\(\frac { 5 }{ 11 } \)

 \(\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } +3\) 
\(\Rightarrow\) \(\frac { 2x }{ 3 } +1=\frac { 7x }{ 15 } - 3\) [transposing \(\frac { 7x }{ 15 } \) to LHS and 1 to RHS]
\(\Rightarrow\)\(\frac { 10x-7x }{ 15 } =2\) [LCM of 3 and 15 = 15]
\(\Rightarrow\) \(\frac { 3x }{ 15 } =2\Rightarrow \frac { 3x }{ 15 } x5=2x5\)
\(\Rightarrow\) \(\frac { 15x }{ 15 } \)=2 x 5
x = 2 x 5 \(\Rightarrow\) x = 10
which is the required solution.

We have 6 = z + 2 
\(\Rightarrow\) 6 - 2 = z  [transposing 2 from RHS to LHS]
\(\Rightarrow\) 4 = z \(\Rightarrow\) z = 4, Which is the Required solution

3 (5z - 7) - 2 (9z - 11) = 4 (8z - 13) - 17
We have
3(5z -7) - 2(9z -11) = 4(8z -13) -17
\(\Rightarrow\)15z - 21 - 18z + 22 = 32z - 52 - 17
\(\Rightarrow\)- 3z + 1 = 32z - 69
\(\Rightarrow\)- 3z - 32z = - 69 - 1
Transposing 32z to LHS and 1 to RHS
\(\Rightarrow\)- 35z = - 70
\(\Rightarrow\)z = -\(\frac{-70}{-35}\)=2
Dividing both sides by - 35 This is the required solution.

Let the three consecutive integers be x, (x + 1)and (x + 2).
According to the question,
\(\Rightarrow\) 2x + 3 (x + 1)+ 4 (x + 2) = 74
\(\Rightarrow\) 2x + 3x + 3 + 4x + 8 = 74
\(\Rightarrow\)9x + 11 = 74 \(\Rightarrow\) 9x = 74 -11 [transposing 11 to RHS]
\(\Rightarrow\) 9x = 63 \(\Rightarrow\) x = \(\frac { 63 }{ 9 } \) = 7 [dividing both sides by 9]
\(\therefore\) First integer = x = 7
Second integer = x+ 1= 7 + 1= 8
and third integer = x + 2 = 7 + 2 = 9
Hence, the required integers are 7,8 and 9.

Let the number be x
After subtracting \(\frac { 1 }{ 2 } \) from this number, we get Number \(\left( x-\frac { 1 }{ 2 } \right) \)
Now, multiply it by \(\frac { 1 }{ 2 } \) We get number = \(\frac { 1 }{ 2 } \) \(\left( x-\frac { 1 }{ 2 } \right) \)
According to the Questions
\(\frac { 1 }{ 2 } \left( x-\frac { 1 }{ 2 } \right) =\frac { 1 }{ 8 } \Rightarrow \frac { 1 }{ 2 } \left( \frac { 2x-1 }{ 2 } \right) =\frac { 1 }{ 8 } \Rightarrow \frac { 2x-1 }{ 4 } =\frac { 1 }{ 8 } \)
\(\Rightarrow\)  \(2x-1=\frac { 1 }{ 8 } \times 4\)  [multiplying both sides by 4]
\(\Rightarrow\) \(2x-1=\frac { 1 }{ 2 } \Rightarrow 2x=\frac { 1 }{ 2 } +1\) [transposing -1 to RHS]
\(\Rightarrow\) \(2x=\frac { 1 }{ 2 } +\frac { 1 }{ 1 } \Rightarrow 2x=\frac { 1+2 }{ 2 } \) [LCM of 2 and 1 is 2]
\(\Rightarrow\) \(2x=\frac { 3 }{ 2 } \Rightarrow x=\frac { 3 }{ 2\times 2 } =\frac { 3 }{ 4 } \)  [dividing both sides by 2]
Hence, the required number is \(\frac { 3 }{ 4 } \)

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.

Let the number of gold coins initially be x 
Then, the number of non-gold coins be 3x. When,
10 more gold coins added
Then, the total number of gold coins = (10 + x)
Then, according to the question , \(\frac { (10+x) }{ 3x } =\frac { 1 }{ 2 } \)
\(\Rightarrow\) 2 (10 + x) = 3x  \(\Rightarrow\) 20 + 2x = 3x \(\Rightarrow\) x = 20
Then, total number of coins at last = 3x + 10 + x  = 4x + 10 = 4 x 20 + 10 = 90

Given, total number of participants is 63.
Let the number of winners be x.
.\(\therefore\)  The number of non-winners = (63 - x)
Prize money given to winners = 100x
Prize money given to non-winners = 25 (63 - x)
According to the question,
Total prize money = 3000
\(\Rightarrow\) 100x + 25 (63 - x) = 3000
\(\Rightarrow\) 100x + 1575 - 25x = 3000
\(\Rightarrow\) 75x + 1575 = 3000
\(\Rightarrow\) 75x = 3000 -1575 [transposing 1575 to RHS]
\(\Rightarrow\) 75x = 1425 \(\Rightarrow\) \(\frac { 1425 }{ 75 } \)
x = 19
Hence, the number of winners are 19