\(k=-17;(10,\frac{1}{17}) and(18,1)\)

The given equation is x - 2y = 4
Put \(x=\sqrt{2}\), y=\(4\sqrt{2}\) in (1), we get
x-2y =\(\sqrt{2}-2(4\sqrt{2})\)
\(=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\) which is not 4.
\((\sqrt{2},\ 4\sqrt{2})\) which is not 4.
\((\sqrt{2}, 4\sqrt{2})\) is not a soluton of (1)

-2x+3y=6
Comparing with ax+by+c=0, we get
a=2, b=3, c=-6

\((x-1)(x+1)(x^2+3x+3)\)

\({ \left[ \frac { 3 }{ 2 } x+1 \right] }^{ 3 }\)
\(={ \left( \frac { 3 }{ 2 } x \right) }^{ 3 }+{ (1) }^{ 3 }+3\left( \frac { 3 }{ 2 } x \right) (1)\left( \frac { 3 }{ 2 } x+1 \right) \) | Using Identity VI
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 9 }{ 2 } x\left( \frac { 3 }{ 2 } x+1 \right) \)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x\)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x+1\)

\(p\left( \frac { 1 }{ 2 } \right) =2\left( \frac { 1 }{ 2 } \right) +1=1+1=2\neq 0\)
\(\therefore \frac { 1 }{ 2 } \)is not a zero of p(x).

\(\frac { 114-41\sqrt { 6 } }{ 30 } \)

\(\frac { 291 }{ 99 } \)

\(\frac { 3 }{ 5 } =\frac { 3\times 10 }{ 5\times 10 } =\frac { 30 }{ 50 } \)
\(\\ \frac { 4 }{ 5 } =\frac { 4\times 10 }{ 5\times 10 } =\frac { 40 }{ 50 } \)
\(\\ \because 30<31<32<33<34<35\)
\(\\ \therefore \frac { 30 }{ 50 } <\frac { 31 }{ 50 } <\frac { 32 }{ 50 } <\frac { 33 }{ 50 } <\frac { 34 }{ 50 } <\frac { 35 }{ 50 } \)
Therefore, five rational numbers between \(\frac { 3 }{ 4 } \) and \(\frac { 4 }{ 5 } \) and be taken as 
\(\frac { 31 }{ 50 } ,\frac { 32 }{ 50 } ,\frac { 33 }{ 50 } ,\frac { 34 }{ 50 } \)and \(\frac { 35 }{ 50 } \)
i.e., \(\frac { 31 }{ 50 } ,\frac { 16 }{ 25 } ,\frac { 33 }{ 50 } ,\frac { 17 }{ 25 } \) and \(\frac { 7 }{ 10 } \) 

105 x 106 = (100 + 5) x (100 + 6)
= (100)2 + (5 + 6) (100) + (5 x 6), using Identity IV
= 10000 + 1100 + 30
= 11130

(i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0.
(ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0.
(iii) 2x = 3 can be written as 2x + 0.y – 3 = 0.
(iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.

2x+3y-4.37=0;
a=2, b=3 and c=-4.37

\(-\frac{3}{2}\)

\(93-24\sqrt { 15 } \)

63^-3

Since we do not know what 0 3 . is , let us call it ‘x’ and so
x = 0.3333
Now here is where the trick comes in. Look at
10 x = 10 x (0.333...) = 3.333
Now, 3.3333 = 3 + x, since x = 0.3333
Therefore, 10 x = 3 + x
Solving for x, we get
\(9 x=3, \text { i.e., } x=\frac{1}{3}\)

4-k(1)=2  ⇒ k=2

x=0, y=4 satisfies y-4=0

Evident

x+y+0=0

definition

\(x-2=0\quad \Rightarrow \quad x=2\)
\(f(x)=5x^2-kx-18\)
\(f(2)=0\)
\(\Rightarrow \ 5(2)^2-k(2)-18=0\)
\(\Rightarrow \ k=1\)

\((x-6)(x-5)=0\quad \Rightarrow\quad\quad x=6, 5\)

Highest power of x=4

Degree of \(x^3+3x^2-4x+3\)​​​​​​​ is 3

\(\frac{1}{y}=y^-1\) has negative exponent so, not a polynomial

(i) 2x + y = 7
When x = 0, 2(0) + y = 7
⇒ 0+ y = 7
⇒ y = 7
∴ Solution is (0, 7).
When x = 1, 2(1) + y = 7
⇒ y = 7 - 2
⇒ y = 5
∴ Solution is (1, 5).
When x = 2, 2(2) + y = 7
⇒ y = 7 - 4
⇒ y = 3
∴ Solution is (2, 3).
When x = 3, 2(3) + y = 7
⇒ y = 7 - 6
⇒ y = 1
∴ Solution is (3, 1).
(ii) πx + y = 9
When x = 0, π(0) + y = 9
⇒ y = 9 - 0 
⇒ y = 9
∴ Solution is (0, 9).
When x = 1, π(1) + y = 9
⇒ y =9-π
∴ Solution is {1, (9 - π)}.
When x = 2, π(2) + y = 9
⇒ y = 9 - 2π
∴ Solution is {2, (9 - 2π)}.
When x = -1, π(-1) + y = 9
⇒ -π+ y = 9
⇒ y=9+π
∴ Solution is {-1, (9 + π)}.
(iii) x = 4y
When x = 0, 4y = 0
⇒ y = 0
∴ Solution is (0, 0).
When x = 1, 4y = 1
⇒ y= \(\frac{1}{4}\)
∴ Solution is (1,\(\frac{1}{4}\)).
When x = 4, 4y = 4
⇒ y = \(\frac{4}{4}\) = 1
∴ Solution is (4, 1).
When x =-4, 4y =-4
y = \(\frac{-4}{4}\) =-1
∴ Solution is (-4, -1).

We have p(x) = 3x³ + 7x and zero of 7 + 3x is \(\frac{-7}{3}\)
[∵7 + 3x = 0 ⇒ x = \(\frac{-7}{3}\)]
∴ p\(\left(\frac{-7}{3}\right)\) = 3\(\left(\frac{-7}{3}\right)\)+7\(\left(\frac{-7}{3}\right)\) = 3\(\left(\frac{-343}{27}\right)\) + \(\left(\frac{-49}{3}\right)\)
= -\(\frac{343}{9}\) - \(\frac{49}{3}\) = \(\frac{-490}{9}\)
Since \(\left(\frac{-490}{9}\right)\) ≠ 0
i.e. the remainder is not 0.
∴ 3x³ - 7x is not divisible by 7 + 3x.
Thus, (7 + 3x) is not a factor of 3x³ - 7x.

756

(i) Given equation is x - 2y= 4.
On putting x = 0 and y = 2 in LHS, we get
LHS = x - 2y = 0 - 2 X 2
=0-4=-4≠4
⇒LHS≠RHS
Hence, (0,2) is not a solution of x - 2y = 4.
(ii) Given equation is x - 2y= 4.
On putting x = 2 and y = 0 in LHS, we get
LHS = x - 2y = 2 - 2 X 0
=2-0=2≠4
⇒LHS≠RHS
Hence, (2, 0) is not a solution of x - 2y = 4.
(iii) Given equation is x - 2y= 4.
On putting x = 4 and y = 0 in LHS, we get
LHS =x - 2y = 4 - 2 X 0
= 4 -0 =RHS
Hence, (4, 0) is a solution of x - 2Y = 4.

\(\therefore \ 0.6+0.\overline { 7 } +0.4\overline { 7 } =\frac { 6 }{ 10 } +\frac { 7 }{ 9 } +\frac { 43 }{ 90 } \)
\(=\frac { 54+70+43 }{ 90 } =\frac { 167 }{ 90 } \)

First, we draw AB = 9.3 units. Now, from B, mark a distance of 1 unit. Let this point be C. Let D be the mid-point of AC. Now, draw a semi-circle with centre D and radius DA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
\(\therefore\)   Distance, BD=\(\sqrt { 9.3 } \)
Draw an arc with centre B and radius BD, which intersects the number line at point E. So point E represents \(\sqrt { 9.3 } \) .

(a) (iii) 2x + 3y = 60
Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is 
2x + 3y = 60.
(b) (i) 10
2x + 3y =60 ⇒ 2(15) + 3y = 60
⇒ 3y = 60 - 30 = 30 
⇒ y = 10
(c) (i) 12
2x + 3y = 60 ⇒ 2x + 3(12) = 60
⇒ 2x 60 - 36 = 24
⇒ x = 12
(d) (iii) 15
On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have 
3 x 5 + 5 x 0 = b
⇒ 15 + 0 = b
⇒ b = 15
(e) (ii) 1.x - 1.y + 5 = 0
y - x = 5 ⇒ y = x + 5
⇒ x - y + 5 = 0
⇒ 1.x - 1.y + 5 = 0

(a) (iii) x2 + x = 1
(b) (ii) x + y = 200
Here, x represents Sita's contribution and y represents Gita's contribution.
(c) (iii) Rs. 124
If x = 76 then 76 + y = 200
y = 200 - 76
y = 124
(d) (ii) Rs. 100, Rs. 100
If x = y then x + x = 200
2x = 200
x = 200/2 = 100
(e) (iii) 1.x + 0.y + 5 = 0
Since, x = -5 ⇒ x + 5 = 0
Thus, standard form of x = -5 is 1.x + 0.y + 5 = 0.

(i) (b) 10x, 5y
Area of one bedroom = 5x sq.m
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
(ii) (d) 54 m
Length of outer boundary = 12 + 15 + 12 + 15 = 54 m
(iii) (d) None of the above
Area of two bedrooms = 10x sq.m
Area of kitchen = 5y sq. m
So, 10x + 5y = 95  2x + y = 19
Also, x + 2 + y = 15 x + y = 13
(iv) (c) x = 6, y = 7
x + y = 6 + 7 = 13
2x + y = 2(6) + 7 = 19
x = 6, y = 7
x + y = 6 + 7 = 13
2x + y = 2(6) + 7 = 19
 x = 6, y = 7
(v) (a) 30 sq. m
Area of living room = (15 x 7) – 30
= 105 – 30 =75 sq. m

Here reason is not true.
\(\sqrt{4}\)= ±2, which is not an irrational number.
Correct option is (c) Assertion (A) is true but reason (R) is false.