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.

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

Complete step-by-step answer:
Let the cost of one chair be x and the cost of one table be y.
Now from the first statement, Two chairs and three tables costs Rs.5650
The equation so formed is, 
2x+3y=5650 …eq1
From second statement, three chairs and two tables cost Rs.7100
The equation so formed is,  3x+2y=7100 …eq2
Now to solve these equations we will use elimination method,
eq1+eq2,
2x+3y+3x+2y=5650+7100
On adding the same terms,
5x+5y=12750
Dividing both sides by 5,
x+y=2550 ….eq3
Now to eliminate any one variable we need it in opposite sign, eq2-eq1,
3x+2y−(2x+3y)=7100−5650
on solving we get,
3x+2y−2x−3y=1450
subtracting the same terms,
x−y=1450….eq4
now to solve the equations eq3+-eq4,
x+y+x−y=2550+1450
Now adding same terms,
2x=40002x=4000
Dividing both sides by 2,
x=2000
This is the cost of one chair.
Now using eq3 we get,
y=x−1450
Putting the value of x,
y=2000−1450
On subtracting we get,
y=550y=550
This is the cost of one table.
Thus the cost of one chair is Rs.2000 and that of one table is Rs.550.
Note: Note that substitution method can also be used to solve the equations. But here the elimination method is easier to solve the equations. But it is important to notice the equations that are formed do not get wrong in any sign. Because that is the only mistake generally students make.

The given system of equations is:3x + 5y = 12 5x + 3y = 4 
These equations are of the forms: a₁x+b₁y+c₁ = 0 and a₂x+b₂y+c₂ = 0 
where a₁ = 3 b₁= 5 c₁ = -12 and a₂ = 5 b₂ = 3 c₂ = -4 
For a unique solution we must have: 
Hence the given system of equations has a unique solution. 
Again the given equations are: 
3x + 5y = 12 …..i 
5x + 3y = 4 …..ii 
On multiplying i by 3 and ii by 5 we get: 
9x + 15y = 36 …….iii 
25x + 15y = 20 ……iv 
On subtracting iii from iv we get: 
16x = -16 ⇒x = -1 
On substituting x = -1 in i we get: 
3-1 + 5y = 12 
⇒5y = 12 + 3 = 15 
⇒y = 3 
Hence x = -1 and y = 3 is the required solution.