4+5(p-1) = 34
\(\Rightarrow\) 4+5p-5 = 34
\(\Rightarrow\) 5p = 35
\(\Rightarrow\) p = 7

We have,
\(\frac { 5 }{ 2 } x=-10\)
\(\Rightarrow \quad \frac { 5 }{ 2 } x\times \frac { 2 }{ 5 } =-10\times \frac { 2 }{ 5 } \)
[Multiplying both sides by \(\frac { 2 }{ 5 } \)]
\(\Rightarrow\) x = -4

The given equations in statement form are as follows:
4 added to 3 times a number pis 25.

4P = 21, Four times a number P is 21.

One-fourth of n is\(\frac { n }{ 4 } \) It  is greater than 6 by 4 means the difference \(\frac { n }{ 4 } \)-6 is 4
Thus, the required equation is\(\frac { n }{ 4 } \) -6 =4

2

\(\frac{40}{19}\)

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 ABC be an isosceles triangle, whose base angles are equal and let measure of base angle be x", Also, vertex angle is 40°.
Since the sum of three angles of a triangle is 180°
\(\therefore \quad \angle A+\angle B+\angle C=180^o\)
\(\Rightarrow\)          40^o + x + x = 180^o
\(\Rightarrow\)              40^o + 2x = 180^o
which is the required equation.
To solve this equation, transposing (+ 40) from LHS to RHS, we get
        2x = 180° - 40°  \(\Rightarrow\)  2x = 140°
On dividing both sides by 2, we get
        \(\frac{2x}{2}=\frac{140^o}{2}=70^o\quad \Rightarrow x=70^o \)
Hence, the base angles of the triangle are of measure 70°.

We have, 4 = 5(p - 2)
We know that an equation remains same when the expression on the LHS and on the RHS are interchanged. So, the given equation can be written as 5{p- 2) = 4
On dividing both sides by 5, we get
         \(\frac{5(p-2)}{5}=\frac{4}{5}\quad \Rightarrow\quad p-2=\frac{4}{5}\)
On transposing (-2) from LHS to RHS, we get
        \(p=\frac{4}{5}+2\quad \Rightarrow \quad p=\frac{4}{5}+\frac{2}{1}\)
\(\Rightarrow \quad p=\frac{4+10}{5}=\frac{14}{5}\) [LCM of 5 and 1 = 5]
Hence, \(p=\frac{14}{5}\) is a solution of the given equation.

(i) Let the number be x.
Now, 1 is subtracted from a number, then
(x -1), the difference is multiplied by \(\frac{1}{2}\)
Here, \(\frac{1}{2}\) (x - 1) gives result 7
\(\therefore \frac{1}{2}(x-1)=7\)
(ii) Let age of Soh an be x yr. Then, the age of
Mohan is (x + 3) yr.
\(\therefore\) Sum of their ages = 43
\(\Rightarrow\)   x+(x+ 3)=43
(iii) Let the number be x. Then, five times of number be 5x.
Since it is increased by 7.
So, 5x + 7 gives result 27.
Hence, the equation is 5x + 7 = 27.

Let x be the age of Amit.
So, age of Sohan Lal, the father of Amit = 4x yr
If the difference of their ages is 27 yr.
Then, 4x- x = 27 \(\Rightarrow\) 3x = 27 \(\Rightarrow x=\frac{27}{3}=9\)
Hence, age of Amit is 9 yr.

Given, \(\frac{1}{6}-x=\frac{1}{6}\Rightarrow \frac{1-6x}{6}=\frac{1}{6}\)
1 - 6x = 1  \(\Rightarrow\)  6x = 1 - 1   \(\Rightarrow\)  6x = 0   \(\Rightarrow\)  x = 0
So, if \(\frac{1}{6}-x=\frac{1}{6}\) then x = 0.

Given, \(\frac{9}{5}x=\frac{18}{5}\quad \Rightarrow x=\frac{18\times 5}{5\times 9}\)
\(\Rightarrow x=\frac{18\times 1}{1\times 9}=2\)
So, if \(\frac{9}{5}x=\frac{8}{5}\) then x = 2.

Given \(\frac{x}{2}=6 \Rightarrow\) = x = 6 x 2 = 12
\(\therefore\)  12 - (-3) = 12 + 3 = 15
So, x - (-3) = 15, When \(\frac{x}{2}=6\)

Given, 4x - 2 = 8
4x = 8 + 2  \(\Rightarrow\) 4x = 10 \(\Rightarrow x=\frac{10}{4}\)
So, in integers 4x - 2 = 8 has no solution.

Given, 3x - 5 = 7
3x = 7 + 5  \(\Rightarrow\)  3x = 12  \(\Rightarrow x=\frac{12}{3}=4\)
\(\therefore\) 4 is the solution of the equation 3x - 5 = 7.