The condition for no solution,
\(\\ \\ \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \Rightarrow \frac { k }{ 12 } =\frac { 3 }{ k } \neq \frac { 1 }{ 2 } \)
When \(\frac { k }{ 12 } =\frac { 3 }{ k } ,we\quad get\quad { k }^{ 2 }=36\)
i.e.,  \(k=\pm 6\)
\(k\neq 6\)
so k = -6   \(\left( \because For\quad k=6,\frac { { a }_{ 2 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \right) \)

For equation, 2x + 3y - 4 = 0
a₁ = 2, b₁ = 3, C₁ = - 4
For equation, (k + 2) x + 6y - (3k + 2)=0
a₂ = k + 2, b₂ = 6, c₂ = - (3k + 2) 
For infinitely many solutions \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { 2 }{ k+2 } =\frac { 3 }{ 6 } =\frac { 4 }{ 3k+2 } \Rightarrow 12=3k+6\)
\(\Rightarrow\) 6 = 3k \(\Rightarrow\) k = 2

Given linear equation is 3x + 4y = 9
(i) Intersecting line is 3x - 5y = 10
(ii) Coincident line is 6x + 8y = 18.

The given equations can be re-written as :
5x-3y-11 = 0
- 10x + 6y + 22 = 0
On comparing with ax + by + c = 0, we have
a₁ = 5, b₁ = - 3, c₁ = - 11
a₂ = - 10, b₂ = 6, c₂ = 22
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 5 }{ -10 } =-\frac { 1 }{ 2 } \)
\(\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ 6 } =\frac { -1 }{ 2 } \)
and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -11 }{ 22 } =\frac { -1 }{ 2 } \)
Thus, \(\frac { -1 }{ 2 } =\frac { -1 }{ 2 } =\frac { -1 }{ 2 } \)
i.e. \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
Hence, the pair of linear equations is consistent.

The given equations can be re-written as
2x-3y-8 = 0
4x-6y-9 = 0
On comparing with ax + by + c = 0, we have
a₁ = 2, b₁ = - 3, C₁ = - 8
a₂ = 4, b₂ = - 6, c₂ = - 9
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } ,\)
\(\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ -6 } =\frac { 1 }{ 2 } \)
and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -8 }{ -9 } =\frac { 8 }{ 9 } \) 
Thus, \(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \neq \frac { 8 }{ 9 } \)
i.e. \(\\ \\ \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
Hence, the pair of linear equations is inconsistent.

(i) 5x-2y+8=0
(ii) 4x-10y-8=0
(iii) 2x+5y+10=0

14 yr, 20yr

90, 30

x=2, y=-3

x=1, y=2

x=4, y=-1

Given that, x, y ane 40⁰ are the angles of a triangle.
\(\therefore\) x+y+40⁰=180⁰  [\(\because\) sum of all the angles of a triangle is 180⁰ ]
\(\Rightarrow\)x+y=140⁰     ...(i)
Also, x-y=30⁰    ...(ii)
On adding Eqs. (i) and (ii), we get
2x=170⁰ \(\Rightarrow\)  x=85⁰
On putting  x=85⁰ in Eqs. (i), we get
85⁰+y=140⁰  \(\Rightarrow\)   y=55⁰
Hence, the required values of x and y are 85⁰ and 55⁰ respectively.

Let the number of students in halls A and B be x and y, respectively.
According to the question,
x-10=y+10 \(\Rightarrow\) x-y=20   ..(i)
and (x+20)=2(y-20) \(\Rightarrow\) x-2y=-60   ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
-y+2y=60+20 \(\Rightarrow\) y=80
On putting y=80 in Eq. (i), we get
x-80=20 \(\Rightarrow\) x=20+80 \(\Rightarrow\) x=100
Hence, there are 100 students in hall A and 80 students in hall B.                                                                                                                                                                        

Points of intersection of lines represented by
3x-2y=10  ..(i)
and 5x+y=8   ..(ii)
is the common solution of this system of equations.
From Eq.(ii), we have
y=8-5x    ...(iii)
On substituting the value of y in Eq.(i), we get
3x-2(8-5x)=10
\(\Rightarrow\) 13x=26
\(\Rightarrow\) x=2
On putting x=2 in Eq. (iii), we get
y=8-5 x 2=-2
So, the point of intersection of the lines (i) and (ii) is (2,-2).
so, this point will satisfy the equations 3x+ay=8.
\(\therefore\) (3 x 2)+[(-2) x a]=8
\(\Rightarrow\) 6-2a=8
\(\Rightarrow\) -2a=2 \(\Rightarrow\) a=-1
Hence, the required value of a is -1.

Given, a pair of linear equations is :
2x + 3y = 11     ......(i)
and 2x - 4y = - 24  ......(ii)
From eqn. (ii), 4y = 2x + 24
\(\\ \Rightarrow \quad y=\frac { x+12 }{ 2 } \)  ......(iii)
On substituting y from eqn. (iii) in eqn. (i), we get
2x + 3 \(\left( \frac { x+12 }{ 2 } \right) \)= 11
\(\Rightarrow\) 4x + 3(x + 12) = 11 \(\times\) 2
\(\Rightarrow\) 4x + 3x + 36 = 22
\(\Rightarrow\) 7x=22 - 36
\(\Rightarrow\) 7x = -14
\(\therefore\) x = -2
From eqn. (iii), \(y=\frac { -2+12 }{ 2 } \)
\(\therefore\)   y = 5
On substituting x = - 2 and y = 5 in the equation
y = mx + 3,
5 = m \(\times\) (- 2) + 3
\(\Rightarrow\) 5 = -2m + 3
\(\Rightarrow\) 2m  = 3 - 5 = -2
\(\therefore\)  m = -1