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

We have,
2q - 6 = 0
\(\Rightarrow\) 2q - 6 + 6 = 0 + 6
[Adding 6 to both sides]
\(\Rightarrow\) 2q = 6
\(\Rightarrow \quad \frac { 2a }{ 2 } =\frac { 6 }{ 2 } \)
[Dividing both sides by 2]
\(\Rightarrow\) q = 3
So, q = 3 is the solution of the given equation.

-1

When, n = 2
then 7n + 5 = 7 \(\times\) 2 + 5 = 14 + 5
= 19
So, n = 2 is the solution of the given equation.

x + 5 = 6, Add x and 5 to get 6.

250

Given, equation 4(4P + 6) =16
Dividing both sides by 4, we get
         \(\frac{4(4p+6)}{4}=\frac{16}{4}\quad \Rightarrow\) 4p + 6 = 4
Now, subtracting 6 from both the sides, we get
           4P + 6 - 6 = 4 - 6  \(\Rightarrow\) 4P = - 2
Now, dividing both sides by 4, we get
           \(\frac{4p}{4}=-\frac{2}{4}\quad \Rightarrow p=-\frac{1}{2}\)
which is the required solution.

Let runs scored by Rahul be x.
Then, runs scored by Sachin = Twice of runs scored by Rahul = 2x
\(\therefore\) Sum of their runs = x + 2x = 3x
Since the sum of their runs be two short of a double century.
Therefore, we get the equation
        3x + 2 = 200
To solve this equation, transposing (+ 2) from LHS to RHS, we get
       3x = 200 - 2  \(\Rightarrow\)  3x = 198
On dividing both sides by 3, we get
          \(\frac{3x}{3}=\frac{198}{3}\)
\(\Rightarrow\)         x = 66
Hence, the runs scored by Rahul is 66 and by Sachin = 2 x 66 = 132.

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.

and given. = 3
From (a), y + y + y + y + y = x + x + 3 + 3
\(\Rightarrow\)  5y = 2x + 6   \(\Rightarrow\)   5y - 2x = 6
\(\Rightarrow\)  2x - 5y = - 6 ...(i)
From (b), x + x = y + y + 3 + 3
\(\Rightarrow\)  2x = 2y + 6   \(\Rightarrow\)   2x - 2y = 6
\(\Rightarrow\)  x - y = 3 [dividing both sides by 2] ...(ii)
From (c), y + y + y + 3 + 3 + 3 = x + x + x
\(\Rightarrow\)  3 y + 9 = 3x  \(\Rightarrow\)  3x - 3y = 9
\(\Rightarrow\)  x - y = 3 [dividing both sides by 3] ...(iii)
From Eq. (iii), x - y = 3  \(\Rightarrow\) x = y + 3
On putting x = y + 3 in Eq. (i). we get
2(Y + 3) - 5Y = -6   \(\Rightarrow\)   2y + 6 - 5Y = -6
-3y + 6 = - 6   \(\Rightarrow\)   -3 y = - 6 - 6 = -12
             \(y=\frac{12}{3}=4\)
On putting y = 4 in Eq. (ii), we get
x - y = 3  \(\Rightarrow\)  x - 4 = 3
\(\Rightarrow\)  x = 3 + 4 = 7   \(\Rightarrow\)  x = 7
:. The value of  \(\Delta\)  = x = 7 and the value of • = y = 4.

Given, number of prizes = 30
Total prize money = Rs 52000
and 1st and 2nd prizes are worth Rs 2000 and Rs 1000 respectively.
(a) If 1stprizes are x in number, the number of 2nd prizes are 30 - x because total number of prizes are 30.
(b) The total value of prizes in terms of x are 2000 x + 1000 (30 - x).
(c) The equation formed is
           1000x + 30000 = 52000
\(\therefore\) From (b) 2000x + 1000 (30 - x) = 52000
       2000x + 30000 -1000x =  52000
                  1000x + 30000 = 52000
(d) The solution of the equation is 52.
\(\therefore\) From (c), 1000x + 30000 = 52000
        1000x = 52000 - 30000 = 22000
                 \(x=\frac{22000}{1000}=22\)
(e) The number of 1st prizes are 22 and the number of 2nd prizes are 8.
\(\therefore\) From (b), 2000x + 1000(30 - x) = 52000
          2 x + 30 - x = 52 [dividing both sides by 1000]
     x + 30 = 52  \(\Rightarrow\) x = 52 - 30 = 22
\(\therefore\) Number of 2nd prizes = 30 - 22 = 8.

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, x + 15 = 19  \(\Rightarrow\) x = 19 - 15  \(\Rightarrow\) x = 4
So, the solution of the equation x + 15 = 19 is 4.

Given, 4x + 5 =-7
4x = -7 - 5  \(\Rightarrow\) 4x = -12 \(\Rightarrow \frac{-12}{4}=-3\)
So, in natural numbers, 4x + 5 = - 7 has no solution.

Given, 3x + 10 = 7
3x = 7 -10  \(\Rightarrow\) 3x = - 3
\(\Rightarrow x=\frac{-3}{3}=-1\)
-1 is the solution of 3x + 10 = 7.

Given, x + 2 = 7  \(\Rightarrow\)  x = 7 - 2 = 5
So, if x + 2 = 7, then x = 5