Since, 5(x-3) = 25
\(\Rightarrow\) 5x - 3 \(\times\) 5 = 25
\(\Rightarrow\) 5x - 15 = 25
\(\Rightarrow\) 5x = 25 + 15
\(\Rightarrow\) 5x = 40
\(\Rightarrow\)  \(\frac { 5x }{ 5 } =\frac { 40 }{ 5 } \)
Thus, x = 8   

\(\frac { y }{ 5 } +3=2\)
\(\Rightarrow \frac { y }{ 5 } =2-3\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\)
\(\Rightarrow \quad \frac { y }{ 5 } =-1\times 5\)
y = -5

3x-5 = 7
\(\Rightarrow\) 3x = 7 + 5
[On transposing 5 to RHS]
\(\Rightarrow\) 3x = 12
\(\frac { 3x }{ 3 } =\frac { 12 }{ 3 } \) [Dividing by 3]
\(\Rightarrow\) x = 4

\(\frac { x }{ 2 } -1=\frac { x }{ 3 } +4\)
\(\Rightarrow \quad \frac { x }{ 2 } -\frac { x }{ 3 } =4+1\)
[ On transposing \(\frac { x }{ 3 } \) to LHS and 1 to RHS]
\(\Rightarrow \quad \frac { 3x-2x }{ 6 } =5\)
\(\Rightarrow\) 3x - 2x = 5 \(\times\) 6
\(\therefore\) x = 30

6s + 24 = 0
\(\Rightarrow\) 6s = - 24
Thus, s = -4

28=4+3t+15
9 = 3t
\(\Rightarrow\) t = 3

10=4+3(t+2)
10=4+3t+6
\(\Rightarrow\) 0=3t
\(\Rightarrow\) t = 0

In order to separate the variable, first multiply both LHS and RHS by 10.
On multiplying by 10,we get 6 \(\times\) 10 in RHS and 3p in LHS.
Now, divide both LHS and RHS by 3, so that p will be separated.
\(\frac { 3p }{ 10 } =6\)
\(\Rightarrow \) 3p = 6 \(\times\) 10
\(\Rightarrow \quad p=\frac { 6\times 10 }{ 3 } \)
\(\Rightarrow \quad p=2\times 10\quad \)
\(\therefore\) p = 20

In order to separate the variable, first multiply both LHS and RHS by 3.
On multiplying by 3, we get 40 \(\times\) 3 in RHS and 20p in LHS.
Now, divide both LHS and RHS by 20, so that p will be separated.
\(\frac { 20p }{ 3 } =40\)
\(\Rightarrow \quad 20p=40\times 3\)
\(\Rightarrow \quad p=\frac { 40\times 3 }{ 20 } \)
\(\Rightarrow\) p = 2 \(\times\) 3
\(\therefore\) p = 6

We have, \(\frac { 3l }{ 2 } =\frac { 2 }{ 3 } \)
\(\Rightarrow \quad \frac { 3l }{ 2 } \times \frac { 2 }{ 3 } =\frac { 2 }{ 3 } \times \frac { 2 }{ 3 } \)
[Multiplying both sides by \(\frac { 2 }{ 3 } \)]
\(\Rightarrow \quad l=\frac { 4 }{ 9 } \)

We have,
6z + 10 = -2
\(\Rightarrow\) 6z = -2 - 10
[Transposing 10 on RHS]
\(\Rightarrow\) 6z = - 12
\(\Rightarrow \quad \frac { 6z }{ 6 } =\frac { -12 }{ 6 } \Rightarrow z=-2\)

According to the question, one-third of z =\(\frac { 1 }{ 3 } \)z
Now, add 3 to one-third of z =\(\frac { 1 }{ 3 } \) z+3 and the result = 30 
Hence, the required equation is \(\frac { 1 }{ 3 } \)z+3 =30

According to the question,
Six times of y = 6y
Now, take away 6 from 6 times y = 6y- 6
and the result = 60
Hence, the required equation is 6y- 6 = 60.

Given equation is n+5 = 19 (n=1)
Here, LHS = n+5
Puting n = 1 in LHS we get
LHS = 1+5 = 6 ≠ 19
∵ LHS ≠ RHS
So, n = 1is not a solution of the given equation

Other forms for equation\(\frac { m }{ 5 } -2=6\) =6 are as follows:
(a) Subtract 2 from one-fifth of a number m to get 6.
(b) One-fifth of m is greater by 2 than 6.