Separate the number 41 and the two parts so that the first number is eight more than twice the second number what are the two numbers

The two numbers are 30 and 11

Solution:

Given that we have to separate the number 41 into two parts

Let the second number be "x"

Given that first number is eight more than twice the second number

first number = eight more than twice the second number

first number = 8 + twice the "x"

first number = 8 + 2x

So we can say first number added with second number ends up in 41

first number + second number = 41

8 + 2x + x = 41

8 + 3x = 41

3x = 41 - 8

3x = 33

x = 11

first number = 8 + 2x = 8 + 2 (11) = 8 + 22 = 30

Thus the two numbers are 30 and 11