7x - 9 =16 \(\Rightarrow\) 7x =16 + 9  [transpoting 9 from LHS to RHS]
7x =25  \(\Rightarrow \frac { 7x }{ 7 } =\frac { 25 }{ 7 } \)
\(\Rightarrow\) \(x=\frac { 25 }{ 7 } \), Which is the required solution

\(x=\frac { -8 }{ 5 } \) 

x = 0, it satisfies equation

We have  3x = 2x + 18 \(\Rightarrow\) 3x - 2x = 18 [transposing 2x to LHS]
\(\therefore\) x = 18, which is the required solution.

-1

Rational number is \(\frac { 1 }{ 12 } \)

If x is a multiple of 11, the next multiple is x + 11. The next to this is x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.
It is given that the sum of these consecutive multiples of 11 is 363. This will give the following equation:
x + (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363
or 3x + 33 = 363
or 3x = 363 – 33
or 3x = 330
\(x=\frac{330}{3}\)
= 110
Hence, the three consecutive multiples are 110, 121, 132

Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x.
The amount Bansi has:
(i) from 5 rupee coins, RS. 5 x x = RS. 5x
(ii) from 2 rupee coins, RS. 2 x 3x = RS. 6x
Hence the total money he has = RS. 11x
But this is given to be  RS.77; therefore,
11x = 77
or \(x=\frac{77}{11}=7\)
Thus, number of five-rupee coins = x = 7
and number of two-rupee coins = 3x = 21

Let the number of currency notes of denominations
Rs 100, Rs. 50 and Rs. 10 be 2x, 3x and 5x, respectively.
The amount she has from Rs.100 notes
= 2x x 100 = 200x
The amount she has from Rs. 50 notes = 3x x 50 = 150x
The amount she has from Rs. 10 notes = 5x x 10 = 50x
According to the question,
Total amount = 400000
\(\Rightarrow\) 200x + 150x + 50x = 400000
\(\Rightarrow\) 400x = 400000
\(\Rightarrow\) \(x=\frac { 400000 }{ 400 } =1000\)
\(\therefore\) Number of Rs.100 notes = 2x = 2 x 1000 = 2000
N umber of Rs. 50 notes = 3x = 3 x 1000 = 3000
and number of Rs. 10 notes = 5x = 5 x 1000 = 5000
Hence, she has 2000, 3000 and 5000 notes of denominations Rs. 100, Rs.  50 and Rs. 10, respectively

Let Shobo's present age be x yr.
Shobe's mother's present age =6x
According to the question, 5 yr later
Shobo's age = \(\frac { 1 }{ 3 } \) Shobo's mother's present age
\(\Rightarrow\) x + 5 = \(\frac { 1 }{ 3 } \) x 6x \(\Rightarrow\) x + 5 = 2x \(\Rightarrow\) 5 = 2x - x 
\(\Rightarrow\) 5 =x \(\Rightarrow\) x = 5
Shobe's present age = x = 5 yr
and Shobo's mother's present age = 6x = 6 x 5 = 30 yr
Hence, Shobe's present age is 5 yr and her mother's age is 30 yr. 

Let the numerator be x. Then, denominator = x + 8 
So, rationl number = \(\frac { x }{ x+8 } \)
According to the question, \(\frac { x+17 }{ (x+8)-1 } =\frac { 3 }{ 2 } \Rightarrow \frac { x+17 }{ (x+8)-1 } =\frac { 3 }{ 2 } \Rightarrow \frac { x+17 }{ x+7 } =\frac { 3 }{ 2 } \)
\(\Rightarrow\) 3 (x + 7) = 2 (x +17  [by cross-multiplication]
\(\Rightarrow\) 3x + 21 = 2x + 34 
\(\Rightarrow\) 3x - 2x = 34 - 21  [transposing 2x to LHS and 21 to RHS]
\(\Rightarrow\) x = 13
Rational number = \(\frac { x }{ x+8 } =\frac { 13 }{ 13+8 } \frac { 13 }{ 21 } \)
Hence, the rational number is \(\frac { 13 }{ 21 } \) 

First Part = 14, Second Part = 20

\(\frac { 1 }{ 3 } \)

Let Abdul buys 2x m of trouser material 
Then, the shirt material bought by him = 3x m
Sale price of 1 m of trouser material
Rs (90 + 12% of 90)
\(\left( 90+\frac { 12\times 90 }{ 100 } \right) =Rs.100.80\)
Sale price of 2x m of trouser material = Rs.(2xx 100.80) = Rs. 201.60x
Sale price of 1m of shirt material
Rs. 50 +20% of Rs 50 = \(\left( 50+\frac { 20\times 50 }{ 100 } \right) =Rs.60\)
Sale price of 3x m of shirt material = Rs 3x x 60
= Rs. 180x
Total sale = Rs. (201.60+180) x = Rs. 381.60x
\(\therefore\)  381.60 = 38160
\(x=\frac { 38160 }{ 38.160 } =100\)
So,Abdul bought 2 x 100 = 200 m of trouser material.

m-\(\frac{m-1}{2}\)=1-\(\frac{m-2}{2}\)
We have m-\(\frac{m-1}{2}\)=1-\(\frac{m-2}{2}\)
It is a linear equation since it involves linear expressions only.
\(\Rightarrow\) m-\(\frac{m}{2}+\frac{1}{2}\)=1-\(\frac{m}{3}+\frac{1}{3}\)
\(\Rightarrow\) m-\(\frac{m}{2}+\frac{m}{3}\)=1+\(\frac{2}{3}+\frac{1}{2}\)
Transposing -\(\frac{m}{3}\)to LHS  and \(\frac{1}{2}\)to RHS 
\(\Rightarrow\) \(\frac { 6m-3m+2m }{ 6 } =\frac { 6+4-3 }{ 6 } \)
Taking LCM
\(\Rightarrow\) \(\frac { 5m }{ 6 } =\frac { 7 }{ 6 } \)
\(\Rightarrow\) m = \(\frac { 7 }{ 6 } \times \frac { 6 }{ 5 } =\frac { 7 }{ 5 } \)
Multiplying both sides by \(\frac{6}{5}\)
This is the required solution.