Given pair of linear equations is 8x + 5y = 11 ...(i)
and x + y = 4  ...(ii)
On multiplying Eq. (ii) by 8 and then subtracting from Eq. (i), we get
\(-3 y=-21 \Rightarrow y=7\) 
Putting y = 7in Eq. (ii), we get
x=4-7=-3
Hence, x = - 3, y = 7.

2(3x - y) = 5xy         .....(i)
2(x + 3y) = 5xy       .....(ii)
Divide eqns. (i) and (ii) by xy,
\(\frac { 6 }{ y } -\frac { 2 }{ x } =5\)          .....(iii)
and  \(\frac { 2 }{ y } +\frac { 6 }{ x } =5\)  ......(iv)  
Let \(\frac { 1 }{ y } \) = a and \(\frac { 1 }{ x } \) = b,
then equations (iii) and (iv) become
6a - 2b = 5              ...(v)
2a + 6b = 5            ....(vi)
Multiplying eqn. (v) by 3 and then adding with eqn. (vi),
20a = 20
\(\therefore\) a = 1
Substituting this value of a in eqn. (v),
b = \(\frac { 1 }{ 2 } \)
Now  \(\frac { 1 }{ y } =a=1\)
\(\Rightarrow\) y = 1
and \(\frac { 1 }{ x } =b=\frac { 1 }{ 2 } \)
\(\Rightarrow\) x = 2

8

\(\frac { 2 }{ x+1 } +\frac { 3 }{ 2(x-2) } =\frac { 23 }{ 5x } \)
\(\Rightarrow \frac { 4(x-2)+3(x+1) }{ 2(x+1)(x-2) } =\frac { 23 }{ 5x } \)
\(\Rightarrow \frac { 4x-8+3x+3 }{ 2(x^{ 2 }-2x+x-2) } =\frac { 23 }{ 5x } \)
\(\Rightarrow \frac { 7x-5 }{ 2(x^{ 2 }-x-2) } =\frac { 23 }{ 5x } \)
\(\Rightarrow 5x(7x-5)=46(x^{ 2 }-x-2)\)
\(\Rightarrow 35x^{ 2 }-25x=46x^{ 2 }-46x-92=0\)
\(\Rightarrow 35x^{ 2 }-25x-46x^{ 2 }+46x+92=0\)
\(\Rightarrow -11x^{ 2 }+21x+92=0\)
\(\Rightarrow 11x^{ 2 }-44x+23x-92=0\)
11x(x-4)+23(x-4)=0
\(\Rightarrow (x-4)=0\quad (11x+23)=0\)
\(\Rightarrow x=4\quad \) or 11x=-23 or x=\(\frac { -23 }{ 11 } \)

\(4x^{ 2 }+4bx+b^{ 2 }-a^{ 2 }\)
\(\Rightarrow (2x+b)^{ 2 }-a^{ 2 }=0\)
\(\Rightarrow (2x+b+a)(2x+b-a)=0\)
\(\Rightarrow x=\frac { \left( a+b \right) }{ 2 } ,x=\frac { a-b }{ 2 } \)

Here a₁ = 2, b₁ = 1, c₁ = 3
and a₂ = 4, b₂ = 2, c₂ = 6
Clearly \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
i.e.\(\frac { 2 }{ 4 } =\frac { 1 }{ 2 } =\frac { 3 }{ 6 } \)
Hence lines are coincident.

(i) k\(\neq\)10
(ii) k=10.
There is no value of k for which given system has infinitely many solutions. i.e, represent coincident lines.

Since x=2 is a root of given quadratic equation 
\(\therefore\) 3(2)²-P(2)-2=0
\(\Rightarrow\)  12-2p-2=0
\(\Rightarrow\)  p=5
Hence, the value of p is 5.

Let first number be x.
Then, second number =x + 4
[∴ difference of two numbers = 4]
According to the question,
\(\frac{1}{x}-\frac{1}{x+4}=\frac{4}{21}\)
- 7, - 3 or 3, 7

For all real values k.

\(\frac { x }{ -80-48 } =\frac { y }{ -4+100 } =\frac { 1 }{ -60-4 } \)
\(\Rightarrow\)  x = 2
and  \(y=\frac { -3 }{ 2 } \)
Alternative method:
Given: 5x + 4y - 4 = 0   .....(i)
x - 12y - 20 = 0            ......(ii)
By cross-multiplication method,
\(\frac { x }{ \left| \begin{matrix} 4 & -4 \\ -12 & -20 \end{matrix} \right| } =\frac { y }{ \left| \begin{matrix} -4 & 5 \\ -20 & 1 \end{matrix} \right| } =\frac { 1 }{ \left| \begin{matrix} 5 & 4 \\ 1 & -12 \end{matrix} \right| } \)
\(\Rightarrow \quad \frac { x }{ -80-48 } =\frac { y }{ -4+100 } =\frac { 1 }{ -60-4 } \)
\(\frac { x }{ -128 } =\frac { 1 }{ -64 } \) and \(\frac { y }{ 96 } =\frac { 1 }{ -64 } \)
\(\Rightarrow \quad x=2\) and \(y=\frac { -3 }{ 2 } \)  

\(\frac { x^{ 2 }+3x+2+x^{ 2 }-3x+2 }{ x^{ 2 }+x-2 } =\frac { 4x-8-2x-3 }{ x-2 } \)
\((2x^{ 2 }+4)(x-2)=(2x-11)(x^{ 2 }+x-2)\)

\(\Rightarrow 5x^{ 2 }+19x-30=0\)
\(\Rightarrow (5x-6)(x+5)=0\)
\(\Rightarrow x=-5,\frac { 6 }{ 5 } \)

The pair of linear equations is :
x + y = 5
\(\Rightarrow\) x + y - 5 = 0       ....(i)
and 2x + 2y = 10
\(\Rightarrow\) 2x + 2y -10 = 0   ...(ii)
On comparing with ax + by + C = 0 , we have
a₁ = 1, b₁ = 1,.c₁ = - 5
a₂ = 2, b₂ = 2, c₂ = -10
\(\therefore \quad \frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ 2 } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { 1 }{ 2 } \)
and  \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -5 }{ -10 } =\frac { 1 }{ 2 } \)
Since \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \) 
So, the pair of linear equations is coincident having many solutions. Thus, the equation is consistent. Now, we need to solve it graphically.
We have x + y = 5
\(\Rightarrow\) y = 5 - x

and 2x + 2y = 10
\(\Rightarrow \quad y=\frac { 10-2x }{ 2 } \)

On plotting these points, we observe that the lines are coincident having infinite solutions.

The given equation is \(\sqrt { 3 } { x }^{ 2 }+11x+6\sqrt { 3 } =0\)
On comparing with \({ ax }^{ 2 }+bx+c=0\)
\(a=\sqrt { 3 } ,b=11\quad c=6\sqrt { 3 } \)
On substituting the values of a,b and c in the quadratic formula.
\(x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ x=\frac { -11\pm \sqrt { { \left( 11 \right) }^{ 2 }-4\left( \sqrt { 3 } \right) \left( 6\sqrt { 3 } \right) } }{ 2(\sqrt { 3 } ) } \\ x=\frac { -11\pm \sqrt { 121-72 } }{ 2\sqrt { 3 } } \\ =\frac { -11\pm \sqrt { 49 } }{ 2\sqrt { 3 } } =\frac { -11\pm 7 }{ 2\sqrt { 3 } } \\ x=\frac { -11+7 }{ 2\sqrt { 3 } } =\frac { -4 }{ 2\sqrt { 3 } } =\frac { -2 }{ \sqrt { 3 } } \\ x=\frac { -11-7 }{ 2\sqrt { 3 } } =\frac { -18 }{ 2\sqrt { 3 } } =\frac { -9 }{ \sqrt { 3 } } \)
Hence, \(\frac { -2 }{ \sqrt { 3 } } \) and \(\frac { -9 }{ \sqrt { 3 } } \) are the required solutions of the given equation.

(i) From the given figure, we see that area of two bedrooms
= 2(5x) = 10x m² 
\(\therefore\) Area of kitchen= 5 \(\times\) y = 5y m²
According to the question,
Area of the two bedrooms and area of kitchen =95m²
\(\therefore\) 10x + 5y =95
\(\Rightarrow\)2x + y =19 (dividing both sidesby 5] ...(i)
Also, length of the home =15 m
\(\therefore\) x+2 + y =15
\(\Rightarrow\) x + y = 13    ....(ii)
Hence, pair of linear equations is
2x + y = 19 and x + y=13
(ii) The length of the outer boundary of the layout
= 2(l + b) =2(15 + 12) = 2(27) = 54 m
(iii) On solving Eqs. (i) and (ii), we get 
x = 6 and y = 7
\(\therefore\) Area of each bedroom =5 \(\times\) x =5 \(\times\)6=30 m²
and area of kitchen=5 \(\times\) y = 5 \(\times\) 7 = 35 m²
(iv) \(\therefore\) Area of living room = 15 \(\times\) (5 + 2) - Area of bedroom 2
= 15 \(\times\)7 - 5 \(\times\)6
= 105 - 30 = 75 m²
(v) Since, area of kitchen = 5 \(\times\) y = 5 \(\times\) 7 = 35 m²
But, it is also given, the cost of laying tiles in kitchen at the rate of Rs 50 per m².
\(\therefore\) Total cost of laying tiles in the kitchen = 35 \(\times\) 50 = Rs 1750

(i) (b):We have \(6 x^{2}+x-2=0\)
\(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \)
\(\Rightarrow \quad(3 x+2)(2 x-1)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\)
(ii) (c): \(2 x^{2}+x-300=0\)
\(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \)
\(\Rightarrow \quad(x-12)(2 x+25)=0 \)
\(\Rightarrow \quad x=12, \frac{-25}{2}\)
(iii) (d): \(x^{2}-8 x+16=0\)
\(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\)
(iv) (d): \(6 x^{2}-13 x+5=0\)
\(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \)
\(\Rightarrow \quad(2 x-1)(3 x-5)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\)
(v) (a): \(100 x^{2}-20 x+1=0\)
\(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)
 

(i) (a) : Suppose two cars meet at point Q. Then,
Distance travelled by car X = A Q,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\(\therefore\) Distance travelled by car X in 9 hours = 9x km
\(\Rightarrow\) AQ=9x
Distance travelled by car Y in 9 hours = 9y km
\(\Rightarrow\) BQ=9y

Clearly, AQ - BQ = AB
\(\Rightarrow\) 9x - 9y = 90
\(\Rightarrow\) x-y =10
(ii) (c): Suppose two cars meet at point P. Then
Distance travelled by car X = AP and
Distance travelled by car Y = BP.
In this case, two cars meet in 917 hours.
\(\therefore\) Distance travelled by car X in 9/7 hours = \(\frac {9}{7}\) x km
\(\Rightarrow A P=\frac{9}{7} x\)
Distance travelled by car Y in 9/7 hours \(\frac {9}{7}\) y km
\(\Rightarrow B P=\frac{9}{7} y\)
Clearly, AP + BP = AB
\(\Rightarrow \quad \frac{9}{7} x+\frac{9}{7} y=90 \Rightarrow \frac{9}{7}(x+y)=90 \Rightarrow x+y=70\)
(iii) (b): We have x - y = 10
\(\Rightarrow x+y=70\)
Adding equations (i) and (ii), we get
2x = 80 \(\Rightarrow\) x = 40
Hence, speed of car X is 40 km/hr.
(iv) (c): We have x - y = 10
\(\Rightarrow\) 40 - Y = 10 \(\Rightarrow\) Y = 30
Hence, speed of car y is 30 km/hr.
(v) (d)