in equation $(1)$:
$x + 4(3) = 27$
$x + 12 = 27$
$x = 15$
Therefore,the fixed charge is $Rs$ $15$ and the charge for each extra day is $Rs$ $3$.

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

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

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

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

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

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

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

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

(B) આપેલ સમીકરણોને પ્રમાણિત સ્વરૂપમાં લખતા:
$3x + y - 1 = 0$
$(2k - 1)x + (k - 1)y - (2k + 1) = 0$
અહીં,$a_1 = 3, b_1 = 1, c_1 = -1$ અને $a_2 = 2k - 1, b_2 = k - 1, c_2 = -(2k + 1)$ છે.
સુરેખ સમીકરણોની જોડને કોઈ ઉકેલ ન હોય તેની શરત છે:
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
તેથી,$\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{-1}{-(2k + 1)}$
$\frac{3}{2k - 1} = \frac{1}{k - 1}$ લેતા:
$3(k - 1) = 1(2k - 1)$
$3k - 3 = 2k - 1$
$3k - 2k = -1 + 3$
$k = 2$
આમ,$k = 2$ માટે આપેલ સમીકરણોની જોડને કોઈ ઉકેલ નથી.

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

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

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

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

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

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

(N/A) Let us write the given pair of equations as:
$2\left(\frac{1}{x}\right) + 3\left(\frac{1}{y}\right) = 13$ $...(1)$
$5\left(\frac{1}{x}\right) - 4\left(\frac{1}{y}\right) = -2$ $...(2)$
These equations are not in the standard form $ax + by + c = 0$. However,if we substitute $\frac{1}{x} = p$ and $\frac{1}{y} = q$ in equations $(1)$ and $(2)$,we get:
$2p + 3q = 13$ $...(3)$
$5p - 4q = -2$ $...(4)$
Now,we solve these linear equations using the elimination method. Multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$8p + 12q = 52$ $...(5)$
$15p - 12q = -6$ $...(6)$
Adding equations $(5)$ and $(6)$:
$23p = 46 \implies p = 2$
Substituting $p = 2$ in equation $(3)$:
$2(2) + 3q = 13 \implies 4 + 3q = 13 \implies 3q = 9 \implies q = 3$
Since $p = \frac{1}{x}$ and $q = \frac{1}{y}$,we have:
$\frac{1}{x} = 2 \implies x = \frac{1}{2}$
$\frac{1}{y} = 3 \implies y = \frac{1}{3}$
Thus,the solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

(A) Let us substitute $\frac{1}{x-1} = p$ and $\frac{1}{y-2} = q$. Then the given equations can be written as:
$5p + q = 2$ $...(1)$
$6p - 3q = 1$ $...(2)$
To solve these,multiply equation $(1)$ by $3$:
$15p + 3q = 6$ $...(3)$
Adding equation $(2)$ and $(3)$:
$(6p - 3q) + (15p + 3q) = 1 + 6$
$21p = 7$
$p = \frac{7}{21} = \frac{1}{3}$
Substitute $p = \frac{1}{3}$ into equation $(1)$:
$5(\frac{1}{3}) + q = 2$
$\frac{5}{3} + q = 2$
$q = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$
Now,substitute back the values of $p$ and $q$:
$\frac{1}{x-1} = \frac{1}{3} \implies x-1 = 3 \implies x = 4$
$\frac{1}{y-2} = \frac{1}{3} \implies y-2 = 3 \implies y = 5$
Thus,the solution is $x = 4, y = 5$.

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

(X=1/2, Y=1/3) Given equations:
$\frac{1}{2x} + \frac{1}{3y} = 2$ ...$(1)$
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$ ...$(2)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$\frac{p}{2} + \frac{q}{3} = 2 \Rightarrow 3p + 2q = 12$ ...$(3)$
$\frac{p}{3} + \frac{q}{2} = \frac{13}{6} \Rightarrow 2p + 3q = 13$ ...$(4)$
Multiply equation $(3)$ by $3$ and equation $(4)$ by $2$:
$9p + 6q = 36$ ...$(5)$
$4p + 6q = 26$ ...$(6)$
Subtracting $(6)$ from $(5)$:
$(9p - 4p) + (6q - 6q) = 36 - 26$
$5p = 10 \Rightarrow p = 2$
Substitute $p = 2$ in equation $(3)$:
$3(2) + 2q = 12$
$6 + 2q = 12 \Rightarrow 2q = 6 \Rightarrow q = 3$
Since $p = \frac{1}{x} = 2$,we get $x = \frac{1}{2}$.
Since $q = \frac{1}{y} = 3$,we get $y = \frac{1}{3}$.

(X=4, Y=9) Let $\frac{1}{\sqrt{x}} = p$ and $\frac{1}{\sqrt{y}} = q$.
Substituting these into the given equations,we obtain:
$2p + 3q = 2$ $...(1)$
$4p - 9q = -1$ $...(2)$
To eliminate $q$,multiply equation $(1)$ by $3$:
$6p + 9q = 6$ $...(3)$
Adding equation $(2)$ and $(3)$:
$(4p - 9q) + (6p + 9q) = -1 + 6$
$10p = 5$
$p = \frac{5}{10} = \frac{1}{2}$
Substitute $p = \frac{1}{2}$ into equation $(1)$:
$2(\frac{1}{2}) + 3q = 2$
$1 + 3q = 2$
$3q = 1$
$q = \frac{1}{3}$
Now,solve for $x$ and $y$:
$p = \frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 4$
$q = \frac{1}{\sqrt{y}} = \frac{1}{3} \implies \sqrt{y} = 3 \implies y = 9$
Thus,the solution is $x = 4, y = 9$.

(A) Given equations are:
$\frac{4}{x} + 3y = 14$ $...(1)$
$\frac{3}{x} - 4y = 23$ $...(2)$
Let $\frac{1}{x} = p$. Substituting this in the equations,we get:
$4p + 3y = 14$ $...(3)$
$3p - 4y = 23$ $...(4)$
Using the elimination method,multiply equation $(3)$ by $4$ and equation $(4)$ by $3$:
$16p + 12y = 56$ $...(5)$
$9p - 12y = 69$ $...(6)$
Adding equations $(5)$ and $(6)$:
$25p = 125$
$p = 5$
Since $p = \frac{1}{x}$,we have $\frac{1}{x} = 5$,which implies $x = \frac{1}{5}$.
Substitute $p = 5$ in equation $(3)$:
$4(5) + 3y = 14$
$20 + 3y = 14$
$3y = 14 - 20$
$3y = -6$
$y = -2$
Thus,the solution is $x = \frac{1}{5}$ and $y = -2$.

(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 $(1)$:
$5(\frac{1}{3}) + v = 2$
$\frac{5}{3} + v = 2$
$v = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$
Now,substitute back the values of $u$ and $v$:
$\frac{1}{x-1} = \frac{1}{3} \implies x-1 = 3 \implies x = 4$
$\frac{1}{y-2} = \frac{1}{3} \implies y-2 = 3 \implies y = 5$
Thus,the solution is $x = 4, y = 5$.

(A) Given equations are:
$1$) $\frac{7x - 2y}{xy} = 5$
$2$) $\frac{8x + 7y}{xy} = 15$
Step $1$: Simplify the equations by dividing each term in the numerator by the denominator $xy$:
Equation $(1)$ becomes: $\frac{7x}{xy} - \frac{2y}{xy} = 5 \implies \frac{7}{y} - \frac{2}{x} = 5$
Equation $(2)$ becomes: $\frac{8x}{xy} + \frac{7y}{xy} = 15 \implies \frac{8}{y} + \frac{7}{x} = 15$
Step $2$: Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$(i)$ $-2u + 7v = 5$
(ii) $7u + 8v = 15$
Step $3$: Solve the system of linear equations using the elimination method.
Multiply $(i)$ by $7$ and (ii) by $2$:
$-14u + 49v = 35$
$14u + 16v = 30$
Adding the two equations: $65v = 65 \implies v = 1$.
Step $4$: Substitute $v = 1$ into $(i)$:
$-2u + 7(1) = 5 \implies -2u = -2 \implies u = 1$.
Step $5$: Find $x$ and $y$:
Since $u = \frac{1}{x} = 1 \implies x = 1$.
Since $v = \frac{1}{y} = 1 \implies y = 1$.
Final Answer: $x = 1, y = 1$.

(A) Given equations are:
$(1)$ $6x + 3y = 6xy$
$(2)$ $2x + 4y = 5xy$
Divide both equations by $xy$:
From $(1)$: $\frac{6x}{xy} + \frac{3y}{xy} = \frac{6xy}{xy} \implies \frac{6}{y} + \frac{3}{x} = 6$
From $(2)$: $\frac{2x}{xy} + \frac{4y}{xy} = \frac{5xy}{xy} \implies \frac{2}{y} + \frac{4}{x} = 5$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$(3)$ $3u + 6v = 6$
$(4)$ $4u + 2v = 5$
Multiply $(4)$ by $3$ to eliminate $v$:
$12u + 6v = 15$ $(5)$
Subtract $(3)$ from $(5)$:
$(12u + 6v) - (3u + 6v) = 15 - 6$
$9u = 9 \implies u = 1$
Substitute $u = 1$ into $(3)$:
$3(1) + 6v = 6 \implies 6v = 3 \implies v = \frac{1}{2}$
Since $u = \frac{1}{x} = 1 \implies x = 1$
Since $v = \frac{1}{y} = \frac{1}{2} \implies y = 2$
Thus,the solution is $(x, y) = (1, 2)$.

(X=3, Y=2) Let $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$.
The equations become:
$10u + 2v = 4$ --- $(1)$
$15u - 5v = -2$ --- $(2)$
Multiply $(1)$ by $5$ and $(2)$ by $2$ to eliminate $v$:
$50u + 10v = 20$
$30u - 10v = -4$
Adding these equations: $80u = 16$,so $u = \frac{16}{80} = \frac{1}{5}$.
Substitute $u = \frac{1}{5}$ into $(1)$:
$10(\frac{1}{5}) + 2v = 4 \implies 2 + 2v = 4 \implies 2v = 2 \implies v = 1$.
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 $(4)$: $2x = 6 \implies x = 3$.
Subtracting $(4)$ from $(3)$: $2y = 4 \implies y = 2$.
Thus,the solution is $x = 3$ and $y = 2$.

(N/A) Let $u = \frac{1}{3x+y}$ and $v = \frac{1}{3x-y}$.
Substituting these into the given equations:
$u + v = \frac{3}{4}$ --- $(1)$
$\frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8}$ --- $(2)$
Multiply equation $(2)$ by $2$ to simplify:
$u - v = -\frac{1}{4}$ --- $(3)$
Adding equation $(1)$ and $(3)$:
$(u + v) + (u - v) = \frac{3}{4} - \frac{1}{4}$
$2u = \frac{2}{4} = \frac{1}{2}$
$u = \frac{1}{4}$
Substitute $u = \frac{1}{4}$ into equation $(1)$:
$\frac{1}{4} + v = \frac{3}{4}$
$v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
Now,solve for $x$ and $y$:
$3x+y = \frac{1}{u} = 4$ --- $(4)$
$3x-y = \frac{1}{v} = 2$ --- $(5)$
Adding $(4)$ and $(5)$:
$6x = 6 \implies x = 1$
Substitute $x = 1$ into $(4)$:
$3(1) + y = 4 \implies y = 1$
Thus,the solution is $x = 1, y = 1$.

(A) Let the speed of Ritu in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Speed of Ritu while rowing:
Upstream $= (x - y)\, km/h$
Downstream $= (x + y)\, km/h$
According to the problem:
For downstream: $2(x + y) = 20 \Rightarrow x + y = 10$ $...(1)$
For upstream: $2(x - y) = 4 \Rightarrow x - y = 2$ $...(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 10 + 2$
$2x = 12 \Rightarrow x = 6$
Substituting $x = 6$ in equation $(1)$:
$6 + y = 10 \Rightarrow y = 4$
Hence, Ritu's speed in still water is $6\, km/h$ and the speed of the current is $4\, km/h$.

(A) Let the number of days taken by a woman and a man be $x$ and $y$ respectively.
Therefore,the work done by a woman in $1$ day $= \frac{1}{x}$.
The work done by a man in $1$ day $= \frac{1}{y}$.
According to the problem:
$4(\frac{2}{x} + \frac{5}{y}) = 1 \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4}$ $(i)$
$3(\frac{3}{x} + \frac{6}{y}) = 1 \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3}$ $(ii)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$2p + 5q = \frac{1}{4} \Rightarrow 8p + 20q = 1$ $(iii)$
$3p + 6q = \frac{1}{3} \Rightarrow 9p + 18q = 1$ $(iv)$
Solving by cross-multiplication:
$\frac{p}{(20)(-1) - (18)(-1)} = \frac{q}{(1)(9) - (1)(8)} = \frac{1}{(8)(18) - (20)(9)}$
$\frac{p}{-20 + 18} = \frac{q}{9 - 8} = \frac{1}{144 - 180}$
$\frac{p}{-2} = \frac{q}{1} = \frac{1}{-36}$
$p = \frac{-2}{-36} = \frac{1}{18}$ and $q = \frac{1}{-36} = -\frac{1}{36}$ (Wait,re-calculating: $p = \frac{1}{18}, q = \frac{1}{36}$).
Thus,$x = 18$ and $y = 36$.
Therefore,$1$ woman alone takes $18$ days and $1$ man alone takes $36$ days.

(N/A) Let the speed of the train be $u \ km/h$ and the speed of the bus be $v \ km/h$.
According to the given information:
Case $1$: Time taken = $4 \ hours$ for $60 \ km$ by train and $240 \ km$ by bus.
$\frac{60}{u} + \frac{240}{v} = 4 \quad ...(1)$
Case $2$: Time taken = $4 \ hours + 10 \ minutes = 4 + \frac{10}{60} = 4 + \frac{1}{6} = \frac{25}{6} \ hours$ for $100 \ km$ by train and $200 \ km$ by bus.
$\frac{100}{u} + \frac{200}{v} = \frac{25}{6} \quad ...(2)$
Let $\frac{1}{u} = p$ and $\frac{1}{v} = q$. The equations become:
$60p + 240q = 4 \quad ...(3)$
$100p + 200q = \frac{25}{6} \implies 600p + 1200q = 25 \quad ...(4)$
Multiply equation $(3)$ by $10$:
$600p + 2400q = 40 \quad ...(5)$
Subtract equation $(4)$ from $(5)$:
$(600p + 2400q) - (600p + 1200q) = 40 - 25$
$1200q = 15 \implies q = \frac{15}{1200} = \frac{1}{80}$
Substitute $q = \frac{1}{80}$ into equation $(3)$:
$60p + 240(\frac{1}{80}) = 4$
$60p + 3 = 4 \implies 60p = 1 \implies p = \frac{1}{60}$
Since $p = \frac{1}{u} = \frac{1}{60}$ and $q = \frac{1}{v} = \frac{1}{80}$,we get:
$u = 60 \ km/h$ and $v = 80 \ km/h$.
Thus,the speed of the train is $60 \ km/h$ and the speed of the bus is $80 \ km/h$.

(A-D) Let the age of Ani be $x$ years and the age of Biju be $y$ years.
Given that the difference between their ages is $3\, years,$ we have two cases:
Case $I$: $x - y = 3$ or Case $II$: $y - x = 3$.
Ani's father Dharam's age $= 2x$ years.
Biju's sister Cathy's age $= y/2$ years.
The difference between the ages of Dharam and Cathy is $30\, years,$ so $2x - y/2 = 30,$ which simplifies to $4x - y = 60$.
Case $I$: $x - y = 3$ and $4x - y = 60$.
Subtracting the first from the second: $(4x - y) - (x - y) = 60 - 3 \implies 3x = 57 \implies x = 19$.
Then $y = 19 - 3 = 16$.
Case $II$: $y - x = 3 \implies y = x + 3$.
Substituting into $4x - y = 60$: $4x - (x + 3) = 60 \implies 3x = 63 \implies x = 21$.
Then $y = 21 + 3 = 24$.
Thus,the ages of Ani and Biju are either $(19, 16)$ years or $(21, 24)$ years.

(A) Let the amounts of capital with the two friends be $Rs$ $x$ and $Rs$ $y$ respectively.
According to the first condition: "Give me a hundred,friend! $I$ shall then become twice as rich as you."
$x + 100 = 2(y - 100)$
$x + 100 = 2y - 200$
$x - 2y = -300$ $...(i)$
According to the second condition: "If you give me ten,$I$ shall be six times as rich as you."
$6(x - 10) = y + 10$
$6x - 60 = y + 10$
$6x - y = 70$ $...(ii)$
To solve the system,multiply equation $(ii)$ by $2$:
$12x - 2y = 140$ $...(iii)$
Subtract equation $(i)$ from equation $(iii)$:
$(12x - 2y) - (x - 2y) = 140 - (-300)$
$11x = 440$
$x = 40$
Substitute $x = 40$ into equation $(i)$:
$40 - 2y = -300$
$-2y = -340$
$y = 170$
Thus,the amounts of their respective capital are $Rs$ $40$ and $Rs$ $170$.

(A) Let the speed of the train be $x\, km/h$ and the time taken be $t\, hours$. The distance $d$ is given by $d = xt$ $...(i)$.
According to the first condition,if the speed is $(x + 10)\, km/h$,the time taken is $(t - 2)\, hours$:
$(x + 10)(t - 2) = d$
$xt - 2x + 10t - 20 = d$
Since $d = xt$,we get $-2x + 10t = 20$,or $-x + 5t = 10$ $...(ii)$.
According to the second condition,if the speed is $(x - 10)\, km/h$,the time taken is $(t + 3)\, hours$:
$(x - 10)(t + 3) = d$
$xt + 3x - 10t - 30 = d$
Since $d = xt$,we get $3x - 10t = 30$ $...(iii)$.
Multiplying equation $(ii)$ by $2$,we get $-2x + 10t = 20$ $...(iv)$.
Adding $(iii)$ and $(iv)$:
$(3x - 10t) + (-2x + 10t) = 30 + 20$
$x = 50\, km/h$.
Substituting $x = 50$ in equation $(ii)$:
$-50 + 5t = 10$
$5t = 60$
$t = 12\, hours$.
Distance $d = xt = 50 \times 12 = 600\, km$.

(36) Let the number of rows be $x$ and the number of students in a row be $y$.
Total number of students in the class $= x \times y = xy$.
According to the first condition:
If $3$ students are extra in a row,there would be $1$ row less.
$(x - 1)(y + 3) = xy$
$xy + 3x - y - 3 = xy$
$3x - y = 3$ $...(i)$
According to the second condition:
If $3$ students are less in a row,there would be $2$ rows more.
$(x + 2)(y - 3) = xy$
$xy - 3x + 2y - 6 = xy$
$-3x + 2y = 6$ $...(ii)$
Adding equation $(i)$ and equation $(ii)$:
$(3x - y) + (-3x + 2y) = 3 + 6$
$y = 9$
Substituting $y = 9$ in equation $(i)$:
$3x - 9 = 3$
$3x = 12$
$x = 4$
Total number of students $= xy = 4 \times 9 = 36$.

(A) Given that,$\angle C = 3 \angle B = 2(\angle A + \angle B)$.
From $3 \angle B = 2(\angle A + \angle B)$,we get $3 \angle B = 2 \angle A + 2 \angle B$,which simplifies to $\angle B = 2 \angle A$,or $2 \angle A - \angle B = 0 \dots (i)$.
We know that the sum of the angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Since $\angle C = 3 \angle B$,we substitute this into the sum: $\angle A + \angle B + 3 \angle B = 180^{\circ}$,which gives $\angle A + 4 \angle B = 180^{\circ} \dots (ii)$.
Multiplying equation $(i)$ by $4$,we get $8 \angle A - 4 \angle B = 0 \dots (iii)$.
Adding equations $(ii)$ and $(iii)$,we get $9 \angle A = 180^{\circ}$,which implies $\angle A = 20^{\circ}$.
Substituting $\angle A = 20^{\circ}$ into equation $(ii)$,we get $20^{\circ} + 4 \angle B = 180^{\circ}$,so $4 \angle B = 160^{\circ}$,which gives $\angle B = 40^{\circ}$.
Finally,$\angle C = 3 \angle B = 3 \times 40^{\circ} = 120^{\circ}$.
Thus,the angles are $\angle A = 20^{\circ}$,$\angle B = 40^{\circ}$,and $\angle C = 120^{\circ}$.

(N/A) For the equation $5x - y = 5$,we can write $y = 5x - 5$. The solution table is:

$x$	$0$	$1$	$2$
$y$	$-5$	$0$	$5$

For the equation $3x - y = 3$,we can write $y = 3x - 3$. The solution table is:

$x$	$0$	$1$	$2$
$y$	$-3$	$0$	$3$

By plotting these points on a graph,we obtain two lines that intersect at the point $(1, 0)$.
The triangle is formed by the intersection of these two lines and the $y$-axis (where $x = 0$).
The vertices of the triangle are the intersection point of the two lines $(1, 0)$ and the points where each line intersects the $y$-axis,which are $(0, -3)$ and $(0, -5)$.
Thus,the coordinates of the vertices are $(1, 0)$,$(0, -3)$,and $(0, -5)$.

(N/A) Given equations are:
$px + qy = p - q \quad \dots(1)$
$qx - py = p + q \quad \dots(2)$
To eliminate $y$,multiply equation $(1)$ by $p$ and equation $(2)$ by $q$:
$p(px + qy) = p(p - q) \implies p^2x + pqy = p^2 - pq \quad \dots(3)$
$q(qx - py) = q(p + q) \implies q^2x - pqy = pq + q^2 \quad \dots(4)$
Adding equations $(3)$ and $(4)$:
$(p^2x + pqy) + (q^2x - pqy) = (p^2 - pq) + (pq + q^2)$
$p^2x + q^2x = p^2 + q^2$
$x(p^2 + q^2) = p^2 + q^2$
$x = \frac{p^2 + q^2}{p^2 + q^2} = 1$
Substituting $x = 1$ in equation $(1)$:
$p(1) + qy = p - q$
$p + qy = p - q$
$qy = -q$
$y = -1$
Thus,the solution is $x = 1$ and $y = -1$.

(N/A) $ax + by = c \dots(1)$
$bx + ay = 1 + c \dots(2)$
Multiplying equation $(1)$ by $a$ and equation $(2)$ by $b$,we obtain:
$a^2x + aby = ac \dots(3)$
$b^2x + aby = b + bc \dots(4)$
Subtracting equation $(4)$ from equation $(3)$:
$(a^2 - b^2)x = ac - bc - b$
$x = \frac{c(a - b) - b}{a^2 - b^2}$
Substituting the value of $x$ in equation $(1)$:
$a \left[ \frac{c(a - b) - b}{a^2 - b^2} \right] + by = c$
$by = c - \frac{ac(a - b) - ab}{a^2 - b^2}$
$by = \frac{c(a^2 - b^2) - (a^2c - abc - ab)}{a^2 - b^2}$
$by = \frac{a^2c - b^2c - a^2c + abc + ab}{a^2 - b^2}$
$by = \frac{abc - b^2c + ab}{a^2 - b^2}$
$by = \frac{bc(a - b) + ab}{a^2 - b^2}$
$y = \frac{c(a - b) + a}{a^2 - b^2}$

Given equations are:
$(a-b)x + (a+b)y = a^2 - 2ab - b^2 \dots(1)$
$(a+b)(x+y) = a^2 + b^2 \dots(2)$
Expanding equation $(2)$:
$(a+b)x + (a+b)y = a^2 + b^2 \dots(3)$
Subtracting equation $(3)$ from $(1)$:
$[(a-b)x + (a+b)y] - [(a+b)x + (a+b)y] = (a^2 - 2ab - b^2) - (a^2 + b^2)$
$(a-b-a-b)x = -2ab - 2b^2$
$-2bx = -2b(a+b)$
$x = a+b$
Substituting $x = a+b$ into equation $(1)$:
$(a-b)(a+b) + (a+b)y = a^2 - 2ab - b^2$
$a^2 - b^2 + (a+b)y = a^2 - 2ab - b^2$
$(a+b)y = -2ab$
$y = \frac{-2ab}{a+b}$

(N/A) Given equations are:
$(1)$ $152x - 378y = -74$
$(2)$ $-378x + 152y = -604$
Step $1$: Add the two equations:
$(152x - 378x) + (-378y + 152y) = -74 - 604$
$-226x - 226y = -678$
Divide by $-226$:
$x + y = 3$ --- $(3)$
Step $2$: Subtract equation $(2)$ from equation $(1)$:
$(152x - (-378x)) + (-378y - 152y) = -74 - (-604)$
$530x - 530y = 530$
Divide by $530$:
$x - y = 1$ --- $(4)$
Step $3$: Solve equations $(3)$ and $(4)$:
Adding $(3)$ and $(4)$: $2x = 4 \implies x = 2$
Subtracting $(4)$ from $(3)$: $2y = 2 \implies y = 1$
Thus,the solution is $x = 2, y = 1$.

(A) Given equations are:
$\frac{x}{a} - \frac{y}{b} = 0 \implies bx - ay = 0 \quad \dots(1)$
$ax + by = a^2 + b^2 \quad \dots(2)$
To solve by elimination,multiply equation $(1)$ by $b$ and equation $(2)$ by $a$:
$b^2x - aby = 0 \quad \dots(3)$
$a^2x + aby = a(a^2 + b^2) = a^3 + ab^2 \quad \dots(4)$
Adding equations $(3)$ and $(4)$:
$(b^2x - aby) + (a^2x + aby) = 0 + a^3 + ab^2$
$x(a^2 + b^2) = a(a^2 + b^2)$
$x = a$
Substitute $x = a$ into equation $(1)$:
$b(a) - ay = 0$
$ab - ay = 0$
$ay = ab$
$y = b$
Thus,the solution is $x = a$ and $y = b$.

(N/A) We know that the sum of the measures of opposite angles in a cyclic quadrilateral is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
$(4y + 20) + (-4x) = 180$
$-4x + 4y = 160$
$-x + y = 40$ $...(i)$
Also,$\angle B + \angle D = 180^{\circ}$.
$(3y - 5) + (-7x + 5) = 180$
$-7x + 3y = 180$ $...(ii)$
Multiplying equation $(i)$ by $3$,we obtain:
$-3x + 3y = 120$ $...(iii)$
Subtracting equation $(iii)$ from equation $(ii)$:
$(-7x + 3y) - (-3x + 3y) = 180 - 120$
$-4x = 60$
$x = -15$
Substituting $x = -15$ in equation $(i)$:
$-(-15) + y = 40$
$15 + y = 40$
$y = 25$
Now,calculating the angles:
$\angle A = 4y + 20 = 4(25) + 20 = 120^{\circ}$
$\angle B = 3y - 5 = 3(25) - 5 = 70^{\circ}$
$\angle C = -4x = -4(-15) = 60^{\circ}$
$\angle D = -7x + 5 = -7(-15) + 5 = 110^{\circ}$