In a two-digit number the units’ digit is 7 more than the tens’ digit. The number with digits reversed is three times as large as the sum of the original number and the two digits. Find the number.

u = t + 7

10u + t = 3(10t + u + t + u)
... = 33t + 6u

4u = 32t

solve the system by substitution or elimination

To find the number, let's assume that the tens digit is "x" and the units digit is "y".

According to the given information, the units' digit is 7 more than the tens' digit, so we can write the equation: y = x + 7.

The number with digits reversed is three times as large as the sum of the original number and the two digits. This can be written as: 10y + x = 3(10x + y + x + y).

Now, let's solve the equations to find the values of x and y.

Substituting the value of y from the first equation into the second equation, we get:

10(x + 7) + x = 3(10x + (x + 7) + (x + 7))

Expanding the equation and simplifying, we get:

10x + 70 + x = 30x + 3x + 42

Combining like terms, we have:

11x + 70 = 33x + 42

Moving the variables to one side and the constants to the other side, we have:

33x - 11x = 70 - 42

22x = 28

Dividing both sides by 22, we get:

x = 28/22

Simplifying, we have:

x = 1.27

Since x represents the tens digit, it cannot be a decimal. Therefore, there is no solution to this problem.

If the number is ab,

b = a+7
10b+1 = 3(10a+b + a+b)

81 = 3(18 + 9)