56.4

x=3

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

\(2y+\frac { 5 }{ 2 } =\frac { 37 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 37 }{ 2 } -\frac { 5 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 37-5 }{ 2 } \)
\(\Rightarrow \quad 2y=\frac { 32 }{ 2 } \)
\(\Rightarrow\) 2y = 16
\(\frac { 2y }{ 2 } =\frac { 16 }{ 2 } \)  [Dividing by 2]
Thus, y = 8

Let the number be x.
According to question,
\(\Rightarrow\) 6x + 10 = 40
\(\Rightarrow\) 6x = 40-10
[On transposing 10 to RHS]
\(\Rightarrow\) 6x = 30
\(\frac { 6x }{ 6 } =\frac { 30 }{ 6 } \)
Thus, x = 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

Since,
\(\frac { x }{ 2 } +5=10\)
\(\Rightarrow \quad \frac { x }{ 2 } =10-5\)
[On transposing 5 to RHS]
\(\Rightarrow \quad \frac { x }{ 2 } =5\)
\(\frac { x }{ 2 } \times 2=5\times 2\)
Multiplying both sides by 2
\(\Rightarrow\) x = 10

Since,
6x + 18 = 8x + 12
\(\therefore\) 6x - 8x + 18 = 12
[On transposing 8x to LHS]
\(\Rightarrow\) 6x - 8x = 12 - 18
[On transposing 18 to RHS]
\(\Rightarrow\) -2x = -6
\(\Rightarrow \quad \frac { -2x }{ -2 } =\frac { -6 }{ -2 } \)
\(\Rightarrow\) x = 3

Let consecutive numbers be x and x + 1.
Hence, according to question,
x + x + 1 = 53
\(\Rightarrow\) 2x + 1 = 53
\(\Rightarrow\) 2x = 53-1
[On transposing 1 to RHS]
\(\Rightarrow\) 2x = 52
\(\Rightarrow \quad \frac { 2x }{ 2 } =\frac { 52 }{ 2 } \)
\(\Rightarrow\) x = 26
\(\therefore\) Smaller number is 26

\(\frac { x }{ 2 } -1=\frac { x }{ 3 } +4\)
\(\Rightarrow \quad \frac { x }{ 2 } -\frac { x }{ 3 } =4+1\)
[ 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

4q-12 = 0
\(\Rightarrow\) 4q = 12
Thus, q = 3

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: 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 ages of A and B are 5x and 3x.
After six years their ages will be
5x + 6 and 3x + 6
So, according to question,
\(\frac { 5x+6 }{ 3x+6 } =\frac { 7 }{ 5 } \)
\(\Rightarrow\) 5(5x+6) = 7(3x+6)
\(\Rightarrow\) 25x + 30 = 21x + 42
\(\Rightarrow\) 25x - 21x = 42 - 30 = 12
\(\Rightarrow\) 4x = 12 \(\Rightarrow\) x = 3
So the present age of A = 5x = 5 \(\times\) 3 = 15 years
and the present age of B = 3x = 3 \(\times\) 3 = 9 years.
(b) Solution of simple equations.
(c) Value: Time waits for none.

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

Since, \(\frac { 2x-1 }{ 3 } =\frac { x-2 }{ 3 } +1\)
[On transposing \(\frac { x-2 }{ 3 } \) to LHS]
\(\Rightarrow \quad \frac { 2x-1 }{ 3 } -\frac { x-2 }{ 3 } =1\)
\(\Rightarrow \quad \frac { (2x-1)-(x-2) }{ 3 } =1\)
\(\Rightarrow \quad \frac { 2x-1-x+2 }{ 3 } =1\)
\(\Rightarrow \quad \frac { x+1 }{ 3 } =1\)
\(\Rightarrow\) x+1 = 3 \(\times\) 1
\(\Rightarrow\) x + 1 = 3
\(\Rightarrow\) x = 3-1
Thus, x = 2

Since, \(3\left( x+\frac { 1 }{ 2 } \right) =18\)
Dividing both sides by 3, we get
\(x+\frac { 1 }{ 2 } =\frac { 18 }{ 3 } \)
\(\Rightarrow \quad x+\frac { 1 }{ 2 } =6\)
\(\Rightarrow x=6-\frac { 1 }{ 2 } \)
[ On transposing to RHS]
\(\Rightarrow \quad x=\frac { 12-1 }{ 2 } =\frac { 11 }{ 2 } \)