Let a = first number, b = 2nd number, and c = 3rd number

(first number exceed twice the second number by 4)

a = 2b + 4

(third number is twice the first)

c = 2a or c = 2(2b +4)

(the sum of the three numbers is 68)

a + b + c = 68

Let's s rewrite the equation using only one variable

a + b + c = 68

2b + 4 + b + 2(2b + 4) = 68

2b + 4 + b + 4b + 8 = 68

7b + 12 = 68

7b = 56

b = 8

Substitute b in the equation to get the value of a and c

a = 2b + 4 c = 2(2b +4)

a = 2(8) + 4 c = 2(2(8) +4)

a = 20 c = 40

Therefore the three numbers are 20, 8, and 40

(The sum of three consecutive integers is 51. Find the integers)

Let 'x' = first number,

(since they are consecutive numbers)

x + 1 = 2nd number

x + 1 = 2nd numberx + 2 = 3rd number

x + x + 1 + x + 2 = 51

3x + 3 = 51

3x = 48

x = 16

x + 1 = 17 x + 2 = 18

Therefore the three numbers are 16, 17, and 18.