Substitution Method Questions in English

(N/A) Step $1:$ We pick either of the equations and write one variable in terms of the other. Let us consider Equation $(2)$.
$x + 2y = 3$
We can write it as $x = 3 - 2y$ $...(3)$
Step $2:$ Substitute the value of $x$ from Equation $(3)$ into Equation $(1)$. We get:
$7(3 - 2y) - 15y = 2$
$21 - 14y - 15y = 2$
$-29y = 2 - 21$
$-29y = -19$
Therefore,$y = \frac{19}{29}$
Step $3:$ Substituting this value of $y$ in Equation $(3)$,we get:
$x = 3 - 2\left(\frac{19}{29}\right)$
$x = 3 - \frac{38}{29}$
$x = \frac{87 - 38}{29} = \frac{49}{29}$
Therefore,the solution is $x = \frac{49}{29}, y = \frac{19}{29}$.

(N/A) Let $s$ and $t$ be the current ages (in years) of Aftab and his daughter,respectively.
According to the first condition,seven years ago:
$(s - 7) = 7(t - 7)$
$s - 7 = 7t - 49$
$s - 7t = -42$ $...(1)$
According to the second condition,three years from now:
$(s + 3) = 3(t + 3)$
$s + 3 = 3t + 9$
$s - 3t = 6$ $...(2)$
From Equation $(2)$,we get $s = 3t + 6$.
Substituting this value of $s$ into Equation $(1)$:
$(3t + 6) - 7t = -42$
$-4t + 6 = -42$
$-4t = -48$
$t = 12$
Now,substitute $t = 12$ into $s = 3t + 6$:
$s = 3(12) + 6 = 36 + 6 = 42$
Thus,the current age of Aftab is $42$ years and his daughter is $12$ years old.

(A) $x+y=14$ $....(1)$
$x-y=4$ $....(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$x=14-y$ $....(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$(14-y)-y=4$
$14-2y=4$
$14-4=2y$
$10=2y$
$y=5$ $....(4)$
Now,substituting the value of $y=5$ into equation $(3)$:
$x=14-5$
$x=9$
Therefore,the solution is $x=9$ and $y=5$.

(S=9, T=6) $s-t=3$ $...(1)$
$\frac{s}{3}+\frac{t}{2}=6$ $...(2)$
From equation $(1)$,we express $s$ in terms of $t$:
$s = t + 3$ $...(3)$
Substituting the value of $s$ from equation $(3)$ into equation $(2)$:
$\frac{t+3}{3} + \frac{t}{2} = 6$
Multiply the entire equation by $6$ (the $LCM$ of $3$ and $2$) to clear the denominators:
$2(t+3) + 3t = 36$
$2t + 6 + 3t = 36$
$5t + 6 = 36$
$5t = 30$
$t = 6$ $...(4)$
Now,substitute the value of $t = 6$ into equation $(3)$:
$s = 6 + 3$
$s = 9$
Therefore,the solution is $s = 9$ and $t = 6$.

(D) Given equations are:
$3x - y = 3$ $...(1)$
$9x - 3y = 9$ $...(2)$
From equation $(1)$,we can express $y$ in terms of $x$:
$y = 3x - 3$ $...(3)$
Substituting the value of $y$ from equation $(3)$ into equation $(2)$:
$9x - 3(3x - 3) = 9$
$9x - 9x + 9 = 9$
$9 = 9$
Since $9 = 9$ is a true statement for all values of $x$,the two equations are coincident lines.
Therefore,the given pair of linear equations has infinitely many solutions.
The relationship between the variables is given by $y = 3x - 3$.

(N/A) Given equations are:
$0.2 x + 0.3 y = 1.3$ $...(1)$
$0.4 x + 0.5 y = 2.3$ $...(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$0.2 x = 1.3 - 0.3 y$
$x = \frac{1.3 - 0.3 y}{0.2}$ $...(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$0.4 \left( \frac{1.3 - 0.3 y}{0.2} \right) + 0.5 y = 2.3$
$2(1.3 - 0.3 y) + 0.5 y = 2.3$
$2.6 - 0.6 y + 0.5 y = 2.3$
$2.6 - 0.1 y = 2.3$
$2.6 - 2.3 = 0.1 y$
$0.3 = 0.1 y$
$y = \frac{0.3}{0.1} = 3$
Now,substitute $y = 3$ into equation $(3)$:
$x = \frac{1.3 - 0.3(3)}{0.2}$
$x = \frac{1.3 - 0.9}{0.2}$
$x = \frac{0.4}{0.2} = 2$
Therefore,the solution is $x = 2$ and $y = 3$.

(A) Given equations are:
$\sqrt{2} x + \sqrt{3} y = 0$ $...(1)$
$\sqrt{3} x - \sqrt{8} y = 0$ $...(2)$
From equation $(1)$,we express $x$ in terms of $y$:
$x = -\frac{\sqrt{3} y}{\sqrt{2}}$ $...(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$\sqrt{3} \left( -\frac{\sqrt{3} y}{\sqrt{2}} \right) - \sqrt{8} y = 0$
$-\frac{3 y}{\sqrt{2}} - 2\sqrt{2} y = 0$
Factoring out $y$:
$y \left( -\frac{3}{\sqrt{2}} - 2\sqrt{2} \right) = 0$
Since the term in the bracket is non-zero,we must have $y = 0$.
Substituting $y = 0$ into equation $(3)$:
$x = -\frac{\sqrt{3} (0)}{\sqrt{2}} = 0$
Thus,the solution is $x = 0$ and $y = 0$.

(A) Given equations are:
$(1) \frac{3x}{2} - \frac{5y}{3} = -2$
$(2) \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Step $1$: Simplify the equations.
Multiply $(1)$ by $6$: $9x - 10y = -12$ --- $(3)$
Multiply $(2)$ by $6$: $2x + 3y = 13$ --- $(4)$
Step $2$: Express $x$ in terms of $y$ from equation $(4)$:
$2x = 13 - 3y$
$x = \frac{13 - 3y}{2}$ --- $(5)$
Step $3$: Substitute $(5)$ into $(3)$:
$9(\frac{13 - 3y}{2}) - 10y = -12$
Multiply by $2$: $9(13 - 3y) - 20y = -24$
$117 - 27y - 20y = -24$
$-47y = -24 - 117$
$-47y = -141$
$y = \frac{141}{47} = 3$
Step $4$: Substitute $y = 3$ into $(5)$:
$x = \frac{13 - 3(3)}{2} = \frac{13 - 9}{2} = \frac{4}{2} = 2$
Thus,the solution is $x = 2, y = 3$.

(N/A) Let the first number be $x$ and the other number be $y$ such that $y > x$.
According to the given information:
$y = 3x$ $...(1)$
$y - x = 26$ $...(2)$
On substituting the value of $y$ from equation $(1)$ into equation $(2)$,we obtain:
$3x - x = 26$
$2x = 26$
$x = 13$ $...(3)$
Substituting the value of $x$ in equation $(1)$,we obtain:
$y = 3(13) = 39$
Hence,the two numbers are $13$ and $39$.

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

(A) Let the larger angle be $x$ and the smaller angle be $y$.
We know that the sum of the measures of angles of a supplementary pair is always $180^{\circ}$.
According to the given information:
$x + y = 180^{\circ}$ $...(1)$
$x - y = 18^{\circ}$ $...(2)$
From equation $(1)$,we obtain:
$x = 180^{\circ} - y$ $...(3)$
Substituting the value of $x$ in equation $(2)$,we get:
$(180^{\circ} - y) - y = 18^{\circ}$
$180^{\circ} - 2y = 18^{\circ}$
$180^{\circ} - 18^{\circ} = 2y$
$162^{\circ} = 2y$
$y = 81^{\circ}$ $...(4)$
Substituting the value of $y$ in equation $(3)$,we get:
$x = 180^{\circ} - 81^{\circ}$
$x = 99^{\circ}$
Hence,the two angles are $99^{\circ}$ and $81^{\circ}$.

(A) Let the cost of a bat be $x$ and the cost of a ball be $y$.
According to the given information:
$7x + 6y = 3800$ $...(1)$
$3x + 5y = 1750$ $...(2)$
From equation $(1)$,we express $y$ in terms of $x$:
$6y = 3800 - 7x$
$y = \frac{3800 - 7x}{6}$ $...(3)$
Substituting this value of $y$ into equation $(2)$:
$3x + 5\left(\frac{3800 - 7x}{6}\right) = 1750$
Multiply the entire equation by $6$ to simplify:
$18x + 5(3800 - 7x) = 10500$
$18x + 19000 - 35x = 10500$
$-17x = 10500 - 19000$
$-17x = -8500$
$x = 500$
Now,substitute $x = 500$ into equation $(3)$:
$y = \frac{3800 - 7(500)}{6}$
$y = \frac{3800 - 3500}{6}$
$y = \frac{300}{6} = 50$
Therefore,the cost of one bat is ₹ $500$ and the cost of one ball is ₹ $50$.

(D) Let the fixed charge be ₹ $x$ and the charge per $km$ be ₹ $y$.
According to the given information:
$x + 10y = 105$ $(1)$
$x + 15y = 155$ $(2)$
From equation $(1)$,we obtain:
$x = 105 - 10y$ $(3)$
Substituting the value of $x$ from equation $(3)$ into equation $(2)$:
$(105 - 10y) + 15y = 155$
$105 + 5y = 155$
$5y = 155 - 105$
$5y = 50$
$y = 10$
Substituting $y = 10$ in equation $(3)$:
$x = 105 - 10(10)$
$x = 105 - 100$
$x = 5$
Thus,the fixed charge is ₹ $5$ and the charge per $km$ is ₹ $10$.
For a distance of $25 \, km$,the total charge is:
$x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255$.

(A) Let the numerator of the fraction be $x$ and the denominator be $y$. The fraction is $\frac{x}{y}$.
According to the first condition: $\frac{x+2}{y+2} = \frac{9}{11}$.
Cross-multiplying gives: $11(x+2) = 9(y+2) \implies 11x + 22 = 9y + 18 \implies 11x - 9y = -4$ (Equation $1$).
According to the second condition: $\frac{x+3}{y+3} = \frac{5}{6}$.
Cross-multiplying gives: $6(x+3) = 5(y+3) \implies 6x + 18 = 5y + 15 \implies 6x - 5y = -3$ (Equation $2$).
From Equation $2$,$5y = 6x + 3 \implies y = \frac{6x+3}{5}$.
Substituting this into Equation $1$: $11x - 9(\frac{6x+3}{5}) = -4$.
Multiply by $5$: $55x - 9(6x+3) = -20 \implies 55x - 54x - 27 = -20 \implies x = 7$.
Now,find $y$: $y = \frac{6(7)+3}{5} = \frac{42+3}{5} = \frac{45}{5} = 9$.
The fraction is $\frac{7}{9}$.

(N/A) Let the present age of Jacob be $x$ years and the present age of his son be $y$ years.
Case $1$: Five years hence,Jacob's age will be $(x + 5)$ and his son's age will be $(y + 5)$.
According to the problem,$(x + 5) = 3(y + 5) \implies x + 5 = 3y + 15 \implies x - 3y = 10$ (Equation $1$).
Case $2$: Five years ago,Jacob's age was $(x - 5)$ and his son's age was $(y - 5)$.
According to the problem,$(x - 5) = 7(y - 5) \implies x - 5 = 7y - 35 \implies x - 7y = -30$ (Equation $2$).
From Equation $1$,$x = 3y + 10$.
Substituting this value into Equation $2$: $(3y + 10) - 7y = -30 \implies -4y = -40 \implies y = 10$.
Substituting $y = 10$ into $x = 3y + 10$: $x = 3(10) + 10 = 40$.
Therefore,the present age of Jacob is $40$ years and the present age of his son is $10$ years.

(A) Let the weight of the father be $x \, kg$ and the weight of the son be $y \, kg$.
According to the problem,the total weight is $x + y = 70$ (Equation $1$).
The weight of the son is one-sixth of the father's weight,so $y = \frac{1}{6}x$,which implies $x = 6y$ or $x - 6y = 0$ (Equation $2$).
Substitute $x = 6y$ into Equation $1$:
$6y + y = 70$
$7y = 70$
$y = 10 \, kg$.
Now,substitute $y = 10$ into $x = 6y$:
$x = 6 \times 10 = 60 \, kg$.
Therefore,the father's weight is $60 \, kg$ and the son's weight is $10 \, kg$.

(C) Given equations are:
$2x + 3y = 11$ .......... $(i)$
$2x - y = -1$ .......... $(ii)$
From equation $(ii)$,we can express $y$ in terms of $x$:
$y = 2x + 1$ .......... $(iii)$
Substitute the value of $y$ from equation $(iii)$ into equation $(i)$:
$2x + 3(2x + 1) = 11$
$2x + 6x + 3 = 11$
$8x + 3 = 11$
$8x = 8$
$x = 1$
Now,substitute $x = 1$ into equation $(iii)$:
$y = 2(1) + 1$
$y = 2 + 1$
$y = 3$
Thus,the solution is $(x, y) = (1, 3)$.

(D) Given equations are:
$3x - 7y = 18$ ..... $(1)$
$6x - 14y = 12$ ..... $(2)$
Divide equation $(2)$ by $2$:
$3x - 7y = 6$ ..... $(3)$
From equation $(1)$,we have $3x = 18 + 7y$. Substituting this into equation $(3)$:
$(18 + 7y) - 7y = 6$
$18 = 6$
This is a contradiction,as $18$ can never be equal to $6$.
Therefore,the given pair of linear equations has no solution,which means the solution set is $\varnothing$.

(A) Let the present age of the father be $x$ years and the present age of the son be $y$ years.
According to the first condition: $x + 2y = 70$ --- $(1)$
According to the second condition: $2x + y = 95$ --- $(2)$
From equation $(1)$,we get $x = 70 - 2y$.
Substituting this value of $x$ into equation $(2)$:
$2(70 - 2y) + y = 95$
$140 - 4y + y = 95$
$140 - 3y = 95$
$-3y = 95 - 140$
$-3y = -45$
$y = 15$
Now,substitute $y = 15$ into $x = 70 - 2y$:
$x = 70 - 2(15)$
$x = 70 - 30 = 40$
Therefore,the father's present age is $40$ years and the son's present age is $15$ years.

(C) Given equations are:
$1) x + y = 7$
$2) 5x + 12y = 7$
From equation $(1)$,we can express $x$ in terms of $y$:
$x = 7 - y$ $(3)$
Substitute equation $(3)$ into equation $(2)$:
$5(7 - y) + 12y = 7$
$35 - 5y + 12y = 7$
$35 + 7y = 7$
$7y = 7 - 35$
$7y = -28$
$y = -4$
Now,substitute $y = -4$ back into equation $(3)$:
$x = 7 - (-4)$
$x = 7 + 4$
$x = 11$
Thus,the solution is $(x, y) = (11, -4)$.

(A) Given equations are:
$1) 5x - 3y = 1$
$2) 2x + 5y = 19$
From equation $(1)$,we get $5x = 1 + 3y$,so $x = \frac{1 + 3y}{5}$.
Substitute this value of $x$ into equation $(2)$:
$2(\frac{1 + 3y}{5}) + 5y = 19$
Multiply by $5$ to clear the denominator:
$2(1 + 3y) + 25y = 95$
$2 + 6y + 25y = 95$
$31y = 93$
$y = 3$
Now,substitute $y = 3$ back into the expression for $x$:
$x = \frac{1 + 3(3)}{5} = \frac{1 + 9}{5} = \frac{10}{5} = 2$.
Thus,the solution is $(x, y) = (2, 3)$.

(A) Given equations are:
$(1)$ $2x - y + 3 = 0$
$(2)$ $y = 2x - \frac{3}{2}$
Substitute the value of $y$ from equation $(2)$ into equation $(1)$:
$2x - (2x - \frac{3}{2}) + 3 = 0$
$2x - 2x + \frac{3}{2} + 3 = 0$
$\frac{3}{2} + 3 = 0$
$\frac{3 + 6}{2} = 0$
$\frac{9}{2} = 0$
Since $\frac{9}{2} \neq 0$,this statement is a contradiction.
Therefore,the given pair of linear equations has no solution,which means the lines are parallel. The correct option is $A$.

(N/A) Given equations are:
$1) \, 2x - 3y = -11$
$2) \, 4x - 6y + 22 = 0$
From equation $(2)$,we can write $4x - 6y = -22$,which simplifies to $2(2x - 3y) = -22$,or $2x - 3y = -11$.
Since both equations are identical,they represent the same line.
This means the system has infinitely many solutions.
Any pair $(x, y)$ satisfying $2x - 3y = -11$ is a solution.
Thus,the solution set is $\{(x, y) \mid 2x - 3y = -11; \, x, y \in R \}$.

(A) Given equations are:
$(1)$ $3x + y + 1 = 0$
$(2)$ $2x - 3y + 8 = 0$
From equation $(1)$,we can express $y$ in terms of $x$:
$y = -3x - 1$ $(3)$
Substitute the value of $y$ from equation $(3)$ into equation $(2)$:
$2x - 3(-3x - 1) + 8 = 0$
$2x + 9x + 3 + 8 = 0$
$11x + 11 = 0$
$11x = -11$
$x = -1$
Now,substitute $x = -1$ into equation $(3)$:
$y = -3(-1) - 1$
$y = 3 - 1$
$y = 2$
Thus,the solution is $(x, y) = (-1, 2)$.

(D) Given equations are:
$1) 8x - 3y = 1$
$2) 34x - 3y = 14$
From equation $(1)$,we can express $3y$ as:
$3y = 8x - 1$
$y = (8x - 1) / 3$
Substitute this value of $y$ into equation $(2)$:
$34x - 3((8x - 1) / 3) = 14$
$34x - (8x - 1) = 14$
$34x - 8x + 1 = 14$
$26x = 13$
$x = 13 / 26 = 1/2$
Now,substitute $x = 1/2$ into the expression for $y$:
$y = (8(1/2) - 1) / 3$
$y = (4 - 1) / 3$
$y = 3 / 3 = 1$
Thus,the solution is $(x, y) = (1/2, 1)$.

(A) Given equations are:
$(1) x + 8y = 19$
$(2) 2x + 11y = 28$
From equation $(1)$,we can express $x$ in terms of $y$:
$x = 19 - 8y$
Substitute this value of $x$ into equation $(2)$:
$2(19 - 8y) + 11y = 28$
$38 - 16y + 11y = 28$
$38 - 5y = 28$
$-5y = 28 - 38$
$-5y = -10$
$y = 2$
Now,substitute $y = 2$ back into the expression for $x$:
$x = 19 - 8(2)$
$x = 19 - 16$
$x = 3$
Thus,the solution is $(x, y) = (3, 2)$.

(B) Given equations are:
$1) \, 2x + 3y = 7$
$2) \, 6x + 9y = 23$
Check the ratio of coefficients:
$\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{7}{23}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other.
Therefore,the system of equations has no solution,denoted by $\varnothing$.

(C) Given equations are:
$(1) x + 11y = 1$
$(2) 8x + 13y = 2$
From equation $(1)$,we get $x = 1 - 11y$.
Substitute this value of $x$ into equation $(2)$:
$8(1 - 11y) + 13y = 2$
$8 - 88y + 13y = 2$
$8 - 75y = 2$
$-75y = 2 - 8$
$-75y = -6$
$y = \frac{-6}{-75} = \frac{2}{25}$.
Now,substitute $y = \frac{2}{25}$ into the expression for $x$:
$x = 1 - 11(\frac{2}{25})$
$x = 1 - \frac{22}{25} = \frac{25 - 22}{25} = \frac{3}{25}$.
Thus,the solution is $(x, y) = (\frac{3}{25}, \frac{2}{25})$.

(A) Given equations are:
$(1)$ $8x + 5y = 9$
$(2)$ $3x + 2y = 4$
From equation $(2)$,we can express $y$ in terms of $x$:
$2y = 4 - 3x$
$y = \frac{4 - 3x}{2}$
Substitute this value of $y$ into equation $(1)$:
$8x + 5(\frac{4 - 3x}{2}) = 9$
Multiply the entire equation by $2$ to eliminate the fraction:
$16x + 5(4 - 3x) = 18$
$16x + 20 - 15x = 18$
$x + 20 = 18$
$x = 18 - 20$
$x = -2$
Now,substitute $x = -2$ back into the expression for $y$:
$y = \frac{4 - 3(-2)}{2}$
$y = \frac{4 + 6}{2}$
$y = \frac{10}{2}$
$y = 5$
Thus,the solution is $(x, y) = (-2, 5)$.

(C) Let the fraction be $\frac{x}{y}$.
According to the first condition: $\frac{x+1}{y+1} = \frac{4}{5}$.
Cross-multiplying gives: $5(x+1) = 4(y+1) \implies 5x + 5 = 4y + 4 \implies 5x - 4y = -1$ (Equation $1$).
According to the second condition: $\frac{x-5}{y-5} = \frac{1}{2}$.
Cross-multiplying gives: $2(x-5) = 1(y-5) \implies 2x - 10 = y - 5 \implies 2x - y = 5$ (Equation $2$).
From Equation $2$,$y = 2x - 5$. Substituting this into Equation $1$:
$5x - 4(2x - 5) = -1$
$5x - 8x + 20 = -1$
$-3x = -21 \implies x = 7$.
Now,find $y$: $y = 2(7) - 5 = 14 - 5 = 9$.
Thus,the fraction is $\frac{7}{9}$.

(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 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 $5$.

(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 $23$.

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

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

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

(C) Given the linear equation $3x + 2y = 8$.
Substituting $x = -2$ into the equation:
$3(-2) + 2y = 8$
$-6 + 2y = 8$
Adding $6$ to both sides:
$2y = 8 + 6$
$2y = 14$
Dividing by $2$:
$y = 7$.

(B) Given the linear equation $2x - 5y = 3$.
Substituting the value $y = -1$ into the equation:
$2x - 5(-1) = 3$
$2x + 5 = 3$
Subtracting $5$ from both sides:
$2x = 3 - 5$
$2x = -2$
Dividing by $2$:
$x = -1$

(A) Given equations are:
$(1)$ $2x + 3y = 10$
$(2)$ $3x - y = 4$
From equation $(2)$,we can express $y$ in terms of $x$:
$y = 3x - 4$
Substitute this value of $y$ into equation $(1)$:
$2x + 3(3x - 4) = 10$
$2x + 9x - 12 = 10$
$11x = 10 + 12$
$11x = 22$
$x = 2$
Thus,the value of $x$ is $2$.

(C) Given the equation: $\frac{4}{x} + \frac{3}{y} = 1$.
Substitute $x = -2$ into the equation:
$\frac{4}{-2} + \frac{3}{y} = 1$
$-2 + \frac{3}{y} = 1$
Add $2$ to both sides:
$\frac{3}{y} = 1 + 2$
$\frac{3}{y} = 3$
Multiply both sides by $y$ and divide by $3$:
$y = \frac{3}{3} = 1$.
Therefore,$y = 1$.

(C) Given equations are:
$y = \frac{1}{2} x$ --- $(1)$
$x + 2y = 8$ --- $(2)$
Substitute the value of $y$ from equation $(1)$ into equation $(2)$:
$x + 2(\frac{1}{2} x) = 8$
$x + x = 8$
$2x = 8$
$x = \frac{8}{2}$
$x = 4$

(A) Let the digit at the ten's place be $x$ and the digit at the unit's place be $y$.
According to the problem,the sum of the digits is $6$,so $x + y = 6$.
Also,the digit at the unit's place is twice the digit at the ten's place,so $y = 2x$.
Substituting $y = 2x$ into the first equation: $x + 2x = 6$.
$3x = 6$,which gives $x = 2$.
Now,find $y$: $y = 2(2) = 4$.
The number is formed by $10x + y$,which is $10(2) + 4 = 24$.

(B) Given the linear equation in two variables: $2x + 3y = 8$.
To express $y$ in terms of $x$,we isolate the term containing $y$ on one side of the equation.
Subtract $2x$ from both sides: $3y = 8 - 2x$.
Now,divide both sides by $3$ to solve for $y$: $y = \frac{8 - 2x}{3}$.
Thus,the correct expression is $\frac{8 - 2x}{3}$.