The sum of the ages of Arturo, Benny, and Carlos is 41 years. Twice Arturo's age exceeds the sum of Benny and Carlos's ages by one year. Five years ago, Benny was 2 years more than twice as old as Carlos. Find their ages now

Arturo is 14 years

Benny is 17 years

Carlos is 10 years

Step-by-step explanation:

Let x represent the age of Arturo.

Let y represent the age of Benny.

Let z represent the age of Carlos.

The sum of the ages of Arturo, Benny, and Carlos is 41 years. It means that

x + y + z = 41- - - - - - - - - - - - -1

Twice Arturo's age exceeds the sum of Benny and Carlos's ages by one year. It means that

2x = y + z + 1- - - - - - - - - 2

Five years ago, Benny was 2 years more than twice as old as Carlos. It means that

y - 5 = 2(z - 5) + 2

y - 5 = 2z - 10 + 2

y = 2z - 8 + 5

y = 2z - 3

Substituting y = 2z - 3 into equation 2, it becomes

2x = 2z - 3 + z + 1

2x = 2z + z - 3 + 1

2x = 3z - 2

x = 3z/2 - 1

Substituting y = 2z - 3 and

x = 3z/2 - 1 into equation 1, it becomes

3z/2 - 1 + 2z - 3 + z = 41

Multiplying through by 2, it becomes

3z - 2 + 4z - 6 + 2z = 82

3z + 4z + 2z - 2 - 6 = 82

9z - 8 = 82

9z = 82 + 8 = 90

z = 90/9

z = 10

x = 3z/2 - 1 = (3 × 10)/2 - 1

x = 14

y = 2z - 3 = 2 × 10 - 3

y = 17

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..