Let basic fare = Rs. B and
Let reservation charge =Rs. R
\(\cfrac { B }{ 2 } +R=half\ ticket\)
B + R = full ticket
B + R = 216     ....(1)
\(\left( \cfrac { B }{ 2 } +R \right) +\left( B+R \right) =327\)
\(\cfrac { B }{ 2 } +R+216=327\)
\(\cfrac { B }{ 2 } +R=111\)
B+2R = 222
B+2R = 222                                ...(1)
(1) \(\Rightarrow\) B+R = 216
(2) \(\Rightarrow\) \(\cfrac { B+2R=222 }{ -R=-6 } \)
R = 6
Substitute R = 6 in (1)
B+6 = 216
B = 216 - 6
B = 210
\(\therefore\) The basic first class full fare = Rs. 210
The half fare = Rs. 105
The reservation charge = Rs. 6

\(\cfrac { x }{ y } =\cfrac { 5 }{ 6 } \)
\(\Rightarrow\) 6x = 5y
6x - 5y = 0 .......(1)
\(\cfrac { x-8 }{ y-8 } =\cfrac { 4 }{ 5 } \)
\(\Rightarrow\) 5(x - 8) = 4(y - 8)
5x-40 = 4y-32
5x -4y = 40 - 32
5x - 4y = 8 ------ (2)
(1) x 5 \(\Rightarrow\) 30x-25y = 0
(2) x 6 \(\Rightarrow\) \(\cfrac { 30x-24y=48 }{ -y=-48 } \)    ...(2)
Substitute y = 48 in (1)
6x- 5(48) = 0
6x-240 = 0
6x = 240
\(x=\cfrac { 240 }{ 6 } =40\)
y = 48
\(\cfrac { x }{ y } =\cfrac { 40 }{ 48 } =\cfrac { 5 }{ 6 } \)
\(\therefore\) The number are in the Ratio 5 : 6

Let the actual price of a T.V. = x
Let the actual price of a Fridge = y
\(\cfrac { 5 }{ 100 } x+\cfrac { 10 }{ 100 } y=2000\)
\(\cfrac { 5 }{ 100 } (x+2y)=2000\)
\(x+2y=2000\times \cfrac { 100 }{ 5 } \)
x+2y = 40000     ...(1)
\(\cfrac { 10 }{ 100 } x-\cfrac { 5 }{ 100 } y=1500\)
\(\cfrac { 5 }{ 100 } (2x-y)=1500\)
\(2x-y=1500\times \cfrac { 100 }{ 5 } \)
2x -y = 30000
(1) X 2 \(\Rightarrow\) 2x + 4y = 80000   
(2) \(\Rightarrow\) \(\cfrac { 2x-y=30000 }{ 5y=50000 } \)
y = 10000
Substitute y = 10000 in (1)
x + 2(10000) = 40000
x + 20000 = 40000
x = 40000 - 20000
x = 20000
\(\therefore\) Actual price of T.V = Rs. 20000
Actual price of Fridge = Rs. 10000

Let the speed of ihe car starts from A is x
Let the speed of the car starts from B is y
When they are moving in same direction, the speed is x - y
7(x - y) = 70· [\(\because \) Speed x Time = Distance]
x - y = 10 ---------- (1)
When they are moving towards each other,the speed is x + y
1(x+y) = 70
x +Y = 70 ---------- (2)
(I) \(\Rightarrow\) x-y = 10
(2) \(\Rightarrow\) \(\cfrac { x+y=70 }{ 2x=80 } \)
x = 40
Substitute x = 40 in (I)
40 - y = 10 
y = 40 -10
y = 30
\(\therefore\) The speed of the car from A is 40 km/hr and the speed of the car from B is 30km/hr

Let the numerator be x
Denominator be y
x +y = 12 ...(1)
\(\cfrac { x }{ y+3 } =\cfrac { 1 }{ 2 } \)
2x = y+3
2x-y = 3 ...(2)
(1) X 2 \(\Rightarrow\)  2x +2y = 24
(2) \(\Rightarrow\) \(\cfrac { 2x-y=3 }{ 3y=21 } \)     ...(2)
y = 7
Substitute y = 7 in (1) .
x + 7 = 12
x = 12-7
x = 5
\(\therefore\) The fraction \(\cfrac { x }{ y } =\cfrac { 5 }{ 7 } \)

Let the time taken by the larger pipe be x hours
and Set the time taken by the smaller pipe be y hours.
\(\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ 24 } \)
In 1.hour the larger pipe can fill it = \(\cfrac { 1 }{ x } \)
In 1hour the smaller pipe can fill it  = \(\cfrac { 1 }{ y } \)
\(\cfrac { 8 }{ x } +\cfrac { 18 }{ y } =\cfrac { 1 }{ 2 } \)
Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
\(a+b\ =\cfrac { 1 }{ 24 } \)
24a + 24b = 1
24a + 24b - 1 = 0    ...(1)
8a + 18b = \(\cfrac { 1 }{ 2 } \)
16a+36b = 1
16a + 36b - 1 = 0    ..(2)
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ -24+36 } =\cfrac { b }{ -16+24 } =\cfrac { 1 }{ 864-384 } \)
\(\cfrac { a }{ 12 } =\cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \)
\(\cfrac { a }{ 12 } =\cfrac { 1 }{ 480 } \ \)  \( \cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \\ \)

\(\therefore\) x = 40, y = 60

Let the monthly income of A and B be 3x and 4x respectively.
Let the monthly expenditure of A and B be 5y and 7y respectively.
\(\therefore\) 3x - 5y = 5000 --------- (1)
4x - 7y = 5000 --------- (2)
(1) \(\times\)4 \(\Rightarrow\) 12x - 20y = 20000
(2) \(\times\) 37 \(\Rightarrow\) \(\cfrac { 12-21y=15000 }{ 1y\quad =\quad 5000 } \)
Substitute y = 5000 in (1)
3x - 5 (5000) = 5000
3x - 25000 = 5000
3x = 5000 + 25000
3x = 30000
x = 10000
\(\therefore\) Monthly income of A is 3x = 3\(\times\)1 0000 = Rs. 30000
Monthly income of B is 4x = 4 \(\times\) 10000 = Rs. 40000

Let Raman's age = x
Let the sum of his two sons age = y
now x = 3y \(\Rightarrow\)x - 3y = 0 .........(1)
After 5 years, x + 5 = 2(y + 10)
x + 5 = 2y+ 20
x - 2y = 20 - 5
x - 2y = 15    ...(2)  
Step (1): From equation (1) x = 3y
Step (2): Substitute x = 3y in (2)
3y- 2y = 15
y = 15
Step (3): Substitute y = 15 in (1)
x = 3y = 3 \(\times\) 15
x = 45
\(\therefore\) Raman's age is 45 years.

Let x, y be the speed of the two cars. If the two cars travel toward's each other they will meet in 1 hr. The distance between them d = 100; \(\frac { d }{ s } \) =t
i.e.,\(\cfrac { 100 }{ x+y } =1\Rightarrow x+y=100\)  .....(1)
If the two cars travel in the same direction they will meet in 2hrs.
    ...(2)
(1) \(\Rightarrow\) x + y = 100
y = -x + 100

(2) \(\Rightarrow\) x - y = 50
-y = -x + 50
y = x - 50

The points to be plotted
(-100, -50), (0, -50),
x + y =100 ...(1)
Put x = 0 in (1), then 0 + y = 100 \(\Rightarrow\)y = 100
A (0, 100) is a point on (1)
Put y = 0 in(1), then x + 0 = 100 \(\Rightarrow\) x = 100
B (100, 0) is another point on (1)
Plot A & B Join them to produce the line (1)
Similarly by x - y = 50
Put x = 0 in (2), then 0 - y = 50 
\(\Rightarrow\)y = 50
P (0, -50) is a point on (2)
Put y = 0 in (2), then x - 0 = 5
\(\Rightarrow\) x = 50
Q (50, 0) is another point on (2)
Plot P & Q Join them to produce the line (2)
The point of intersection (75, 25) of the two lines (1)& (2) is the solution.
\(\therefore\)The solution i.e., the speed of the two cars x and y is given by x = 75 km and y = 25 km

y = 5x + 2, y = 10 +2x

Given linear equations are
8x + 5y = 9
kx + 10y = 15
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = 8, b₁ = 5, c₁ = 9, a₂ = k, b₂ = 10, c₂ = 15
For no solution, we know that \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \) and so, \(\frac { 8 }{ k } =\frac { 5 }{ 10 } \neq \frac { 9 }{ 15 } \)
80 = 5k
k = 16

Given two linear equations are
2x - 3y = 7;
(k + 2)x - (2k + 1)y = 3(2k - 1)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = 2, b₁ = −3, a₂ = (k + 2), b₂ = −(2k +1), c₁ = 7, c₂ = 3(2k −1)
For infinite number of solution we consider \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\frac { 2 }{ k+2 } =\frac { -3 }{ -(2k+1) } =\frac { 7 }{ 3(2k-1) } \)
\(\frac { 2 }{ k+2 } =\frac { -3 }{ -(2k+1) } \)
2(2k +1) = 3(k + 2)
4k + 2 = 3k + 6
k = 4
\(\frac { -3 }{ -(2k+1) } =\frac { 7 }{ 392k-1) } \)
9(2k −1) = 7(2k +1)
18k − 9 = 14k + 7
4k = 16
k = 4

Given linear equations are
kx + 2y = 3 ......(1)
2x - 3y = 1  .......(2)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = k, b₁ = 2, a₂ = 2, b₂ = −3 ;
For unique solution we take \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \); therefore \(\frac { k }{ 2 } \neq \frac { 2 }{ -3 } \);\(k\neq \frac { 4 }{ -3 } \), that is \(k\neq -\frac { 4 }{ 3 } \)

The given system of equations are 3x + 5y − 21 = 0; −7x − 6y + 49 = 0
Now using the coefficients for cross multiplication, we get,

\(\Rightarrow \frac { x }{ (5)(49)-(-6)(-21) } =\frac { y }{ (-21)(-7)-(49)(3) } =\frac { 1 }{ (3)(-6)-(-7)(5) } \)
\(\frac { x }{ 119 } =\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow \frac { x }{ 119 } =\frac { 1 }{ 17 } ,\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow x=\frac { 119 }{ 17 } ,y=\frac { 0 }{ 17 } \)
\(\Rightarrow\) x = 7, y = 0
Verification :
3x + 5y = 21 ...(1)
3(7) + 5(0) = 21
21 + 0 = 21
21 = 21 True
-7x - 6y = -49 ...(2)
-7(7) - 6(0) = - 49
-49 = -49
-49 = -49 True

Given, 2x + 3y = 14 ...(1)
            3x − 4y = 4 ...(2)
To eliminate y:

Therefore, x = 4
Substitute x = 4 in (1) to get 2x + 3y = 14
2(4) + 3y = 14
8 + 3y = 14
y = 2
Thus the solution is x = 4, y = 2.
Verification :
2x+3y = 14 ...(1)
2(4)+3(2) = 14
8 + 6 = 14
14 = 14 True
3x – 4y = 4 ...(2)
3(4) – 4(2) = 4
12 – 8 = 4
4 = 4 True

5b = 75  which gives b = 15
Put b = 15 in (2):
a + 2(15) = 35 which simplifies to a = 5
Thus the solution is a = 5, b = 15.
Verification :
4a+3b = 65 ...(1)
4(5)+3(15) = 65
20 + 45 = 65
65 = 65 True
a + 2b = 35 ...(2)
5 + 2(15) = 35
5+30 = 35
35 = 35 True

Given x + 3y = 16 ... (1)
2x – y = 4 ... (2)

Given x – 2y = 7 …(1)
2x + 3y = 7 …(2)
When x = 5, y = −1 we get
From (1) x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1)
From (2) 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2)
Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1) is a solution of the given equations.

The given equation is 2y – 3x = 12
\(\Rightarrow\) 2y = +3x + 12
\(\Rightarrow \ \frac { 2y }{ 2 } =\frac { 3x+12 }{ 2 } \)
\(\Rightarrow \ y=\frac { 3x }{ 2 } +\frac { 12 }{ 2 } \)
\(\Rightarrow \ y=\frac { 3 }{ 2 } x+6\)
compare with, y=mx + c
Slope m = \(\frac { 3 }{ 2 } \), y-intercept c = 6

Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3 }{ 5 } \)
Slope of CD = \(\frac { change\ in\ y }{ change\ in\ x } =-\frac { 3 }{ 2 } \)
Slope of EF = \(\frac { 4 }{ 7 } \)
Slope of PQ = Undefined Slope of Rs. = 0.

The given system of equations are
3x − 4y = 10 \(\Rightarrow\) 3x − 4y −10 = 0 .....(1)
  4x + 3y = 5 \(\Rightarrow\) 4x + 3y − 5 = 0 .....(2)
For the cross multiplication method, we write the co-efficients as

\(\frac { x }{ (-4)(-5)-(3)(-10) } =\frac { y }{ (-10)(4)-(-5)(3) } =\frac { 1 }{ (3)(3)-(4)(-4) } \)
\(\frac { x }{ (20)-(-30) } =\frac { y }{ (-40)-(-15) } =\frac { 1 }{ (9)-(-16) } \)
\(\frac { x }{ 20+30 } =\frac { y }{ -40+15 } =\frac { 1 }{ 9+16 } \)
\(\frac { x }{ 50 } =\frac { y }{ -25 } =\frac { 1 }{ 25 } \)
Therefore, we get \(x=\frac { 50 }{ 25 } ;y=\frac { -25 }{ 25 } \)
x = 2; y = -1
Thus the solution is x = 2, y = –1.
Verification :
3x–4y = 10 ...(1)
3(2)–4(–1) = 10
6 + 4 = 10
10 = 10 True
4x + 3y = 5 ...(2)
4(2) + 3(–1) = 5
8–3 = 5
5 = 5 True

The given system of equations are 8x − 3y = 5xy ...(1)
6x − 5y = −2xy ...(2)
Observe that the given system is not linear because of the occurrence of xy term. Also note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other solution would have both x \(\neq \) 0 and y \(\neq \) 0
Let us take up the case where x \(\neq \) 0 and y \(\neq \) 0
Dividing both sides of each equation by xy,
\(\frac { 8x }{ xy } -\frac { 3y }{ xy } =\frac { 5xy }{ xy } \)
\(\frac { 6x }{ xy } -\frac { 5y }{ xy } =\frac { -2xy }{ xy } \)
Let \(a=\frac { 1 }{ x } ,b=\frac { 1 }{ y } \)
We get, \(\frac { 8 }{ y } -\frac { 3 }{ x } =5\)  .....(3)
\(\frac { 6 }{ y } -\frac { 5 }{ x } =-2\)  .......(4)
(3)&(4) respectively become, 8b − 3a = 5 ...(5)
b − 5a = −2 ...(6)
which are linear equations in a and b.
To eliminate a, we have,  (5) \(\times\) 5\(\Rightarrow\) 40b −15a = 25 .....(7)
(6) × 3\(\Rightarrow\) 18b −15a = −6 .....(8)
Now proceed as in the previous example to get the solution\(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \).
Thus, the system have two solutions \(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \) and (0,0).

Let x be the digit at ten’s place and y be the digit at unit place.
Given that x + y = 5 …… (1)

Given, Original number − reversing number = 27
(10x + y) − (10y + x) = 27
10x − x + y −10y = 27
9x − 9y = 27
\(\Rightarrow\) x − y = 3 ... (2)
Also from (1), y = 5 – x ... (3)
Substitute (3) in (2) to get x − (5 − x) = 3
x − 5 + x = 3
2x = 8
x = 4
Substituting x = 4 in (3), we get y = 5 − x = 5 − 4
y = 1
Thus, 10x + y = 10 × 4 +1 = 40 +1 = 41.
Therefore, the given two-digit number is 41.
Verification :
sum of the digits = 5
x + y = 5
4 + 1 = 5
5 = 5 true
Original number – reversed number = 27
41 - 14 = 27
27 = 27 true

Let us form equations for the given statement.
Let us consider l and b as the length and breadth of the rectangle respectively.
Now let us frame the equation for the first statement
Perimeter of rectangle = 36
2(l+b) = 36
\(l+b=\frac { 36 }{ 2 } \)
l = 18 - b  .....(1)

Points: (2,16), (4,14), (5,13), (8,10)
The second statement states that the length is 2 metres more than three times the width which is a straight line written as l = 3b + 2 ... (2)
Now we shall form table for the above equation (2).

Points: (2, 8), (4, 14), (5,17), (8, 26)
The solution is the point that is common to both the lines. Here we find it to be (4,14).
We can give the solution to be b = 4, l = 14.

Verification :
2(l + b) = 36 ...(1)
2(14 + 4) = 36
2 \(\times\) 18 = 36
36 = 36 true
l = 3b + 2 ...(2)
14 = 3(4) +2
14 = 12 + 2
14 = 14 true

Let us form table of values for each line and then fix the ordered pairs to be plotted.
Graph of 3x + 2y = 6

Points to be plotted:
(−2,6), (0,3), (2,0)
Graph of 6x + 4y = 8

Points to be plotted:
(-2, 5), (0, 2), (2, -1)
When we draw the graphs of these two equations, we find that they are parallel and they fail to meet to give a point of intersection. As a result there is no ordered pair that can be common to both the equations. In this case there is no solution to the system.

Given x + y = 5 ...(1)
2x – y = 4 ...(2)
To draw the graph (1) is very easy. We can find the x and y intercepts and thus two of the points on the line (1).
When x = 0, (1) gives y = 5.
Thus A(0,5) is a point on the line.
When y = 0, (1) gives x = 5.
Thus B(5,0) is another point on the line.
Plot A and B; join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, (2) gives y = −4.
Thus P(0,−4) is a point on the line.
When y = 0, (2) gives x = 2.
Thus Q(2,0) is another point on the line.
Plot P and Q; join them to produce the line (2).
The point of intersection (3, 2) of lines (1) and (2) is a solution.
The solution is the point that is common to both the lines. Here we find it to be (3,2). We can give the solution as x = 3 and y = 2.

We have already come across one method: forming a table of values, listing and plotting ordered pairs and joining the points.
But, to fix a line, after all, how many points do we need? Just two! These can easily be obtained when a line is given in the form y = mx + c.
The given line y = 4x – 3
put x = 0 to get y-intercept
y = 4(0)–3
y = –3
point is (0, –3) and y-intercept = – 3
put y = 0 to get x-intercept
0 = 4x – 3
3 = 4x
\(\frac { 3 }{ 4 } =x\)
point is \(\left( \frac { 3 }{ 4 } ,0 \right) \) and x-intercept = \(\frac { 3 }{ 4 } \)
The graph may be drawn through two points (0,−3) and \(\left( \frac { 3 }{ 4 } ,0 \right) \)