Let the unit's place digit be x and ten's place digit be y.
Original number=10y+x
Number obtained by reversing the order of digits
=10x+y
Sum of both number=99  [given]
(10y+x)+(10x+y)=99
11x+11y=99
x+y=9  .....(i) [dividing both sides by 11]
Also, given that the digits of the numbers differ by 3.
If y>x, then y=-x=3  ..(ii)
and if y On adding Eqs. (i) and (ii), we get
2y=12
\(\Rightarrow\)  \(y= \frac {12}{2}\)=6
Then, from Eq. (i),
x=9-y=9-6=3  [\(\therefore\)y=6]
\(\therefore\)Number=10y+x=10(6)+3=63
Now, adding Eqs. (i) and (iii), we get
2x=12
x=6
Then, from Eq. (i), we get
y=9-x=9-6=3  [\(\therefore\)x=6]
\(\therefore\)Number=10y+x=10(3)+6=36
Hence, the required number 63 and 36

Given linear equation is 2x=7=13-x.
On substituting x=2 in the given equation, we get
LHS=2 \(\times\) 2+7=11 and RHS=13-2=11
\(\therefore\)  LHS=RHS​​​​​
Hence, x=2 is a solution of the equation.

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

Let income of A = 8x and income of B = 7x.
Also their expenditures be 19y and 16y.
\(\Rightarrow\)  8x - 19y = 2550 .....(i)
and 7x - 16y = 2550 .....(ii)
Solving the equations
x = 1530 and y = 510
\(\therefore\) Salary of A = 12240
Salary of B = 10710
Importance of saving in life is responsibility and indirectly helping the nation.

Given, a pair of linear equation is:
\(\sqrt { 2 } x+\sqrt { 3 } y=0\)   ......(i)
and \(\sqrt { 3 } x-\sqrt { 8 } y=0\quad or\quad y=\frac { \sqrt { 3 } x }{ \sqrt { 8 } } \)  ......(ii)   
On substituting y from eqn. (ii) in eqn. (i),
\(\sqrt { 2 } x+\sqrt { 3 } \times \left( \frac { \sqrt { 3 } x }{ \sqrt { 8 } } \right) =0\)
\(\Rightarrow \quad \sqrt { 2 } x\times \sqrt { 8 } +3x=0\)
\(\Rightarrow \quad \sqrt { 2 } x+\frac { 3x }{ \sqrt { 8 } } =0\)
\(\Rightarrow \quad \sqrt { 2 } x\times \sqrt { 8 } +3x=0\)
\(\\ \Rightarrow \)  4x + 3x = 0
\(\\ \Rightarrow \)  7x = 0
\(\therefore\) x = 0
On substituting x = 0 in eqn. (ii),
\(y=\frac { \sqrt { 3 } \times 0 }{ \sqrt { 8 } } =0\)
\(\therefore\)y = 0
Hence, x = 0, y = 0

We know that, in a cyclic quadrilateral, the sum of two opposite angles is 180°.
\(\therefore \quad \angle B+\angle D={ 180 }^{ 0 } and \quad \angle A+\angle C={ 180 }^{ 0 }\)
\(\Rightarrow \) 3y-5-7x+5=180 and 4y+20-4x=180
\(\Rightarrow \) 3y-7x=180   ...(i)
and 4y-4x=160 
\(\Rightarrow \) y-x=4. [dividing both sides by 4]  ...(ii)
On multiplying Eq. (ii) by 7 and then subtracting from Eq. (i), we get
-4y=180-280
\(\Rightarrow \) -4y=-100 y=25
On putting y=25 in Eq. (ii), we get
25-x=40 x=-15
On putting the values of x and y, we calculate the angles as
\(\angle A=4y+20=100+20={ 120 }^{ 0 }.\) 
\(\\ \angle B=3y-5=75-5={ 70 }^{ 0 }.\) 
\(\\ \angle C=-4x=-4(-15)={ 60 }^{ 0 }.\) 
and \( \angle D=-7x+5=105+5={ 110 }^{ 0 }.\)
Hence, the angles are \(\angle A={ 120 }^{ 0 },\angle B={ 70 }^{ 0 },\angle C={ 60 }^{ 0 }\) and \( \angle D={ 110 }^{ 0 }.\)

Step 1 : Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal. Then we get the equations as
4x + 6y = 16    (3)
4x + 6y = 7       (4)
Step 2 : Subtracting Equation (4) from Equation (3),
(4x - 4x) + (6y - 6y) = 16 - 7
i.e., 0 = 9, which is a false statemnt.
Therefore, the pair of equations has no solution.

Two straight paths are parallel to each other. Hence, they do not cross each other.

x=-2, y=-1

Pair of linear equations kx - 4y - 3 = 0 and 6x - 12y - 9 = 0
Condition for infinite solutions: - \(\frac { { a }_{ 1 } }{ a_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\frac { -k }{ 6 } =\frac { -4 }{ -12 } =\frac { 3 }{ 9 } \)
\(\Rightarrow\) k = 2

The given equations can be re-written as :
5x-3y-11 = 0
- 10x + 6y + 22 = 0
On comparing with ax + by + c = 0, we have
a₁ = 5, b₁ = - 3, c₁ = - 11
a₂ = - 10, b₂ = 6, c₂ = 22
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 5 }{ -10 } =-\frac { 1 }{ 2 } \)
\(\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ 6 } =\frac { -1 }{ 2 } \)
and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -11 }{ 22 } =\frac { -1 }{ 2 } \)
Thus, \(\frac { -1 }{ 2 } =\frac { -1 }{ 2 } =\frac { -1 }{ 2 } \)
i.e. \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
Hence, the pair of linear equations is consistent.

The equation of one line is 4x + 3y = 14. We know that if two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel, then
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ c_{ 2 } } \)
\(\Rightarrow \ \ \ \frac { 4 }{ { a }_{ 2 } } =\frac { 3 }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \Rightarrow \frac { { a }_{ 2 } }{ { b }_{ 2 } } =\frac { 4 }{ 3 } =\frac { 12 }{ 9 } \)
Hence one of the possible, second parallel line is 12x + 9y = 5

\(ad\neq bc\Rightarrow \frac { a }{ c } \neq \frac { b }{ d } \)
Hence, the pair of given linear equations has unique solution.

The pair of equations y = 0 and y = -5 has no solution.

By property of rectangle, we know that its opposite sides are of equal lengths.
i.e. DC=AB \(\Rightarrow\) x+3y=13   ...(i)
and AD=BC \(\Rightarrow\) 3x+y=7   .....(ii)
On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

\(\Rightarrow\) x=1
On putting x=1 in Eq. (i), we get
3y=12 \(\Rightarrow\) y=4
Hence, x=1 and y=4.

x=14, y=9

x=1, y=2

\(x=5,\quad y=\frac { 2 }{ 13 } \)

Given that, x, y ane 40⁰ are the angles of a triangle.
\(\therefore\) x+y+40⁰=180⁰  [\(\because\) sum of all the angles of a triangle is 180⁰ ]
\(\Rightarrow\)x+y=140⁰     ...(i)
Also, x-y=30⁰    ...(ii)
On adding Eqs. (i) and (ii), we get
2x=170⁰ \(\Rightarrow\)  x=85⁰
On putting  x=85⁰ in Eqs. (i), we get
85⁰+y=140⁰  \(\Rightarrow\)   y=55⁰
Hence, the required values of x and y are 85⁰ and 55⁰ respectively.

Given equation are 7x-5y=3 and 14x-10y=5
Here, a₁=7, b₁=-5, c₂=-3
and a₂=14, b₂=-10, c₂=-5
Now, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } ,\quad \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -5 }{ -10 } =\frac { 1 }{ 2 } \) and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { 3 }{ 5 } \)
Thus, for the given equations, we have
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
So, we conclude that the two paths are parallel to each other. Hence, the paths do not cross each other. 

Given, pair of equations is
3x-(a+1)y=2b-1
and 5x+(1-2a)y=3b
Here, a₁=3, b₁=-(a+1), c₂=-(2b-1)
and a₂=5, b₂=1-2a, c₂=-3b
For infinitely many solutions, we must have
\(\underset { I }{ \frac { 3 }{ 5 } } =\underset { II }{ \frac { -(a+b) }{ 1-2a } } =\underset { III }{ \frac { 2b-1 }{ 3b } } \)
Now, take I and II terms, \(\frac { 3 }{ 5 } =\frac { -(a+b) }{ 1-2a } \)
\(\Rightarrow \quad 3-6a=-5a-5\)
\(\Rightarrow \quad a=8\)
Now, take I and III terms,
\(\frac { 3 }{ 5 } =\frac { 2b-1 }{ 3b } \)
\(\Rightarrow \quad 9b=10b-5\quad \Rightarrow \quad b=5\)

 By cross-multiplication method,
\(\frac { x }{ -2{ b }^{ 2 }+8a-2{ a }^{ 2 }-8ab } =\frac { y }{ -2ab+8{ b }^{ 2 }-2ab+8{ a }^{ 2 } } \)
\(x=-\frac { 1 }{ 2 } ,\quad y=2\)

(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.

Let x and y be length and breadth of rectangle.
Then, its area=xy
According to the questions,
9x-5)(y+2)=xy-80 \(\Rightarrow\) 2x-5y=-70
(x+10)(y-5)=xy+50  \(\Rightarrow\) -5x+10y=100
Length=40 units, breadth=30 units