Given pair of linear equations is
41x+53y=135   ..(i)
and 53x+41y=147  ...(ii)
On adding Eqs. (i) and (ii), we get
94x+94y=282
\(\Rightarrow\) x+y=3 [dividingboth sides by 94] ...(iii)
On subtracting Eq. (i) from Eq. (ii), we get
12x-12y=12
\(\Rightarrow\) x-y=1 [dividing both sides by 12]   ...(iv)
Now, on adding Eqs.(iii) and (iv), we get
2x=4 \(\Rightarrow\) x=2
On substituting x=2 in Eq. (iii), we get
y=3-2=1
Hence, x=2 and  y=1 is the required solution.

Given, pair of linear equations is
kx-5y-2=0 and 6x+2y-7=0
Here a₁=k, b₁=-5, c₁=-2
and a₂=6, b₂=2, c₂=-7
For no solution,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \neq \frac { -2 }{ -7 } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \)
\(\Rightarrow \quad k=-15\)

Given equations are x+4y=27xy and x+2y=21xy
On dividing both sides of the above equations by xy, we get
\(\frac { 1 }{ y } +\frac { 4 }{ x } =27\) and \(\frac { 1 }{ y } +\frac { 2 }{ x } =21\)
On putting \(\frac { 1 }{ y } =u\) and \(\frac { 1 }{ x } =v\),  we get
u+4v=27  ..(i)
and u+2v=21  ...(ii)
On subtractin Eq. (ii) from Eq. (i), we get
2v=6 \(\Rightarrow\) v=3
On putting the value of v in Eq. (i), we get
u+12=27 \(\Rightarrow\) u=15 
Now, \(v=3\Rightarrow \frac { 1 }{ x } =3\Rightarrow x=\frac { 1 }{ 3 } \)
and \(u=15\Rightarrow \frac { 1 }{ y } =15\Rightarrow y=\frac { 1 }{ 15 } \)
Hence, \(x=\frac { 1 }{ 3 } \) and \(y=\frac { 1 }{ 15 } \)  is the required solution.

Given, system of equations is
\(\frac { 1 }{ 2(2x+3y) } +\frac { 12 }{ 7(2x+3y) } =\frac { 1 }{ 2 } \quad \quad ...(i)\)
and \(\frac { 7 }{ 2x+3y } +\frac { 12 }{ 3x-2y } =2\quad \quad ...(ii)\)
Put  \(\frac { 7 }{ 2x+3y } =u\)  and \(\frac { 12 }{ 3x-2y } =v\)
Then, given system of equations becomes
\(\frac { u }{ 2 } +\frac { 12 }{ 7 } v=\frac { 1 }{ 2 } \)
7u+24v=7   ...(iii)
and 7u+4v=2  ...(iv)
Hence, x=2 and y=1

\(x=\frac { a }{ { a }^{ 2 }+{ b }^{ 2 } } ,\quad x=\frac { b }{ { a }^{ 2 }+{ b }^{ 2 } } \)

x=20, y=30; \(\angle A=100^{ 0 },\angle B={ 100 }^{ 0 },\angle C={ 50 }^{ 0 },\angle D=80^{ 0 }\)

Let x and y be length and breadth of rectangle.
Then, its area=xy
According to the questions,
9x-5)(y+2)=xy-80 \(\Rightarrow\) 2x-5y=-70
(x+10)(y-5)=xy+50  \(\Rightarrow\) -5x+10y=100
Length=40 units, breadth=30 units

For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
k=-6

Let the speed of train be x km/h and that of car y km/h, then
\(\Rightarrow \frac { 320 }{ x } +\frac { 280 }{ y } =\frac { 26 }{ 3 } \quad \quad ...(i)\)
\(\Rightarrow \frac { 200 }{ x } +\frac { 400 }{ y } =\frac { 55 }{ 6 } \quad \quad ...(ii)\)
Solve Eq. (i) and E₁. (ii) to get the speed of the train and car.
Speed of the train=80 km/h
Speed of car=60 km/h.

Let the man finishes the work in x days and the boy in y
According to question,
\(\frac { 8 }{ x } +\frac { 12 }{ y } =\frac { 1 }{ 10 } \quad ...(i)\)
and \(\frac { 6 }{ x } +\frac { 8 }{ y } =\frac { 1 }{ 14 } \quad \quad ...(ii)\)  
Solve Eq. (i) and Eq. (ii) to get value of x.
Man alone finish the work in 140 days.

Given equation are 7x-5y=3 and 14x-10y=5
Here, a₁=7, b₁=-5, c₂=-3
and a₂=14, b₂=-10, c₂=-5
Now, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } ,\quad \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -5 }{ -10 } =\frac { 1 }{ 2 } \) and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { 3 }{ 5 } \)
Thus, for the given equations, we have
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
So, we conclude that the two paths are parallel to each other. Hence, the paths do not cross each other. 

Points of intersection of lines represented by
3x-2y=10  ..(i)
and 5x+y=8   ..(ii)
is the common solution of this system of equations.
From Eq.(ii), we have
y=8-5x    ...(iii)
On substituting the value of y in Eq.(i), we get
3x-2(8-5x)=10
\(\Rightarrow\) 13x=26
\(\Rightarrow\) x=2
On putting x=2 in Eq. (iii), we get
y=8-5 x 2=-2
So, the point of intersection of the lines (i) and (ii) is (2,-2).
so, this point will satisfy the equations 3x+ay=8.
\(\therefore\) (3 x 2)+[(-2) x a]=8
\(\Rightarrow\) 6-2a=8
\(\Rightarrow\) -2a=2 \(\Rightarrow\) a=-1
Hence, the required value of a is -1.

Let us put \(\frac{1}{x-1}=p\) and \(\frac{1}{y-2}=q\) .Then the given equations
\(5\left(\frac{1}{x-1}\right)+\frac{1}{y-2}=2\)   ...(1)
\(6\left(\frac{1}{x-1}\right)-3\left(\frac{1}{y-2}\right)=1\)  ...(2)
can be written as : 5p + q = 2 .. (3)
6p – 3q = 1  ...(4)
Equations (3) and (4) form a pair of linear equations in the general form. Now,you can use any method to solve these equations. We get \(p=\frac{1}{3}\) and \(q=\frac{1}{3}\) .
Now, substituting \(\frac{1}{x-1}\) ,for p, we have
\(\frac{1}{x-1}=\frac{1}{3}\) 
i.e., x – 1 = 3, i.e., x = 4.
Similarly, substituting \(\frac{1}{y-2}\) ,for q, we get
\(\frac{1}{y-2}=\frac{1}{3}\) 
i.e., 3 = y – 2, i.e., y = 5
Hence, x = 4, y = 5 is the required solution of the given pair of equations.

Let  \(\frac{1}{\sqrt{x}}=p\) then the equations changes as below:
2p + 3q = 2 ... (i)
4p - 9q = -1 ... (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 ... (iii)
Adding equation (ii) and (iii), we get
10p = 5
p = 1/2 ... (iv)
Putting in equation (i), we get
\(2 \times \frac{1}{2}+3 q=2\) 
3q = 1
\(q=\frac{1}{3}\) 
\(p=\frac{1}{\sqrt{x}}=\frac{1}{2}\) 
\(\sqrt{x}=2\) 
x = 4 and \(q=\frac{1}{\sqrt{y}}=\frac{1}{3}\) 
\(\sqrt{y}=3\) 
y = 9
Hence, x = 4, y = 9

Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information
\(\frac{60}{u}+\frac{240}{v}=4\)  ...(i)
\(\frac{100}{u}+\frac{200}{v}=\frac{25}{6}\)  ...(ii)
Putting \(\frac{1}{u}=p\) in the equations, we get
60p + 240q = 4 ... (iii)
100p + 200q = 25/6
600p + 1200q = 25 ... (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 .... (v)
Subtracting equation (iv) from (v), we get1200q = 15
\(q=\frac{15}{200}=\frac{1}{80}\)   ...(vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/6
\(p=\frac{1}{u}=\frac{1}{60}\) 
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.