a positive number is 5 times …

Let the numbers be x and 5x.
If 21 is added to both the numbers, then first new number = x + 21
and, second new number = 5x + 21
Case I. When first new number is twice the second new number.
Then,
x + 21 = 2(5x + 21)
∴ x + 21 = 10x + 42
∴ x – 10x = 42 – 21 . . . . [Transposing 10x to L.H.S. and 21 to R.H.S.]
∴ – 9x = 21
∴ {tex}x = - \frac{{21}}{9}{/tex}. . . . [Dividing both sides by –9]
∴ {tex}x = \frac{{ - 21 \div 3}}{{9 \div 3}}{/tex}
∴ {tex}x = - \frac{7}{3}{/tex}
∴ {tex}5x = \left( { - \frac{7}{3}} \right) \times 5 = - \frac{{35}}{3}{/tex} hence the numbers are {tex} - \frac{7}{3}{/tex} and {tex} - \frac{{35}}{3}{/tex}.
This case is inadmissible as the required numbers are positive.
Case II. When second new number is twice the first new number.
Then, 5x + 21 = 2(x + 21)
∴  5x + 21 = 2x + 42
∴  5x – 2x = 42 – 21 . . . . [Transposing 2x to L.H.S. and 21 to R.H.S.]
∴ 3x = 21
∴ {tex}x = \frac{{21}}{3}{/tex} . . . . [Dividing both sides by 3]
∴ x = 7
∴ 5x = 5 × 7 = 35  hence, the numbers and 7 and 35.