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

(A) $x+y=7$
$\therefore y=7-x$
For $x=0, y=7-0=7$
For $x=7, y=7-7=0$

$x$	$0$	$7$
$y$	$7$	$0$

$\therefore$ Plot the ordered pairs $(0, 7)$ and $(7, 0)$ of the solution set of $x+y=7$ on the graph paper and draw the line by joining them.
$5x+2y=20$
$\therefore 2y=20-5x \quad \therefore y=\frac{20-5x}{2}$
For $x=0, y=\frac{20-0}{2}=10$
For $x=4, y=\frac{20-20}{2}=0$

$x$	$0$	$4$
$y$	$10$	$0$

$\therefore$ Plot the ordered pairs $(0, 10)$ and $(4, 0)$ of the solution set of $5x+2y=20$ on the graph paper and draw the line by joining them.
The intersection point (common point) of these two lines is $(2, 5)$,which satisfies both equations.
Thus,the solution of the pair of linear equations is $(2, 5)$.

(N/A) For the equation $3x + 6y = 3900$:
Dividing by $3$,we get $x + 2y = 1300$,which implies $y = \frac{1300 - x}{2}$.
If $x = 0$,$y = 650$. If $x = 1300$,$y = 0$.

$x$	$0$	$1300$
$y$	$650$	$0$

Plot the points $(0, 650)$ and $(1300, 0)$ and join them to draw the line.
For the equation $x + 3y = 1300$:
This implies $y = \frac{1300 - x}{3}$.
If $x = 1300$,$y = 0$. If $x = 100$,$y = 400$.

$x$	$1300$	$100$
$y$	$0$	$400$

Plot the points $(1300, 0)$ and $(100, 400)$ and join them to draw the line.
The intersection point of these two lines is $(1300, 0)$,which satisfies both equations.
Thus,the solution is $x = 1300, y = 0$.

(D) For the equation $3x + 4y = 6$:
$4y = 6 - 3x$
$y = \frac{6 - 3x}{4}$
If $x = -2$,$y = \frac{6 - 3(-2)}{4} = \frac{12}{4} = 3$.
If $x = 2$,$y = \frac{6 - 3(2)}{4} = \frac{0}{4} = 0$.

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

Plot the points $(-2, 3)$ and $(2, 0)$ and join them to draw the line.
For the equation $3x + 4y = 19$:
$4y = 19 - 3x$
$y = \frac{19 - 3x}{4}$
If $x = 5$,$y = \frac{19 - 3(5)}{4} = \frac{4}{4} = 1$.
If $x = 1$,$y = \frac{19 - 3(1)}{4} = \frac{16}{4} = 4$.

$x$	$5$	$1$
$y$	$1$	$4$

Plot the points $(5, 1)$ and $(1, 4)$ and join them to draw the line.
Since the lines are parallel and do not intersect,the system of equations has no solution. The solution set is $\varnothing$.

(N/A) Here,dividing each term of $3x+3y=15$ by $3$,we get the equation $x+y=5$.
Thus,both the equations of the pair are identical.
Hence,we say that both the lines are the same.
So,they are coincident. It means that there are infinitely many solutions. To draw the graph,we make the following table:
$x+y=5$
$\therefore y=5-x$
For $x=0, y=5-0=5$
For $x=5, y=5-5=0$
and
$3x+3y=15$
$\therefore 3y=15-3x$
$\therefore y=\frac{15-3x}{3}$
$\therefore y=5-x$
$\therefore$ Both the tables are the same.

$x$	$0$	$5$
$y$	$5$	$0$

$\therefore$ Plot the ordered pairs $(0, 5)$ and $(5, 0)$ of the solution set of $x+y=5$ (or $3x+3y=15$,i.e.,$x+y=5$) on the graph paper and draw the line by joining them.
Here,the graphs of both the equations are the same. Also,we can see that there are infinite points on the line,and they all make the solution set. Thus,the solution set of the pair of linear equations is ${(x, y) \mid x+y=5, x, y \in R}$.

(A) Given equations are:
$1) \, y + x = 5 \implies y = 5 - x$
$2) \, y - x = 9 \implies y = 9 + x$
To plot the graph, we find points for each line:
For $y = 5 - x$: If $x = 0, y = 5$; If $x = 5, y = 0$. Points are $(0, 5)$ and $(5, 0)$.
For $y = 9 + x$: If $x = 0, y = 9$; If $x = -9, y = 0$. Points are $(0, 9)$ and $(-9, 0)$.
Solving algebraically for intersection:
Adding the two equations: $(y + x) + (y - x) = 5 + 9 \implies 2y = 14 \implies y = 7$.
Substituting $y = 7$ in $y + x = 5$: $7 + x = 5 \implies x = -2$.
The point of intersection is $(-2, 7)$.

(D) Given equations are:
$1) 3x + 6y = 4$
$2) 2x + 4y = \frac{8}{3}$
Multiply equation $(2)$ by $\frac{3}{2}$:
$\frac{3}{2}(2x + 4y) = \frac{3}{2} \times \frac{8}{3}$
$3x + 6y = 4$
Since both equations represent the same line,they are coincident lines.
Therefore,the system has infinitely many solutions. The solution set is $\{(x, y) \mid 3x + 6y = 4; x, y \in R \}$.

(A) Given equations are:
$1) 2x + y = 7 \implies y = 7 - 2x$
For $x = 0, y = 7$. For $x = 2, y = 3$. For $x = 3, y = 1$.
$2) x - 2y = 6 \implies x = 6 + 2y$
For $y = 0, x = 6$. For $y = -1, x = 4$. For $y = -2, x = 2$.
Plotting these points on a graph,the two lines intersect at the point $(4, -1)$.
Thus,the solution is $x = 4$ and $y = -1$.

(D) Given equations are:
$1) y = 2$
$2) x + 4y = 10$
Step $1$: The first equation $y = 2$ represents a horizontal line passing through all points where the $y$-coordinate is $2$.
Step $2$: For the second equation $x + 4y = 10$,substitute $y = 2$ into the equation:
$x + 4(2) = 10$
$x + 8 = 10$
$x = 10 - 8$
$x = 2$
Step $3$: The point of intersection of the two lines is $(2, 2)$.
Graphically,the line $y = 2$ is a horizontal line,and $x + 4y = 10$ is a line passing through $(10, 0)$ and $(2, 2)$. Both lines intersect at the point $(2, 2)$.

(A) Given equations are:
$1) x + y = 5$
$2) 5x - 2y = 4$
For equation $(1)$,if $x = 0$,then $y = 5$. If $y = 0$,then $x = 5$. Points are $(0, 5)$ and $(5, 0)$.
For equation $(2)$,if $x = 0$,then $-2y = 4 \implies y = -2$. If $y = 0$,then $5x = 4 \implies x = 0.8$. Points are $(0, -2)$ and $(0.8, 0)$.
By plotting these lines on a graph,the point of intersection is $(2, 3)$.
Verification: $2 + 3 = 5$ (Correct) and $5(2) - 2(3) = 10 - 6 = 4$ (Correct).
Thus,the solution is $(2, 3)$.

(B) Given equations are:
$1) \, 2x - 3y = 5$
$2) \, 4x - 6y = -13$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 2, b_1 = -3, c_1 = -5$
$a_2 = 4, b_2 = -6, c_2 = 13$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-5}{13}$
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.
Parallel lines do not intersect at any point,which means the system of equations has no solution $(\varnothing)$.

(C) Given equations are:
$1) 4x + y = 7$
$2) 16x + 4y = 28$
Divide equation $(2)$ by $4$:
$4x + y = 7$
Since both equations are identical,they represent the same line on the graph.
For any value of $x$,we can find a corresponding value of $y$ such that $y = 7 - 4x$.
Therefore,the system has infinitely many solutions.
The solution set is ${(x, y) \mid 4x + y = 7; x, y \in R}$.

(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:
$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$. If $y = 0$,then $x = -4$. The line passes through $(0, -2)$ and $(-4, 0)$.
$2$. For the second equation $3x + 4y = -6$,if $x = 0$,then $y = -1.5$. If $y = 0$,then $x = -2$. The line passes through $(0, -1.5)$ and $(-2, 0)$.
$3$. Solving the equations algebraically: Multiply the first equation by $2$: $2x + 4y = -8$. Subtract this from the second equation: $(3x + 4y) - (2x + 4y) = -6 - (-8)$,which gives $x = 2$.
$4$. Substitute $x = 2$ into $x + 2y = -4$: $2 + 2y = -4 \implies 2y = -6 \implies y = -3$.
$5$. The intersection point is $(2, -3)$.

(A) To solve the system of equations $x + y = 8$ and $x - y = 2$ graphically:
$1$. For the equation $x + y = 8$,we find two points: If $x = 0, y = 8$; if $x = 8, y = 0$. The line passes through $(0, 8)$ and $(8, 0)$.
$2$. For the equation $x - y = 2$,we find two points: If $x = 0, y = -2$; if $x = 2, y = 0$. The line passes through $(0, -2)$ and $(2, 0)$.
$3$. Plotting these two lines on a Cartesian plane,they intersect at the point $(5, 3)$.
$4$. Therefore,the solution is $x = 5$ and $y = 3$.

(B) Given the equations:
$1) \ 2x + 3y = 12$
$2) \ 2x + 3y = 6$
Comparing these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we have:
$a_1 = 2, b_1 = 3, c_1 = -12$
$a_2 = 2, b_2 = 3, c_2 = -6$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{2} = 1$
$\frac{b_1}{b_2} = \frac{3}{3} = 1$
$\frac{c_1}{c_2} = \frac{-12}{-6} = 2$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines are parallel to each other.
Parallel lines do not intersect at any point,which means the system of equations has no solution.
Therefore,the solution set is $\varnothing$.

(C) $1$. For the equation $x+y=10$,if $x=0, y=10$; if $y=0, x=10$. The line passes through $(0,10)$ and $(10,0)$.
$2$. For the equation $x-y=4$,if $x=0, y=-4$; if $y=0, x=4$. The line passes through $(0,-4)$ and $(4,0)$.
$3$. Solving the equations simultaneously: Adding $(x+y=10)$ and $(x-y=4)$ gives $2x=14$,so $x=7$. Substituting $x=7$ into $x+y=10$ gives $y=3$. The intersection point is $(7,3)$.
$4$. The triangle is formed by the intersection of the two lines $(7,3)$ and their respective intersections with the $X$-axis,which are $(4,0)$ and $(10,0)$.
$5$. Thus,the vertices of the triangle are $(7,3), (4,0),$ and $(10,0)$.

(D) $1$. For the equation $x + 3y = 6$: If $x = 0$,$3y = 6 \implies y = 2$. If $y = 0$,$x = 6$. Points are $(0, 2)$ and $(6, 0)$.
$2$. For the equation $2x - 3y = 12$: If $x = 0$,$-3y = 12 \implies y = -4$. If $y = 0$,$2x = 12 \implies x = 6$. Points are $(0, -4)$ and $(6, 0)$.
$3$. Solving the system: Adding the two equations,$(x + 3y) + (2x - 3y) = 6 + 12 \implies 3x = 18 \implies x = 6$. Substituting $x = 6$ into $x + 3y = 6$,we get $6 + 3y = 6 \implies y = 0$. The intersection point is $(6, 0)$.
$4$. The triangle is formed by the lines and the $Y$-axis. The vertices are the intersection of the two lines $(6, 0)$,the $Y$-intercept of the first line $(0, 2)$,and the $Y$-intercept of the second line $(0, -4)$.
$5$. Thus,the vertices are $(6, 0), (0, 2), (0, -4)$.

$(A)$ Let the number of boys be $x$ and the number of girls be $y$.
According to the problem,the total number of students is $6$,so $x + y = 6$.
The number of girls is $2$ less than the number of boys,so $y = x - 2$,which can be written as $x - y = 2$.
Solving the system of equations:
$1) x + y = 6$
$2) x - y = 2$
Adding the two equations: $(x + y) + (x - y) = 6 + 2 \rightarrow 2x = 8 \rightarrow x = 4$.
Substituting $x = 4$ into the first equation: $4 + y = 6 \rightarrow y = 2$.
Thus,there are $4$ boys and $2$ girls.

(A) To check if the pair of equations is consistent,we compare the ratios of the coefficients:
$a_1 = 2, b_1 = 4, c_1 = -9$
$a_2 = 3, b_2 = 6, c_2 = -\frac{27}{2}$
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2}{3}$
$\frac{b_1}{b_2} = \frac{4}{6} = \frac{2}{3}$
$\frac{c_1}{c_2} = \frac{-9}{-27/2} = \frac{9 \times 2}{27} = \frac{18}{27} = \frac{2}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$,the lines are coincident.
Coincident lines have infinitely many solutions,therefore the system is consistent.

(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 numerator of the fraction be $x$ and the denominator be $y$.
According to the first condition,if $2$ is subtracted from the numerator and $3$ is added to the denominator,the fraction becomes $\frac{1}{4}$.
$\frac{x-2}{y+3} = \frac{1}{4}$
$4(x-2) = y+3$
$4x - 8 = y + 3$
$4x - y = 11$ .......... $(1)$
According to the second condition,if $6$ is added to the numerator and the denominator is multiplied by $3$,the fraction becomes $\frac{2}{3}$.
$\frac{x+6}{3y} = \frac{2}{3}$
$3(x+6) = 2(3y)$
$3x + 18 = 6y$
$3x - 6y = -18$
Dividing by $3$,we get $x - 2y = -6$ .......... $(2)$
From equation $(1)$,$y = 4x - 11$. Substituting this into equation $(2)$:
$x - 2(4x - 11) = -6$
$x - 8x + 22 = -6$
$-7x = -28$
$x = 4$
Now,substitute $x = 4$ into $y = 4x - 11$:
$y = 4(4) - 11 = 16 - 11 = 5$
Therefore,the original fraction is $\frac{x}{y} = \frac{4}{5}$.

(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 $(x, y) = (-2, 5)$.

(B) Given equations are:
$1) 2x + y = 40$
$2) x + y = 30$
Subtracting equation $(2)$ from equation $(1)$:
$(2x + y) - (x + y) = 40 - 30$
$x = 10$
Substituting $x = 10$ in equation $(2)$:
$10 + y = 30$
$y = 20$
So,the solution is $(x, y) = (10, 20)$.
Now,substitute these values into the equation $y = mx + 6$:
$20 = m(10) + 6$
$20 - 6 = 10m$
$14 = 10m$
$m = \frac{14}{10} = 1.4$.

(A) The sum of angles in a triangle is $180^{\circ}$. Therefore,$x + 3x + y = 180$,which simplifies to $4x + y = 180$.
From this,we get $y = 180 - 4x$.
Substituting this into the given equation $3y - 5x = 30$:
$3(180 - 4x) - 5x = 30$
$540 - 12x - 5x = 30$
$540 - 17x = 30$
$17x = 510$
$x = 30^{\circ}$.
Now,find $y$:
$y = 180 - 4(30) = 180 - 120 = 60^{\circ}$.
Wait,checking the angles: $m \angle A = 30^{\circ}$,$m \angle B = 3(30) = 90^{\circ}$,and $m \angle C = 60^{\circ}$.
Since one angle is $90^{\circ}$,the triangle is right-angled.
The angles are $30^{\circ}, 90^{\circ}, 60^{\circ}$.

(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 $\frac{7}{9}$.

(A) Let the cost of a ticket from Ahmadabad to Nadiad be Rs. $x$ and the cost of a ticket from Ahmadabad to Vadodara be Rs. $y$.
According to the problem,we have two equations:
$1) 5x + 10y = 1050$
Dividing by $5$,we get: $x + 2y = 210$
$2) x + y = 130$
Subtracting equation $(2)$ from equation $(1)$:
$(x + 2y) - (x + y) = 210 - 130$
$y = 80$
Substituting $y = 80$ into equation $(2)$:
$x + 80 = 130$
$x = 130 - 80 = 50$
Therefore,the cost of a ticket from Ahmadabad to Nadiad is Rs. $50$ and the cost of a ticket from Ahmadabad to Vadodara is Rs. $80$.

(A) Let the numerator of the fraction be $x$ and the denominator be $y$.
According to the problem,the denominator is $6$ more than the numerator: $y = x + 6$.
If $1$ is added to the numerator and $3$ is subtracted from the denominator,the fraction becomes $\frac{3}{4}$: $\frac{x + 1}{y - 3} = \frac{3}{4}$.
Substitute $y = x + 6$ into the second equation: $\frac{x + 1}{(x + 6) - 3} = \frac{3}{4}$.
$\frac{x + 1}{x + 3} = \frac{3}{4}$.
Cross-multiply: $4(x + 1) = 3(x + 3)$.
$4x + 4 = 3x + 9$.
$4x - 3x = 9 - 4$.
$x = 5$.
Now,find $y$: $y = x + 6 = 5 + 6 = 11$.
Therefore,the fraction is $\frac{x}{y} = \frac{5}{11}$.

(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 $(x, y) = (2, 3)$.

(A) Suppose there are $x$ coins of $50$ paise and $y$ coins of $25$ paise in the bag.
The total number of coins is $x + y = 150$ ....... $(1)$
The value of $x$ coins of $50$ paise is $50x$ paise and the value of $y$ coins of $25$ paise is $25y$ paise.
The total amount is $50x + 25y$ paise.
Since the total amount is Rs. $55$,which is $5500$ paise,we have:
$50x + 25y = 5500$
Dividing the equation by $25$,we get:
$2x + y = 220$ ....... $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$(2x + y) - (x + y) = 220 - 150$
$x = 70$
Substituting $x = 70$ in equation $(1)$:
$70 + y = 150$
$y = 80$
Thus,the number of $50$ paise coins is $70$ and the number of $25$ paise coins is $80$.

(C) Let the digit at the ten's place be $y$ and the digit at the unit's place be $x$.
The original number is $10y + x$.
According to the problem,the sum of the digits is $6$:
$x + y = 6$ ............ $(1)$
When the digits are interchanged,the new number is $10x + y$.
According to the problem,the new number is $18$ less than the original number:
$(10y + x) - (10x + y) = 18$
$9y - 9x = 18$
Dividing by $9$,we get:
$y - x = 2$ ............ $(2)$
Adding equation $(1)$ and equation $(2)$:
$(x + y) + (y - x) = 6 + 2$
$2y = 8$
$y = 4$
Substituting $y = 4$ into equation $(1)$:
$x + 4 = 6$
$x = 2$
Therefore,the original number is $10(4) + 2 = 42$.

(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 $(x, y) = (5, 1)$.