1. If the units digit of a two digit number is six more than the tens digit and the number is eight more than twice the sum of the digits, what is the number?

let the tens digit be x

then the units digit is x+6

then the number is 10x + (x+6) = 11x + 6
sum of the digits = x + x+6 = 2x+6

11x + 6 = 2(2x+6) + 8
7x = 14
x = 2

the number is 28

To solve this problem, we'll need to carefully translate the statement into equations.

Let's assume the tens digit of the number is represented by "x," and the units digit is represented by "y."

According to the problem statement, the units digit is six more than the tens digit. This can be translated as: y = x + 6.

It is also given that the number is eight more than twice the sum of the digits. So, the number can be expressed as: 10x + y = 2(x + y) + 8.

Now, we have a system of two equations:
y = x + 6
10x + y = 2(x + y) + 8

To solve this system, we can substitute the value of y from the first equation into the second equation:

10x + (x + 6) = 2(x + (x + 6)) + 8

Simplifying the equation, we get:

10x + x + 6 = 2(2x + 6) + 8

Combining like terms:

11x + 6 = 4x + 12 + 8
11x + 6 = 4x + 20

Subtracting 4x from both sides:

7x + 6 = 20

Subtracting 6 from both sides:

7x = 14

Dividing both sides by 7:

x = 2

Now, we can substitute the value of x back into the first equation to find y:

y = x + 6
y = 2 + 6
y = 8

Therefore, the number is 28.