The given system of equations are 3x + 5y − 21 = 0; −7x − 6y + 49 = 0
Now using the coefficients for cross multiplication, we get,

\(\Rightarrow \frac { x }{ (5)(49)-(-6)(-21) } =\frac { y }{ (-21)(-7)-(49)(3) } =\frac { 1 }{ (3)(-6)-(-7)(5) } \)
\(\frac { x }{ 119 } =\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow \frac { x }{ 119 } =\frac { 1 }{ 17 } ,\frac { y }{ 0 } =\frac { 1 }{ 17 } \)
\(\Rightarrow x=\frac { 119 }{ 17 } ,y=\frac { 0 }{ 17 } \)
\(\Rightarrow\) x = 7, y = 0
Verification :
3x + 5y = 21 ...(1)
3(7) + 5(0) = 21
21 + 0 = 21
21 = 21 True
-7x - 6y = -49 ...(2)
-7(7) - 6(0) = - 49
-49 = -49
-49 = -49 True

Given, A = {2,5,6,7} and B = {3,5,7,8}
\(A\cup B\) = {2,3,5,6,7,8}  ...........(1)
\(B\cup A\) = {2,3,5,6,7,8} ..............(2)

(4m-3m)² = (4m)² - 2(4m) (3m) + (3m)² = 16m² - 24mn + 9n²

\(-2\sqrt [ 4 ]{ 768 } +5\sqrt [ 4 ]{ 1875 } +3\sqrt [ 4 ]{ 48 } \) = \(-2\sqrt [ 4 ]{ 256\times 3 } +5\sqrt [ 4 ]{ 625\times 3 } +3\sqrt [ 4 ]{ 16\times 3 } \)
=\(-2\sqrt [ 4 ]{ { 4 }^{ 4 }\times 3 } +5\sqrt [ 4 ]{ { 5 }^{ 4 }\times 3 } +3\sqrt [ 4 ]{ { 2 }^{ 4 }\times 3 } \)
=\(-2\times 4\sqrt [ 4 ]{ 3 } -5\times 5\sqrt [ 4 ]{ 3 } +3\times 2\sqrt [ 4 ]{ 3 } \)
=\(-8\sqrt [ 4 ]{ 3 } +25\sqrt [ 4 ]{ 3 } +6\sqrt [ 4 ]{ 3 } =23\sqrt [ 4 ]{ 3 } \)

(i)\(\angle\)CAB = 25⁰
\(\therefore\) \(\angle\)BDC = 25⁰
(ii) \(\angle\)DBA = \(\angle\)DCA =180-(90+25⁰)
= 180⁰ - 115⁰
= 65
(iii)\(\angle\)COB = 2\(\angle\)CAB = 2\(\times\)25⁰ = 50⁰

\(\sqrt [ 3 ]{ (1024)^{ -2 } } =\left[ \sqrt [ 3 ]{ ({ 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 3 }\times 2 } )^{ -2 } \right] \)
=\(\left[ \sqrt [ 3 ]{ ({ 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 3 }\times 2 } )^{ -2 } \right] \)  [Laws of radicals - (i)]
=\(\left[ \sqrt [ 3 ]{ 2^{ 3 } } \times \sqrt [ 3 ]{ { 2 }^{ 3 } } \times \sqrt [ 3 ]{ { 2 }^{ 3 } } \times \sqrt [ 3 ]{ { 2 } } \right] ^{ -2 }\) [Laws of radicals – (ii)]
= \(\left[ 2\times 2\times 2\times \sqrt [ 3 ]{ 2 } \right] ^{ -2 }\) [Laws of radicals – (i)]
=\(\left[ 8\times \sqrt [ 3 ]{ 2 } \right] ^{ -2 }=\left[ \frac { 1 }{ 8 } \right] ^{ 2 }\times \left( \frac { 1 }{ \sqrt [ 3 ]{ 2 } } \right) ^{ 2 }\)
=\(\frac { 1 }{ 64 } \sqrt [ 3 ]{ \frac { 1 }{ 4 } } \)
order = 3 ; radicand = \(\frac{1}{4}\); coefficient = \(\frac{1}{64}\)
(These results can also be obtained using index notation).

n(U) = 125
y = 2x
z = x + 10
ஃ x + y + z + 4 + 17 + 6 + 3 + 5 = 125
x + 2x + x + 10 + 35 = 125
4x + 45 = 125
4x = 125 - 45
4x = 80
x = 20
ஃ y = 2x = 2 \(\times\) 20 = 40
z = x +10 = 20 + 10 = 30
Hence x = 20; y = 40; z = 30

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac { { 360 }^{ 0 } }{ 30 } \)
=12⁰
3x =3 \(\times\) 12 = 36°
5x = 5 \(\times\) 12 = 60°
9x = 9 \(\times\) 12 = 108°
13x = 13 \(\times\) 12 = 156°
The required angles of quadrilateral are 36°, 60° 108° and 156°.

ㄥACD = ㄥA + ㄥB
(An exterior angle of a triangle is sum of its interior opposite angles)
120° = 50° + x⁰
x⁰ = 120° - 50°
= 70°
In the the triangle ABC
ㄥA + ㄥB + ㄥACB = 180° (Sum of the angles of a Δ)
50° +x + ㄥACB = 180°
50° + 70° + ㄥACB = 180°
ㄥACB = 180° - 120°
y = 60°    (OR)
ㄥACD + ㄥACB = 180⁰(Angles of a linear pair)
ㄥACB = 180° - 120°
= 60°
The value ofx = 70° andy = 60°.

P (- 4,4), Q (8, 2), R (6, -3) and S (-2, -2)

\(\frac { 23 }{ 10 } \)is 2.3 and\(\frac { 12 }{ 5 } \) is 2.4
You need an irrational number greater than 2.3 but less than 2.4
One such irrational number is
2.301001000100001000001000000100000001……..
observe that it is non-terminating and non-recurring.

Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AB =\(\sqrt { (-6+3)^{ 2 }+(-7-1)^{ 2 } } \)
= \(\sqrt { (-3)^{ 2 }+(-8)^{ 2 } } =\sqrt { 9+64 } =\sqrt { 73 } \)
BC = \(\sqrt { (3+6)^{ 2 }+(-9+7)^{ 2 } } \)
= \(\sqrt { { 9 }^{ 2 }+(-2)^{ 2 } } =\sqrt { 81+4 } =\sqrt { 85 } \)
CD = \(\sqrt { (6-3)^{ 2 }+(-1+9)^{ 2 } } \)
= \(\sqrt { (3)^{ 2 }+({ 8 })^{ 2 } } =\sqrt { 9+64 } =\sqrt { 73 } \)
AD = \(\sqrt { (6+3)^{ 2 }+(-1-1)^{ 2 } } \)
= \(\sqrt { (9)^{ 2 }+(-2)^{ 2 } } =\sqrt { 81+4 } =\sqrt { 85 } \)
AB = CD =\(\sqrt { 73 } \) and BC = AD = \(\sqrt { 85 } \)(Opposite sides sre equal)
∴ ABCD is a parallelogram.

Radius of the circle = 30 units. The point O is (0, 0).
Let a intersect the x - axis and b intersect the y - axis
∴ The point A is (a, 0) and B is (0, b)
Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
OA =\(\sqrt { (a-0)^{ 2 }+(0-0)^{ 2 } } \)
30 = \(\sqrt { a^{ 2 } } \)
Squaring on both sides
30² = a²
∴ a = 30
The point A is (30, 0)
OB = \(\sqrt { (0-0)^{ 2 }+(b-0)^{ 2 } } \)
= \(\sqrt { 0+{ b }^{ 2 } } \)
30 = \(\sqrt {b^{ 2 } } \)
Squaring on both sides
30² = b²
∴ b = 30
The point B is (0, 30)
Distance AB =\(\sqrt { (30-0)^{ 2 }+(0-30)^{ 2 } } \)
= \(\sqrt { 30^{ 2 }+30^{ 2 } } =\sqrt { 900+900 } \)
= \(\sqrt { 1800 } =\sqrt { 2\times 900 } =30\sqrt { 2 } \)
∴  Distance between the two points = 30\(\sqrt { 2 } \)

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°

P (x) = 4x³ + 6x² + 7x + 2
d(x) = x - 2
Standard form of p (x) 4x³ + 6x² + 7x + 2 and
d(x) x - 2

4x² + 6x² + 7x + 2 = (x - 2)(4x³ + 14x + 35) + 72
Hence the quotient is 4x³ + 14x + 35 and remainder is 72

Distance = \(\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(6-3)^2+(5-7)^2}\)
\(\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}\)
BC=\(\sqrt{(15-6)^2+(-1-5)^2}=\sqrt{(9)^2+(-6)^2}\)
\(\sqrt{81+36}=\sqrt{117}\)
\(\sqrt{9\times 13}-3\sqrt{13}\)
AC =\(\sqrt{(15-3)^2+(-1-7)^2}\)
\(\sqrt{12^2+(-8)^2}=\sqrt{144+64}\)
\(=\sqrt{208}=\sqrt{16\times 13}=4\sqrt{13}\)

AB + BC =AC\(\Rightarrow \sqrt{13}+3\sqrt{13}=4\sqrt{13}\) 
\(\therefore\)The points A,B,C are collinear.

\({6\over 7}=0.\overline {857142}\)
\({12\over 13}=0.\overline{923076}\)

The two irrational numbers are  0.866142 ...and 0.903076.......

\(\angle A+\angle B=\angle ACD\) (An exterior angle of a triangle is sum of its interior opposite angles)
x + 35 + 2x - 5 = 4x - 15
3x + 30 = 4x- 15
30+15 = 4x-3x
45° = x
\(\angle A=x+35^0\)
= 45°+ 35°
= 80°
\(\angle B=2x-5\)
= 2(45°) - 5°
= 90° - 5° = 85°
\(\angle ACD=4x-15\)
= 4(45°)-15°
= 180°-15° = 165°
\(\angle ACB=180^0-\angle ACD\)
=180° - 165° = 15°
\(\angle A=80^0,\angle B=85^0,\angle C=15^0\).

\(G\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) =\left( \frac { 2+5+9 }{ 3 } ,\frac { -5+11+9 }{ 9 } \right) =\left( \frac { 16 }{ 3 } ,5 \right) \)

2x- 3y = 7
(s + 2) x - (2s + 1)y == 3 (2s - 1)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } =\cfrac { 7 }{ 3(2s-1) } \)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } \)
2(2s + 1 ) = 3 (s + 2)
4s + 2 = 3s + 6
4s - 3s = 6- 2
s = 4

"The centroid of the triangle obtained by joining the mid points of the sides of a triangle is the same as the centroid of the original triangle."
\(\therefore\) The mid points of the sides of the triangle are given as
  x₁y_{1 ,} x₂y, x₃y₃
\(\left( \frac { 3 }{ 2 } ,5 \right) \left( 7,\frac { -9 }{ 2 } \right) \left( \frac { 13 }{ 2 } ,\frac { -13 }{ 2 } \right) \)
\(\therefore\)Centroid \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 3 }{ 2 } +\frac { 7 }{ 1 } +\frac { 13 }{ 2 } }{ 3 } ,\frac { 5+\left( \frac { -9 }{ 2 } \right) +\frac { (-13) }{ 2 } }{ 3 } \right) =\left( \frac { \frac { 3+14+13 }{ 2 } }{ 3 } ,\frac { \frac { 10-9-13 }{ 2 } }{ 3 } \right) \)
\(=\left( \frac { \frac { 30 }{ 2 } }{ 3 } ,\frac { \frac { -12 }{ 2 } }{ 3 } \right) =\left( \frac { 15 }{ 3 } ,\frac { -6 }{ 3 } \right) =(5,-2)\)

\(\frac { 1 }{ 4 } \) = 0.25 and \(\frac { 1 }{ 3 } \) = 0.3333….. = \(0.\bar3\)
In between 0.25 and \(0.\bar3\), there are infinitely many irrational numbers.
Four irrational numbers between 0.25 and \(0.\bar3\) are
0.2601001000100001…..
0.2701001000100001…..
0.2801001000100001…..
0.3101001000100001…..

\(M(x,y)=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((-4,2)=\left( \frac { -3+{ x }_{ 2 } }{ 2 } ,\frac { 7+{ y }_{ 2 } }{ 2 } \right) \left( { x }_{ 1 }{ y }_{ 1 } \right) \)
\(\frac { -3+{ x }_{ 2 } }{ 2 } =-4\quad \frac { 7+{ y }_{ 2 } }{ 2 } =2\)
-3 + x₂ = -8
7 +y₂ = 4
x₂ = -8 +3
y₂ = 4 -7 = -3
x₂ = -5
The other end is (-5, -3)

\(\left( 5\sqrt { 125 } \right) ^{ 7 }\) = \({ 125 }^{ \frac { 7 }{ 5 } }\)

\(\sqrt { 27 } \) =\(\sqrt { 3 } \times \sqrt { 3 } \times \sqrt { 3 } =\left( { 3 }^{ \cfrac { 1 }{ 2 } } \right) ^{ 3 }=\left( 3 \right) ^{ \left( \cfrac { 3 }{ 2 } \right) }\)

\(\angle \)QPR = \(\cfrac { 1 }{ 2 } \) \(\angle \)QPR =\(\cfrac { 1 }{ 2 } \) X 70^o = 35^o
In\(\triangle\)QPR
\(\angle \)R + \(\angle \)P + \(\angle \)Q = 180^o
75^o + 35^o +\(\angle \)Q =180^o
\(\angle \)Q =180^o-110^o = 70^o
\(\angle \)Q = x^o + 55^o = 70^o
x^o = 70^o - 55^o = 15^o

The distance between the two chord
FE = OF + OE
OE = \(\sqrt { { 5 }^{ 2 }-3^{ 2 } } =\sqrt { 25-9 } \)
= \(\sqrt { 16 } \) = 4 cm
OF = \(\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } =\sqrt { 25-16 } =\sqrt { 9 } \) = 3 cm
\(\therefore\) DE = 4cm + 3cm = 7cm
Distance between the chord is 7 cm.

We know that (a+b)³ ≡ a³+3a²b+3ab²+b³
(2a+3b)³ = (2a)³+3(2a)²(3b)+3(2a)(3b)²+(3b)³
= 8a³+36a²b+54ab²+27b³

22

{ }

p(x) = x² - \(2\sqrt { 2 } +1\)
\(p\left( 2\sqrt { 2 } \right) ={ \left( 2\sqrt { 2 } \right) }^{ 2 }-2\sqrt { 2 } \left( 2\sqrt { 2 } \right) +1\)
= 8 - 8 + 1
= 0 + 1 = 1

It is a trapezium