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

(i) 8x- 3y = 12      ...(1)
5x-2y = 7      ..(2)
8x- 3y-12 = 0
5x-2y- 7 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ (-3)(-7)(-2)(-12) } =\cfrac { y }{ (-12)(5)-(-7)(8) } =\cfrac { 1 }{ (8)(-2)-(5)(-3) } \)
\(\cfrac { x }{ 21-24 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\cfrac { x }{ -3 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\therefore \ \cfrac { x }{ 3 } =\cfrac { 1 }{ -1 } \ \cfrac { y }{ -4 } =\cfrac { 1 }{ -1 } \)
x = 3 , y = 4
\(\therefore\)  Solutions: x = 3; y = 4
(ii) 6x+ 7y-11 = 0
5x+ 2y-13 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { x }{ -91-(-22) } =\cfrac { y }{ -55-(-68) } =\cfrac { 1 }{ 12-35 } \)
\(\cfrac { x }{ -91+22 } =\cfrac { y }{ -55+78 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { 1 }{ -23 } \quad \cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)

x = 3, y= -1
\(\therefore\)  x = 3; y= -1
(iii) 

\(\cfrac { 2 }{ x } +\cfrac { 3 }{ y } -5=0\)
\(\cfrac { 3 }{ x } -\cfrac { 1 }{ y } +9=0\)
In (1), (2) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(1) \(\Rightarrow\) 2a + 3b - 5 = 0
(2) \(\Rightarrow\) 3a - b + 9 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ (3)(9)-(-1)(-5) } =\cfrac { b }{ (-5)(3)-(9)(2) } =\cfrac { 1 }{ (2)(-1)-(3)(3) } \)
\(\cfrac { a }{ 27-5 } =\cfrac { b }{ -15-18 } =\cfrac { 1 }{ -2-9 } \)
\(\cfrac { a }{ 22 } =\cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)
\(\therefore \ \cfrac { a }{ 22 } =\cfrac { 1 }{ -11 } \cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)

a = -2    b = 3
\(a=\cfrac { 1 }{ x } =-2\quad b=\cfrac { 1 }{ y } =3\)
\(\therefore \ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)
solution \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

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 – 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.

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.

(i) Let p(x) = x³ - 7x + 13x - 7
Sum of coefficients = 1 − 7 + 13 − 7 = 0
Thus (x -1) is a factor of p (x)
(ii) Let q(x) = x³ + 7x² + 13x + 7
Sum of coefficients of even powers of x with constant = 7 + 7 = 14
Sum of coefficients of odd powers of x = 1 + 13 = 14
Hence, (x + 1) is a factor of q (x)

Given, \(\left( y-\frac { 1 }{ y } \right) ^{ 3 }\)= 729
Take cube root on both sides
\( \sqrt [ 3 ]{ \left( \left( y-\frac { 1 }{ y } \right) ^{ 3 } \right) } =\sqrt [ 3 ]{ 729 } =\sqrt [ 3 ]{ 9^{ 3 } } \)
Therefore, \(y-\frac { 1 }{ y } \) = 9
\(\left( y-\frac { 1 }{ y } \right) ^{ 3 }=\left( y-\frac { 1 }{ y } \right) ^{ 3 }+3\left( y-\frac { 1 }{ y } \right) \) [∵ a³-b³ = (a-b)³+3ab(a-b]
= 9³ + 3(9)
\({ y }^{ 3 }-\frac { 1 }{ { y }^{ 3 } } \)= 756

We have
f(x) = x³– ax² + 6x – a
f(a) = a³- a(a)²+ 6a - a
=a³- a³+ 5a = 5a
Hence the required remainder is 5a
Hint:
Let g(x) = x–a
g(x) = 0
x – a = 0
x = a

\(\therefore f(\frac{2}{3}) =(\frac{2}{3})^{3}-\frac{2}{3}-1\)
\(=\frac{8}{27}-\frac{2}{3}+1\)
=\(\frac{8-18+27}{27}=\frac{17}{27}\ne0\)
⇒ f(x) is not multiple of g(x)
Hint:
g(x) = 2–3x = 0
gives x = \(\frac{2}{3}\)