Let the cost of one bat be Rs x and one ball be Rs y.Then, the algebraic representation is given by the following equations:
3x + 6y = 3,900         .....(i)
and x + 2y = 1,300    .....(ii)
To obtain the equivalent geometric representation we find two points on the line representing each equation i.e., we find two solutions of each equation. These solutions 'are given below in the table: 
For 3x + 6y = 3,900
\(\Rightarrow \quad y=\frac { 1,300-x }{ 2 } \)

for x + 2y = 1,300
\(\Rightarrow \quad y=\frac { 1,300-x }{ 2 } \)

Weplot the points A (0, 650) and B(1,300, 0) to obtain the geometric representation of 3x + 6y =3,900 and C(500,400) and D(100, 600) to obtain the geometric representation of x + 2y = 1,300. We observe these lines are coincident.

Given linear equation is 2x+3y-8=0.   ...(i)
(i) For intersecting lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
\(\therefore\) Any line intersecting with Eq. (i) may be taken as
3x+2y-9=0 or 3x+2y-7=0 
(ii) For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore\) Any line parallel to Eq. (i) may be taken as
6x+9y+7=0 or 2x+3y-12=0
(iii) For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore\) Any line coincident to Eq. (i) may be taken as
4x+6y-16=0
There can be several linear equations in each of (i), (ii) and (iii)

The given system of equations is
\(\frac { 2xy }{ x+y } =\frac { 3 }{ 2 } \) and \(\frac { xy }{ 2x-y } =\frac { -3 }{ 10 } \)
\(\Rightarrow\) \(\frac { x+y }{ 2xy } =\frac { 4 }{ 3 } \) and  \(\frac { 2x-y }{ xy } =\frac { -10 }{ 3 } \)
\(\frac { 1 }{ y } +\frac { 1 }{ x } =\frac { 4 }{ 3 } \) and \(\frac { 2 }{ y } -\frac { 1 }{ x } =\frac { -10 }{ 3 } \)
Put \(\frac { 1 }{ x } =u\)  and \(\frac { 1 }{ y } =v\)  then the system of equations becomes u+v=\(\frac { 4 }{3 } \) and  -u+2v=\(\frac { -10 }{3 } \)
Now, solve these equation.
x=\(\frac {1}{2}\) and y=\(\frac {-3}{2}\)

Let unit's digit be y and ten's digit be x. Then, x+y=8 and (10x+y)-(10y+x)=18
\(\Rightarrow\) 9x-9y=18
x=5, y=3, Required number=53

Two straight paths are parallel to each other. Hence, they do not cross each other.

x=3, y=4, vertices of triangle are (3,2), (0,-1) and (0,11).

x=3, y=4, vertices of triangle are (3,4), (-1,0) and (5,0).

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.

Let the number of Rs.100 and Rs.50 notes be x and y, respectively. Then, according to the question, x+y=14100x+50y=1000
Rs.100 notes=6, Rs.50 notes=8

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