5(x - 3) = 25
or 5(x - 3) = \(\frac{25}{5}\) [Dividing both sides by 5]
or x - 3 = 5
or x = 5 + 3 [Transposing -3 to R.H.S.]
=8
Thus, x = 8 is the required solution.

Subtracting 6 from both sides, we have
2q + 6 - 6 = 12 - 6
or 2q = 6
Dividing both sides by 2, we have
\(\frac{2q}{2}=\frac{6}{2}\)or q=3
Thus, q = 3 is the required solution.

Let the breadth of the rectangle be x em.
\(\therefore\)Length = (x + 18) em
Since, the perimeter of a rectangle
= 2 [Length + Breadth]
\(\therefore\)According to the condition,
2[(x + 18) + x] = 84
\(\Rightarrow \)2[x + x + 18] = 84
\(\Rightarrow \)2[2x + 18] = 84
\(\Rightarrow \)4x + 36 = 84
\(\Rightarrow \)4x = 84 - 36
[Transposing 36 to R.H.S.]
\(\Rightarrow \)4x = 48
\(\Rightarrow \) x=\(\frac{48}{4}\)=12
i.e. Breadth of the rectangle = 12 em and length of the rectangle = 12 + 18 = 30 cm Thus, the required length is 30 em and breadth is 12 em.

Age of Prashant = 40 years.
Let the age of son be x years.
 \(\therefore\)According to the condition,
5 x (Age of son) + 5 = Prashant's age
\(\Rightarrow \)5[x] + 5 = 40
\(\Rightarrow \)5x + 5 = 40
Transposing 5 to R.H.S., we have
5x = 40 - 5
\(\Rightarrow \)5x = 35
\(\Rightarrow \) x=\(\frac{35}{5}\)=7
Hence, the required age of son is 7 years.

We have:
0.6x + 0.8 = 0.56x + 2.32
Transposing 0.8 to R.H.S. and 0.56 to L.H.S.,
we have
0.6x - 0.56x = 2.32 - 0.8
\(\Rightarrow \) 0.60x - 0.56x = 2.32 - 0.80
\(\Rightarrow \)(0.60 - 0.56)x = (2.32 - 0.80)
\(\Rightarrow \)0.04x = 1.52
\(\Rightarrow \) x=\(\frac{1.52}{0.04}\)
\(\Rightarrow \) x=38
Thus, the required solution of the given equation is x = 38.

We have: 5x +\(\frac{1}{3}\)=2-3x
\(\Rightarrow \) 5x+3x=2-\(\frac{1}{3}\)
[Transposing (-3x) to L.H.S. and 1. to R.H.S.]
\(\Rightarrow \) 8x=\(\frac{6-1}{3}=\frac{5}{3}\)
\(\Rightarrow \) x=\(\frac{5}{3}\times \frac{1}{8}=\frac{5}{24}\)
Thus, x=\(\frac{5}{24}\)
is the required solution of the given equation.

(a) Let Hari Babu's property be x.
So, according to question,
\(\frac { 1 }{ 3 } x+\frac { 1 }{ 4 } x+18,000=x\)
\(\Rightarrow \quad \frac { 4x+3x }{ 12 } +18,000=x\)
or, x-\(\frac { 7x }{ 12 } =18,000\)
\(\Rightarrow \quad \frac { 12x-7x }{ 12 } =18,000\)
\(\Rightarrow\) 5x = 18,000 \(\times\) 12
\(\Rightarrow \quad x=\frac { 18,000\times 12 }{ 5 } \)
= 3600 \(\times\) 12
= Rs 43,200
(b) Solution of simple equations.
(c) Value: Girls have equal right in father's property and there should be no discrimination between girls and boys.  

Since,
\(\frac { 2x-1 }{ 3 } -\frac { 6x-2 }{ 5 } =\frac { 1 }{ 3 } \)
\(\therefore \quad \frac { 5(2x-1) }{ 3\times 5 } -\frac { 3(6x-2) }{ 3\times 5 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5 }{ 15 } -\frac { (18x-6) }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 10x-5-18x-6 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { -18x+1 }{ 15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \) -18x+1=\(\frac { 15 }{ 3 } =5\)
\(\Rightarrow\) -18x = 5 - 1 = 4
\(\Rightarrow \quad x=\frac { 4 }{ -18 } \)
Thus, x=\(\frac { -2 }{ 9 } \)

Let the number be x.
According to question,
\(x-\frac { 2 }{ 3 } x=20\)
\(\Rightarrow \quad \frac { 3x }{ 3 } -\frac { 2 }{ 3 } x=20\)
\(\Rightarrow \quad \frac { 3x-2x }{ 3 } =20\)
\(\Rightarrow \quad \frac { x }{ 3 } =20\)
\(\Rightarrow\) x = 60

Given, \(\frac{m}{4}=16\)
Taking LHS, \(\frac{m}{4}=4\times \frac{m}{4}=m\)  [multiplying by 4]
Taking RHS, 16 = 16 x 4 = 64 [multiplying by 4]
\(\therefore\) m = 64, which is the required solution.

Let the number be x.
Three-fourths of the number = \(\frac{3}{4}x\)
According to the question,
On adding 3 to it, we get 2l.
i.e.   \(\frac{3}{4}x+3=21\)
which is the required equation.
Now, to solve this equation, transposing (+3) from LHS to RHS, we get
       \(\frac{3}{4}x=21-3\quad \Rightarrow \frac{3}{4}x=18\)
On multiplying both sides by 4, we get
       \(\frac{3}{4}x\times 4=18\times 4\Rightarrow 3x=72\)
Again, dividing both sides by 3, we get
       \(\frac{3x}{3}=\frac{72}{3}\quad \Rightarrow x=24\)
Hence, the required number is 24.

Let the number be x.
\(\therefore\) One-fifth of the number = \(\frac{x}{5}\)
According to the question,
One-fifth of a number minus 4 = 3, i.e. \(\frac{x}{5}\) -4 = 3.
which is the required equation.
Now, we find the value of unknown number i.e. the value of x.
On transposing (-4) from LHS to RHS, we get
         \(\frac{x}{5}=3+4\quad \Rightarrow \frac{x}{5}=7\)
On multiplying both sides by 5, we get
         \(\frac{x}{5}\times 5=7\times 5\quad \Rightarrow x=35\)
Hence, the required number is 35.

Let the number be x.
\(\therefore\) Eight times of the number = 8x
Now, add 4 to eight times a number, we get 8x + 4
According to the question,
        8x + 4 = 60
which is the required equation.
Now, we find the value of unknown number, i.e. the value of x.
We have, 8x + 4 = 60
On transposing (+ 4) from LHS to RHS, we get
       8x = 60 - 4  \(\Rightarrow\)  8x = 56
On dividing both sides by 8, we get
       \(\frac{8x}{8}=\frac{56}{8}\quad \Rightarrow x=7\)
Hence, the required number is 7.

Given equation is 3m - 14 = 4.
When m = 0, then LHS = 3\(\times\)0 - 14 = 0 -14 = -14
and  RHS = 4
∴  LHS ≠ RHS
When m = 1, then LHS = 3\(\times\)1 - 14 = 3 - 14 = - 11
and  RHS = 4
∴  LHS ≠ RHS
When m = 2, then LHS = 3\(\times\)2 - 14 = 6 - 14 = - 8
and  RHS = 4  ⇒  LHS ≠ RHS
When m = 3, then LHS = 3\(\times\)3 - 14 = 9 - 14 = -5
and  RHS = 4  ⇒  LHS ≠ RHS
When m = 4, then LHS = 3\(\times\)4 - 14 = 12 - 14 = -2
and  RHS = 4  ⇒  LHS ≠ RHS
When m = 5, then then LHS = 3\(\times\)5 - 14 = 15 - 14 = 1
and  RHS = 4  ⇒  LHS ≠ RHS
When m = 6, then LHS = 3\(\times\)6 - 14 = 18 -14
and RHS = 4  ⇒  LHS = RHS
Hence, m = 6 is the solution of the given equation.

Given equation is 5P + 2 = 17.
When P = 0, then LHS = 5\(\times\)0 + 2 = 0 + 2 = 2
and    RHS = 17
∴ LHS ≠ RHS
When p = 1, then LHS = 5\(\times\)1+ 2 = 5 + 2 = 7
and RHS = 17
∴  LHS ≠ RHS
When p = 2, then LHS = 5\(\times\)2 + 2 = 10 + 2 = 12
and  RHS =17
∴  LHS ≠ RHS
When p = 3, then LHS = 5\(\times\)3 + 2 = 15 + 2 = 17
∴  LHS = RHS
So, p = 3 is the solution of the given equation.