\(\frac{3}{2} x+\frac{5}{3} y=7,9 x-10 y=14\) 
Here,\(\frac{a_{1}}{a_{2}}=\frac{3}{2 \times 9}=\frac{1}{6}, \frac{b_{1}}{b_{2}}=-\frac{5}{3 \times 10}=\frac{-1}{6}, \frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{1}{2}\) 
\(\because \quad \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) 
\(\therefore\) Pair of equations is consistent with unique solution
 

The given equations can be rewritten as
2x - 3y - 8 = 0 and 4x - 6y - 9 = 0
On comparing with standard form of pair of linear equations, we get
a₁ = 2, b₁ = -3, c₁ = -8
and a₂ = 4, b₂ = -6, c₂ = -9
Now, \(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2},\frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2} and \frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}\)
Thus, \(\frac{1}{2}=\frac{1}{2}\neq \frac{8}{9}\) i.e.,\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, the pair of linear equations is inconsistent.

Given linear equation can be written as y = 2 + x                 ...(i)
When x = - 2, then from Eq. (i), we get y = 2 - 2 = 0
When x = 0, then from Eq. (i), we get y = 2
When x = 1, then from Eq. (i), we get y = 2 + 1 = 3
Thus, we get the table 

Draw the coordinate axes XOX' and YOY', and plot the points A( -2,0), B (0, 2) and C( 1, 3 ) by taking a suitable scale.

On joining the points A, Band C, we get a straight line AC. Thus, the line AC represents the required graph of given linear equation in two variables.

The given equations can be rewritten as
3x + 2y - 5 = 0 and 2x - 3y - 7 = 0
On comparing with standard form of pair of linear equations, we get a₁ = 3, b₁ = 2, c₁ = -5
and a₂ = 2, b₂ = -3, c₂ = -7
Now, \(\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=-\frac{2}{3}\) and \(\frac{c_{1}}{c_{2}}=\frac{5}{7}\)
Thus, \(\frac{3}{2}\neq -\frac{2}{3},i.e.\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)
Hence, the pair of linear equations is consistent.

Let the speed of steam be x km/h, then speed of person going upstream=(x+5) km/h and speed of person going downstream=(5-x)h
According to question
\(\frac { 4 }{ 5-x } =3\left( \frac { 4 }{ 5-x } \right) \)
2.4 km/h 
Lete the speed of boat in still water be x km/h and that of stream be y km/h, then according to question
\(\frac { 30 }{ x-y } +\frac { 28 }{ x+y } =7\quad \quad ...(i)\)
and \(\frac { 21 }{ x+y } +\frac { 21 }{ x-y } =5\quad \quad ...(ii)\)  
Solve Eq. (i) and Eq. (ii) to get the speed of boat and speed of stream.
Speed of motorboat in still water=10 km/h
Speed of stream=4 km/h

x = 11, y = 8

The given pair oflinear equations is
4x+3y-7=0 ...(i)
and 12x+9y-21=0 ...(ii)
On comparing the above equations with standard form of pair of linear equations, we get
a₁ = 4, b₁ = 3,c₁ = - 7
and  a₂ =12,b₂ =9, c₂ =- 21
Now,  \( \frac{a_{1}}{a_{2}}=\frac{4}{12}=\frac{1}{3},\frac{b_{1}}{b _{2}}=\frac{3}{9}=\frac{1}{3} \)
and  \(\frac{c_{1}}{c _{2}}=\frac{-7}{-21}=\frac{1}{3}\)
\(\therefore\)  \(\frac{a _{1}}{a _{2}}=\frac{b_{1}}{b _{2}}=\frac{c_{1}}{c _{2}}\)
\(\therefore\) The lines representing the given pair of linear equations will coincide.

Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is
s – 7 = 7 (t – 7), i.e., s – 7t + 42 = 0  ...(1)
and s + 3 = 3 (t + 3), i.e., s – 3t = 6   ...(2)
Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get
(3t + 6) – 7t + 42 = 0,
i.e., 4t = 48, which gives t = 12.
Putting this value of t in Equation (2), we get
s = 3 (12) + 6 = 42
So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.

2x − 2y − 2 = 0
4x − 4y − 5 = 0
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{-2}{-4}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{2}{5}\) 
Since \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) 
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent

Let the supplementary angles be x and y(x > y).
Then, x + y = 180 ...(i)
Now, according to the question,
x - y = 18              ...(ii)
From Eq. (ii), we have
y = x - 18              ...(iii)
On substituting the value of y from Eq. (iii) in Eq. (i),
we get
x + x - 18 = 180
\(\Rightarrow\) 2x = 198
\(\Rightarrow\) x = 99
On substituting x = 99 in Eq. (iii) we get
y = 99 - 18
\(\Rightarrow\) y = 81
Hence, the required angles are 99° and 81°.

Given equation of lines are
3x - y = 3    ...(i)
2x-3y=2   ...(ii)
and x + 2y = 8 ....(iii)
Let lines (i), (ii) and (iii) represent the sides of \(\triangle\)ABC,
say AB, BC and CA, respectively.

On multiplying Eq. (i) by 3 and then subtracting Eq. (ii) from Eq. (i), we get
7x=7 \(\Rightarrow\) x=1
On putting the value of x in Eq. (i), we get
3 x 1-y=3
\(\Rightarrow\)   y=0
So, the coordinate of point or vertex B is (1, 6). Similarly, on solving Eq. (ii) and Eq. (iii), we get the coordinate of point or vertex Cis (4, 2). 
And on solving Eq. (i) and Eq. (iii), we get the coordinate of point or vertex A is (2, 3). 
Hence, the vertices of \(\triangle\)ABC formed by the given lines are A (2,3), B (1, 0) and C( 4, 2).

\(x=\frac{9}{13}, y=\frac{-5}{13}\)

Let us write the given pair of equations as
\(2\left(\frac{1}{x}\right)+3\left(\frac{1}{y}\right)=13\)   ...(1) 
\(5\left(\frac{1}{x}\right)-4\left(\frac{1}{y}\right)=-2\)   ...(2)
These equations are not in the form ax + by + c = 0. However, if we substitute
\(\frac{1}{x}=p\) and \(\frac{1}{y}=q\) in Equations (1) and (2), we get
2p + 3q = 13  ...(3)
5p – 4q = – 2  ...(4)
So, we have expressed the equations as a pair of linear equations. Now, you can use any method to solve these equations, and get p = 2, q = 3.
You know that \(p=\frac{1}{x}\) and \(q=\frac{1}{y}\) 
Substitute the values of p and q to get
\(\frac{1}{x}=2, \text { i.e., } x=\frac{1}{2}\) and \(\frac{1}{y}=3, \text { i.e., } y=\frac{1}{3}\) 
Verification : By substituting \(x=\frac{1}{2}\) and \(y=\frac{1}{3}\) in the given equations, we find that both the equations are satisfied.

Let fixed charge be Rs x and charge per km be Rs y.
We are given that,
x + 10y = 105      .....(i)
x + 15y = 155      ......(ii)
From eqn. (i), x = 105 - 10y ... (iii)
On substituting x from eqn. (iii) in eqn. (ii),
105 - 10y + 15y = 155
\(\Rightarrow\) 5y = 155 - 105 = 50
\(\therefore\) y = 10
On substituting y = 10 in eqn. (iii),
x = 105 - 10 \(\times\) 10
= 105 - 100 = 5
\(\therefore\) x = 5
Hence, fixed charges = Rs 5
Rate per km = Rs 10
Amount to be paid for travelling 25 km
= Rs 5 + Rs 10 \(\times\) 25
= Rs 5 + Rs 250
= Rs 255

Given, a pair of linear equations is :
0.2x + 0.3y = 1.3        .....(i)
and 0.4 x + 0.5y= 2.3  ....(ii)
Multiply both sides of Eq. (i) and Eq. (ii) by 10, we get
2x + 3y = 13                ....(iii)
and 4x + 5y = 23          ....(iv)
On substituting y from eqn. (iii) in eqn. (ii),
\(\frac { 4 }{ 10 } x+\frac { 5 }{ 10 } \times \frac { (13-2x) }{ 3 } =\frac { 23 }{ 10 } \)
\(\Rightarrow \quad 4x+\frac { 5 }{ 3 } (13-2x)=23\)
\(\Rightarrow\) 12x + 5 (13 - 2x) = 3 \(\times\) 23
\(\Rightarrow\) 12x + 65 - 10x = 69
\(\Rightarrow\) 2x = 69 - 65 = 4
\(\therefore\) x = 2
On substituting x = 2 in eqn. (iii), we get
\(\\ y=\frac { 13-2\times 2 }{ 3 } =\frac { 9 }{ 3 } \)
i.e., y = 3
Hence, x = 2,y = 3

(i) For batch I,
20x + 5y = 9000
For batch II,
5x + 25y = 26000
(ii) We have,
20x + 5y = 9000
\(\Rightarrow\) 4x + y = 1800  ...(i)
and 5x + 25y = 26000
\(\Rightarrow\)  x + 5y = 5200 ...(ii)
Multiplying Eq. (i) by 5 and subtracting Eq. (ii) from it.
5(4x + y) - (x + 5y) = 5 \(\times\) 1800 - 5200
\(\Rightarrow\)  19x = 3800
\(\Rightarrow\) x = 200
Or
On substituting x = 200 in Eq. (i), we get
4(200) + y = 1800
\(\Rightarrow\) 800 + y = 1800
\(\Rightarrow\) y = 1000
\(\therefore\) Difference in the monthly fee paid by a poor child and a rich child = y - x = Rs (1000 - 200) = Rs 800
(iii) Total monthly collection of fees, if there are 10 poor and 20 rich children = 10x + 20 y
= 10 \(\times\) 200 + 20 \(\times\)1000
= 2000 + 20000
= Rs 22000

Speed of boat in upstream = (x - y)km/hr and speed of boat in downstream = (x + y)km/hr.
(i) (a): 1^st situation can be represented algebraically as \(\frac{24}{x-y}+\frac{36}{x+y}=6\)
(ii) (b): 2nd situation can be represented algebraically as \(\frac{36}{x-y}+\frac{24}{x+y}=\frac{13}{2}\)
(iii) (c) : Putting \(\frac{1}{x-y}=u \text { and } \frac{1}{x+y}=v\)
we get,
24u + 36v = 6 and 36u + 24v = 13/2
Solving the above equations, we get u \(=\frac{1}{8}, v=\frac{1}{12}\)
(iv) (d): \(\because u=\frac{1}{8}=\frac{1}{x-y} \Rightarrow x-y=8\) ........(i)
\(\text { and } v=\frac{1}{12}=\frac{1}{x+y} \Rightarrow x+y=12\) .........(ii)
Adding equations (i) from (ii), we get 2x = 20 \(\Rightarrow\) x = 10
\(\therefore\) Speed of boat in still water = 10 km/hr -.
(v) (c): From equation (i), 10 - y = 8 \(\Rightarrow\) y = 2
\(\therefore\) Speed of stream = 2 km/hr.

(i) (a): 1^st situation can be represented as x + 7y = 650 ...(i) and
2^nd situation can be represented as x + 11y = 970 ...(ii)
(ii) (b): Subtracting equations (i) from (ii), we get 
\(4 y=320 \Rightarrow y=80\)
\(\therefore\) Proportional expense for each person is Rs 80.
(iii) (c): Puttingy = 80 in equation (i), we get
x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90
\(\therefore\) Fixed expense for the party is Rs 90
(iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290
(v) (a): We have a₁ = 1, b₁ = 7, c₁ = -650 and 
\(a_{2}=1, b_{2}=11, c_{2}=-970 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\)
\(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution.