(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 \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

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

From (1) and (2), we get A -(BUC) = (A-B)∩(A-C). Hence it is verified.

(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)

\(\frac { \sqrt { 7 } -2 }{ \sqrt { 7 } +2 } =a\sqrt{7}+b\)
L.H.S = \(\frac{\sqrt{7}-2\times\sqrt{7}-2}{\sqrt{7}+2\times\sqrt{7}-2}=\frac{(\sqrt{7}-2)^2}{\sqrt{7}^2-2^2}=\frac{\sqrt{7}^2-2\sqrt{7}\times2+2^2}{7-4}\)
=\(\frac{7-4\sqrt{7}+4}{3}=\frac{11-4\sqrt{7}}{3}=\frac{11}{3}-\frac{4\sqrt{7}}{3}\)
\(\frac{-4\sqrt{7}}{3}+\frac{11}{3}=a\sqrt{7}+b\)
ஃa √7=\(\frac{-4\sqrt{7}}{3}\)
a = \(\frac{-4}{3}\)
b = \(\frac{11}{3}\)

A = {11,13,14,15,16,18}
B = {11,12,15,16,17,19}, C = {13,15,16,17,18,20}
(BUC) = {11,12,13,15,16,17,18,19,20}
A⋂(BUC) = {11,13,14,15,16,18}⋂{11,12,13,15,16,17,18,19,20}
= {11,13,15,16,18}...(1)
A⋂B = {11,15,16}; A⋂C = {13,15,16,18}
(A⋂B)U(A⋂C) = {11,13,15,16,18}...(2)
From (1) and (2) it is verified that A⋂(BUC) = (A⋂B)U(A⋂C).

From (1) and (2), \(A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\) is verified.

f(x) = x³ + 3x² + 3x + 1
f(–1) = (–1)³+3(–1)²+3(–1)+1
= –1+3–3+1 = 0
Hence, the remainder is 0
Hint:
g(x) = x+1
g(x) = 0
x +1 = 0
x = –1

p(x) = x²⁰¹⁸ + 2018
When it is divided by x - 1,
p( 1) = 1²⁰¹⁸+ 2018
= 1 + 2018
= 2019
The Remainder : 2019

\(\therefore\) The Quotient: 8z²–6z+2
The Remainder: 10

 \(\therefore\) The Quotient: 4x²–6x–5
The Remainder : 33

f(x) = x³ + 1, g(x) = x+1

Quotient = x² – x + 1 and Remainder = 0

(5x²-7x+2) ÷ (x-1)

∴ Quotient = 5x–2 Remainder = 0
(i) \(\frac{5x^{2}}{x}=5x\)
(ii) 5x(x-1)=5x²-5x
(iii) \(-\frac{2x}{x}=-2\)
(iv) -2(x-1)=(-2x+2)

Let x = -21.213777 ...... \(\rightarrow\) (1)
Here period of decimal is 4,multiply equation (1) by 10000
10000 x = -21213, 7.777 .............. (2)
(2) - (1)
9999 x = 212116.5640
\( x =\frac{-212116.5640}{9999} \)
\( x =\frac{-2121165640}{99990000} \)
\( =\frac{-190924}{9000}=\frac{-47731}{2250} \)

2x⁴ + 4x² - 3x + 7 - (3x³ - x² + 2x + 1)
= 2x⁴ + 4x² - 3x + 7 - 3x³ + x² - 2x - 1
= 2x⁴ - 3x³ + 4x² + x2 - 3x - 2x + 7 - 1
= 2x⁴ - 3x³ + 5x² - 5x + 6
2x⁴ - 3x³ + 5x² - 5x + 6 must be subtracted to get 3x³ - x² + 2x + 1.

h(z) - f(z) = z⁵ - 6z⁴ + z - (6z² + 10z -7)
= z⁵ -6z⁴ + z - 6z² - 10z + 7
= z⁵ -6z⁴ -6z² + z - 10z + 7
= z⁵– 6z⁴– 6z²– 9z+ 7
The degree of the polynomial is 5

p(x) - q(x) = 7x² + 6x - 1 - (6x - 9)
= 7x² + 6x - 1 - 6x + 9
= 7x² + 6x - 6x - 1 + 9
= 7x²+8
The degree of the polynomial is 2

The standard form is \({ 9y }^{ 4 }+\sqrt { 5 } { y }^{ 3 }+{ y }^{ 2 }-\frac { 7 }{ 3 } y-11\) or \(-11-\frac { 7 }{ 3 } y+{ y }^{ 2 }+\sqrt { 5 } { y }^{ 3 }+{ 9y }^{ 4 }\)

The standard form is \(-\frac { 7 }{ 2 } { x }^{ 4 }-5{ x }^{ 3 }+\sqrt { 2 } { x }^{ 2 }+x\) or \(x+\sqrt { 2 } { x }^{ 2 }-{ 5x }^{ 3 }-\frac { 7 }{ 2 } { x }^{ 4 }\)

In the given figure AC and BD bisect each other at O.
\(\therefore\) OA = OC (Given) ; OB = OD (Given)
\(\angle AOB=\angle COD\) (vertically opposite angles)
By SAS congruency
\(\triangle AOB\cong\triangle OCD\)

Using the distance formula, we have
\(AB=\sqrt { \left( 6-3 \right) ^{ 2 }+\left( 4-1 \right) ^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
\(BC=\sqrt { \left( 8-6 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
\(AC=\sqrt { \left( 8-3 \right) ^{ 2 }+\left( 6-1 \right) ^{ 2 } } =\sqrt { 25+25 } =\sqrt { 50 } =5\sqrt { 2 } \)
\(AB+BC=3\sqrt { 2 } +3\sqrt { 2 } =5\sqrt { 2 } =AC\)
Therefore the points lie on a straight line.

Draw TE || AB.
\(\angle ABT+\angle ETB=180^0(AB||TE)\)
\(48^0+\angle ETB=180^0\)
\(\angle ETB=180^0-48^0=132^0\)
Similarly \(\angle CDT+\angle DTE=180^0\)
\(24^0+\angle DTE=180^0\)
\(\therefore \angle DTE=180^0-24^0\) = 156⁰
\(\therefore \angle BTE+\angle ETD=132^0+156^0\) = 288°
x = 288°

We see that p(x) + q(x) is also a polynomial. Hence the sum of any two polynomials is also a polynomial