Elimination Method Questions in English

(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 $25$.

(D) Step $1$: Multiply Equation $(1)$ by $2$ and Equation $(2)$ by $1$ to make the coefficients of $x$ equal. Then we get the equations as:
$4x + 6y = 16$ $...(3)$
$4x + 6y = 7$ $...(4)$
Step $2$: Subtracting Equation $(4)$ from Equation $(3)$:
$(4x - 4x) + (6y - 6y) = 16 - 7$
i.e.,$0 = 9$,which is a false statement.
Therefore,the pair of linear equations has no solution.

(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 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 $x = 4, y = 5$.

(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: $x = 1, y = 1$.

(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) 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 $(1)$:
$a \left[ \frac{c(a - b) - b}{a^2 - b^2} \right] + by = c$
$by = c - \frac{ac(a - b) - ab}{a^2 - b^2}$
$by = \frac{c(a^2 - b^2) - (a^2c - abc - ab)}{a^2 - b^2}$
$by = \frac{a^2c - b^2c - a^2c + abc + ab}{a^2 - b^2}$
$by = \frac{abc - b^2c + ab}{a^2 - b^2}$
$by = \frac{bc(a - b) + ab}{a^2 - b^2}$
$y = \frac{c(a - b) + a}{a^2 - b^2}$

(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 $y = b$.

(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 $y = \frac{1}{4}$.

(A) Given pair of linear equations is:
$\frac{x}{a} + \frac{y}{b} = a + b \quad .....(i)$
$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b \neq 0 \quad .....(ii)$
Multiply equation $(i)$ by $\frac{1}{a}$:
$\frac{x}{a^{2}} + \frac{y}{ab} = 1 + \frac{b}{a} \quad .....(iii)$
Subtract equation $(iii)$ from equation $(ii)$:
$\left( \frac{x}{a^{2}} + \frac{y}{b^{2}} \right) - \left( \frac{x}{a^{2}} + \frac{y}{ab} \right) = 2 - \left( 1 + \frac{b}{a} \right)$
$\frac{y}{b^{2}} - \frac{y}{ab} = 1 - \frac{b}{a}$
$y \left( \frac{a - b}{ab^{2}} \right) = \frac{a - b}{a}$
Since $a \neq b$ (assuming general case),we can cancel $(a - b)$:
$\frac{y}{ab^{2}} = \frac{1}{a}$
$y = b^{2}$
Substitute $y = b^{2}$ into equation $(ii)$:
$\frac{x}{a^{2}} + \frac{b^{2}}{b^{2}} = 2$
$\frac{x}{a^{2}} + 1 = 2$
$\frac{x}{a^{2}} = 1$
$x = a^{2}$
Thus,the solution is $x = a^{2}$ and $y = b^{2}$.

(B) Given pair of equations is:
$\frac{x}{10} + \frac{y}{5} = 1 \dots (i)$
$\frac{x}{8} + \frac{y}{6} = 15 \dots (ii)$
Multiplying equation $(i)$ by $10$,we get $x + 2y = 10 \dots (iii)$.
Multiplying equation $(ii)$ by $24$,we get $3x + 4y = 360 \dots (iv)$.
To eliminate $y$,multiply equation $(iii)$ by $2$:
$2x + 4y = 20 \dots (v)$
Subtracting equation $(v)$ from equation $(iv)$:
$(3x + 4y) - (2x + 4y) = 360 - 20$
$x = 340$
Substituting $x = 340$ into equation $(iii)$:
$340 + 2y = 10$
$2y = -330$
$y = -165$
Given the relation $y = \lambda x + 5$,substitute $x = 340$ and $y = -165$:
$-165 = \lambda(340) + 5$
$-170 = 340\lambda$
$\lambda = -\frac{170}{340} = -\frac{1}{2}$
Thus,the solution is $x = 340, y = -165$ and $\lambda = -\frac{1}{2}$.

(B) Given equations are:
$2x + y = 4$ --- $(1)$
$x + 3y = 7$ --- $(2)$
To eliminate $x$,multiply equation $(2)$ by $2$:
$2(x + 3y) = 2(7) \implies 2x + 6y = 14$ --- $(3)$
Now,subtract equation $(1)$ from equation $(3)$:
$(2x + 6y) - (2x + y) = 14 - 4$
$5y = 10$
$y = 2$
Substitute $y = 2$ in equation $(1)$:
$2x + 2 = 4$
$2x = 2$
$x = 1$
Thus,the solution is $(x, y) = (1, 2)$.

(C) First,we multiply the equation $\frac{x}{4} + \frac{y}{3} = 2$ by $12$ to convert it into an equation with integer coefficients:
$3x + 4y = 24$ .......... $(1)$
$4x + 3y = 25$ .......... $(2)$
To eliminate $x$,multiply equation $(1)$ by $4$ and equation $(2)$ by $3$:
$12x + 16y = 96$ .......... $(3)$
$12x + 9y = 75$ .......... $(4)$
Subtracting equation $(4)$ from equation $(3)$:
$(12x - 12x) + (16y - 9y) = 96 - 75$
$7y = 21$
$y = 3$
Substituting $y = 3$ into equation $(1)$:
$3x + 4(3) = 24$
$3x + 12 = 24$
$3x = 12$
$x = 4$
Thus,the solution is $(x, y) = (4, 3)$.

(D) Given equations are:
$5x + 6y = 14$ ............ $(1)$
$3x - 2y = -14$ ............ $(2)$
To eliminate $y$,multiply equation $(2)$ by $3$:
$3(3x - 2y) = 3(-14) \implies 9x - 6y = -42$ ............ $(3)$
Now,add equation $(1)$ and equation $(3)$:
$(5x + 6y) + (9x - 6y) = 14 + (-42)$
$14x = -28$
$x = \frac{-28}{14} = -2$
Substitute $x = -2$ into equation $(1)$:
$5(-2) + 6y = 14$
$-10 + 6y = 14$
$6y = 14 + 10$
$6y = 24$
$y = \frac{24}{6} = 4$
Thus,the solution is $(x, y) = (-2, 4)$.

(A) Multiplying both equations by $10$ to remove decimals:
$4x + 3y = 17$ ............. $(1)$
$7x - 2y = 8$ ............. $(2)$
To eliminate $y$,multiply equation $(1)$ by $2$ and equation $(2)$ by $3$:
$8x + 6y = 34$
$21x - 6y = 24$
Adding these two equations:
$(8x + 21x) + (6y - 6y) = 34 + 24$
$29x = 58$
$x = \frac{58}{29} = 2$
Substituting $x = 2$ in equation $(1)$:
$4(2) + 3y = 17$
$8 + 3y = 17$
$3y = 17 - 8$
$3y = 9$
$y = 3$
Thus,the solution is $(x, y) = (2, 3)$.

(D) Given equations are:
$(1) x + 3y = 6$
$(2) 2x - y = 5$
To eliminate $y$,multiply equation $(2)$ by $3$:
$3(2x - y) = 3(5) \implies 6x - 3y = 15$ $(3)$
Now,add equation $(1)$ and equation $(3)$:
$(x + 3y) + (6x - 3y) = 6 + 15$
$7x = 21$
$x = 3$
Substitute $x = 3$ into equation $(2)$:
$2(3) - y = 5$
$6 - y = 5$
$y = 1$
Thus,the solution is $(x, y) = (3, 1)$.

(A) Given equations are:
$(1) \quad 4x - 3y = 5$
$(2) \quad \frac{5}{2}x - y = 4$
Multiply equation $(2)$ by $3$ to eliminate $y$:
$3 \times (\frac{5}{2}x - y) = 3 \times 4$
$\frac{15}{2}x - 3y = 12 \quad (3)$
Subtract equation $(1)$ from equation $(3)$:
$(\frac{15}{2}x - 3y) - (4x - 3y) = 12 - 5$
$\frac{15}{2}x - 4x = 7$
$\frac{15x - 8x}{2} = 7$
$\frac{7x}{2} = 7$
$x = 2$
Substitute $x = 2$ into equation $(2)$:
$\frac{5}{2}(2) - y = 4$
$5 - y = 4$
$y = 1$
Thus,the solution is $(x, y) = (2, 1)$.

(B) Given equations are:
$1) \frac{x}{2} + \frac{3y}{5} = -1$
$2) \frac{x}{2} + \frac{y}{3} = \frac{1}{3}$
Subtracting equation $(2)$ from equation $(1)$ to eliminate $x$:
$(\frac{x}{2} + \frac{3y}{5}) - (\frac{x}{2} + \frac{y}{3}) = -1 - \frac{1}{3}$
$\frac{3y}{5} - \frac{y}{3} = -\frac{4}{3}$
$\frac{9y - 5y}{15} = -\frac{4}{3}$
$\frac{4y}{15} = -\frac{4}{3}$
$y = -\frac{4}{3} \times \frac{15}{4} = -5$
Substitute $y = -5$ into equation $(2)$:
$\frac{x}{2} + \frac{-5}{3} = \frac{1}{3}$
$\frac{x}{2} = \frac{1}{3} + \frac{5}{3} = \frac{6}{3} = 2$
$x = 4$
Thus,the solution is $(x, y) = (4, -5)$.

(C) Given equations are:
$(1) \ 9x + 7y = 55$
$(2) \ 7x + 9y = 57$
Adding equations $(1)$ and $(2)$:
$(9x + 7x) + (7y + 9y) = 55 + 57$
$16x + 16y = 112$
Dividing by $16$:
$x + y = 7 \quad (3)$
Subtracting equation $(2)$ from $(1)$:
$(9x - 7x) + (7y - 9y) = 55 - 57$
$2x - 2y = -2$
Dividing by $2$:
$x - y = -1 \quad (4)$
Adding equations $(3)$ and $(4)$:
$(x + y) + (x - y) = 7 + (-1)$
$2x = 6 \implies x = 3$
Substituting $x = 3$ in equation $(3)$:
$3 + y = 7 \implies y = 4$
Thus,the solution is $(x, y) = (3, 4)$.

(D) Given equations are:
$(1) \ 11x + 13y = 61$
$(2) \ 13x + 11y = 59$
Adding equations $(1)$ and $(2)$:
$(11x + 13x) + (13y + 11y) = 61 + 59$
$24x + 24y = 120$
Dividing by $24$:
$x + y = 5$ --- $(3)$
Subtracting equation $(1)$ from $(2)$:
$(13x - 11x) + (11y - 13y) = 59 - 61$
$2x - 2y = -2$
Dividing by $2$:
$x - y = -1$ --- $(4)$
Adding equations $(3)$ and $(4)$:
$(x + y) + (x - y) = 5 + (-1)$
$2x = 4 \implies x = 2$
Substituting $x = 2$ in equation $(3)$:
$2 + y = 5 \implies y = 3$
Thus,the solution is $(x, y) = (2, 3)$.

(A) First,simplify the given equations:
$1$) $5(x-2) + 3y = 1 \implies 5x - 10 + 3y = 1 \implies 5x + 3y = 11$ (Equation $1$)
$2$) $3(x+1) + 5(y-3) = 1 \implies 3x + 3 + 5y - 15 = 1 \implies 3x + 5y = 13$ (Equation $2$)
To eliminate $y$,multiply Equation $1$ by $5$ and Equation $2$ by $3$:
$25x + 15y = 55$ (Equation $3$)
$9x + 15y = 39$ (Equation $4$)
Subtract Equation $4$ from Equation $3$:
$(25x - 9x) + (15y - 15y) = 55 - 39$
$16x = 16 \implies x = 1$
Substitute $x = 1$ into Equation $1$:
$5(1) + 3y = 11$
$5 + 3y = 11$
$3y = 6 \implies y = 2$
Thus,the solution is $(x, y) = (1, 2)$.

(B) First,simplify the given equations:
Equation $1$: $7y + 21 - 2x - 4 = 14 \implies -2x + 7y = -3$
Equation $2$: $4y - 8 + 3x - 9 = 2 \implies 3x + 4y = 19$
To eliminate $x$,multiply Equation $1$ by $3$ and Equation $2$ by $2$:
$-6x + 21y = -9$
$6x + 8y = 38$
Adding the two equations: $29y = 29 \implies y = 1$
Substitute $y = 1$ into Equation $2$: $3x + 4(1) = 19 \implies 3x = 15 \implies x = 5$
Thus,the solution is $(x, y) = (5, 1)$.

(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 $(x, y) = (\frac{3}{2}, -\frac{2}{3})$.

(A) 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 $\left(\frac{1}{2}, \frac{1}{2}\right)$.

(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.

(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.

(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 $u = \frac{1}{x-1}$ 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 $(x, y) = (4, 5)$.

(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 $(x, y) = (1, 3)$.

(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,$(a, b) = \left(-1, \frac{1}{2}\right)$.

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

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

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