A number of two digits is increased by 54 when the digits are reversed. The ten digits is three times the unit digit. Find the number​

Let n be the unknown number. We can write it as

n = 10a + b

with a and b integers between 1 and 9 (either with positive or negative sign).

Reversing the digits gives another number

m = 10b + a

The first number is increased by 54 when the digits are reversed, which means

m = n + 54   →   10b + a = 10a + b + 54   →   9b - 9a = 54   →   b - a = 6

The digit in the tens place of n is 3 times the digit in the ones place, so

a = 3b

Substitute this into the previous equation and solve for b :

b - a = b - 3b = -2b = 6   →   b = -3

Solve for a :

a = 3b = 3(-3) = -9

Then the original number is n = 10a + b = 10(-9) + (-3) = -93