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

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

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

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 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

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