The pair of linear equations is :
x - y = 8
\(\Rightarrow\) x - y - 8 = 0       ....(i)
and 3x - 3y = 16
\(\Rightarrow\) 3x - 3y - 16 = 0 ... (ii)
On comparing with ax + by + C = 0, we have
a₁ = 1, b₁ = - 1, c₁ = - 8
a₂ = 3, b₂ = - 3, c₂ = - 16
\(\therefore \quad \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ 3 } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { 1 }{ 3 } ,\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { 1 }{ 2 } \)
As \( \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \) the lines are parallel having no solution.
So, the pair of linear equations is inconsistent.

The pair of linear equations is :
x + y = 5
\(\Rightarrow\) x + y - 5 = 0       ....(i)
and 2x + 2y = 10
\(\Rightarrow\) 2x + 2y -10 = 0   ...(ii)
On comparing with ax + by + C = 0 , we have
a₁ = 1, b₁ = 1,.c₁ = - 5
a₂ = 2, b₂ = 2, c₂ = -10
\(\therefore \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ 2 } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { 1 }{ 2 } \)
and  \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -5 }{ -10 } =\frac { 1 }{ 2 } \)
Since \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \) 
So, the pair of linear equations is coincident having many solutions. Thus, the equation is consistent. Now, we need to solve it graphically.
We have x + y = 5
\(\Rightarrow\) y = 5 - x

and 2x + 2y = 10
\(\Rightarrow \quad y=\frac { 10-2x }{ 2 } \)

On plotting these points, we observe that the lines are coincident having infinite solutions.

Given equation of lines are
3x - y = 3    ...(i)
2x-3y=2   ...(ii)
and x + 2y = 8 ....(iii)
Let lines (i), (ii) and (iii) represent the sides of \(\triangle\)ABC,
say AB, BC and CA, respectively.

On multiplying Eq. (i) by 3 and then subtracting Eq. (ii) from Eq. (i), we get
7x=7 \(\Rightarrow\) x=1
On putting the value of x in Eq. (i), we get
3 x 1-y=3
\(\Rightarrow\)   y=0
So, the coordinate of point or vertex B is (1, 6). Similarly, on solving Eq. (ii) and Eq. (iii), we get the coordinate of point or vertex Cis (4, 2). 
And on solving Eq. (i) and Eq. (iii), we get the coordinate of point or vertex A is (2, 3). 
Hence, the vertices of \(\triangle\)ABC formed by the given lines are A (2,3), B (1, 0) and C( 4, 2).

Given, pair oflinear equations is
x+y-2=0   ..(i)
and 2x - y - 1= 0    ..(ii)
Now, table for x + y = 2
or y=2-x is

Table for 2x - y -1 = 0
or y = 2x -1 is 

Plot the points A (0, 2) and B (2, 0) and join them to get the straight line AB. Similarly, plot the points C (0, - 1) and D (2, 3) and join them to get the straight line CD. The lines AB and CD intersect at E (1, 1). So, the solution of the given pair oflinear equations is (1, 1). 

It is clear from the graph that infinite lines can pass through the intersection point of linear equations x + y = 2 and 2x - y = 1, i.e. point E (1, 1) satisfy the many linear equations such as y = x, 2x +Y = 3, x + 2y = 3 and so on.

Let the present age of Sagar be x yr and the age of Tiru be y yr.
5 yr ago, Sagar's age = (x - 5) yr and Tiru's age = (y - 5) yr
According to the given condition,
(x-5)=2(y-5)
\(\Rightarrow\) x-5=2y-10
\(\Rightarrow\) x-2y+5=0
After 10 yr, Sagar's age = (x + 10) yr and Tiru's age = (y + 10) yr
According to the given condition,
x+10=(y+10)=10
\(\Rightarrow\)x+10=y+20
\(\Rightarrow\) x-y-10=0
Thus, we get the following pair of linear equations
x-2y+5=0  ..(i)
x-y-10=0   ..(ii)
Now, let us draw the graphs of Eqs.(i) and (ii), by finding at least two solutions for each of these equations: The solutions of the equations are given in tables.
Table for x-2y+5=0 or \(y=\frac { x+5 }{ 2 } \) is

Table for x-y-10=0 or y=x-10 is

Plot the points A (5, 5) and B (-5,0) and join them to get the line AB. Similarly, plot the points C (5,-5) and D (10, 0) and join-them to get the line CD.

It is clear from the graph that, lines AB and CD intersect each other at point E (25,15).
So, x= 25 and y=15 is the required solution.
Hence, Sagar's present age = 25 yr and Tiru's present age = 15 yr.

Table for line 3x +y - 5 = 0
or y = 5 -3xis

Now, plot all these points on a graph paper and join them to get a straight line BC.
Table for line 2x - y - 5 = 0 or y = 2x - 5 is

Now, plot all these points on a graph paper and join them to get a straight line EF. Thus, we get the following graph and it is clear from the graph that two lines intersect each other at point C(2, -1). 

Line 3x + y - 5 = °intersects Y-axis at the point B(O,5) and line 2x - y - 5 = 0 intersects Y-axis at the point E (0, - 5).

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

x=a², y=b²

Given lines are 2ax-by=2a²-b²       ...(i)      
and ax+2by=a²+2b²         ...(ii)
On multiplying Eq. (ii) by w and then subtracting from Eq. (i), we get
-5by=-5b²
\(\Rightarrow\) y=b
On putting the value of y in Eq. (i), we get
2ax-b²=2a²-b²  
2ax=2a² \(\Rightarrow\) x=a
So, the solution of the given system is x=a and y=b or the lines intersect at (a,b).
Now, we have to prove that point (a,b) lies on the line
(a-b)x+(a+b)y=a²+b²
On putting x=a and y=b in LHS, we get
LHS=(a-b)a+(a+b)b
=a²-ab+ab+b²
=a²+b²=RHS
so, the lines are concurrent.

Let the unit's place digit be x and ten's place digit be y.
Original number=10y+x
Number obtained by reversing the order of digits
=10x+y
Sum of both number=99  [given]
(10y+x)+(10x+y)=99
11x+11y=99
x+y=9  .....(i) [dividing both sides by 11]
Also, given that the digits of the numbers differ by 3.
If y>x, then y=-x=3  ..(ii)
and if y On adding Eqs. (i) and (ii), we get
2y=12
\(\Rightarrow\)  \(y= \frac {12}{2}\)=6
Then, from Eq. (i),
x=9-y=9-6=3  [\(\therefore\)y=6]
\(\therefore\)Number=10y+x=10(6)+3=63
Now, adding Eqs. (i) and (iii), we get
2x=12
x=6
Then, from Eq. (i), we get
y=9-x=9-6=3  [\(\therefore\)x=6]
\(\therefore\)Number=10y+x=10(3)+6=36
Hence, the required number 63 and 36

Given pair of equations is
2x+3y-4=0
and (k+2)x+6y-(3k+2)=0
On comparing the given equations with standard form
a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, we get
a₁=2, b₁=3, c₁=4
and a₂=k+2, b₂=6, c₂=-(3k+2)=0
For infinitely many solutions
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore \ \underset { I }{ \frac { 2 }{ k+2 } } =\underset { II }{ \frac { 3 }{ 6 } } =\underset { III }{ \frac { -4 }{ -(3k+2) } } \)
On taking I and II terms, we get
\(\frac { 2 }{ k+2 } =\frac { 3 }{ 6 } \ \Rightarrow \ \frac { 2 }{ k+2 } =\frac { 1 }{ 2 } \)
\(\Rightarrow \quad k+2=4\quad \Rightarrow \quad k=2\)
Which also satisfy the last two terms of Eq. (i)
Hence, the required value of k is 2.

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,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \neq \frac { -2 }{ -7 } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \)
\(\Rightarrow \quad k=-15\)

The given equation can be rewritten as 2x+ky-1=0 and 3x-5y-7=0
On comparing with
a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0, we get
a₁=2, b₁=k, c₁=-1
and a₂=3, b₂=-5, c₂=-7
A pair of linear equations has unique solution, if
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
\(\Rightarrow \quad \frac { 2 }{ 3 } \neq \frac { k }{ -5 } \)
\(\Rightarrow \quad k\neq \frac { -10 }{ 3 } \)
So, given lines have unique solution for all real values of k, except \(\frac { -10 }{ 3 } \)

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, linear equation is 2x + 3y - 8 = 0     .....(i)
(i) For intersecting lines, we know that \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
One of the possible line may be taken as
5x + 2y-9 =0
(ii) For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore\)  One of the possible line parallel to eqn. (i) may be taken as
6x + 9y + 7 = 0
(iii) For coincident lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
One of the possible line coincident to eqn. (i) may be taken as
4x + 6y - 16 = 0