The ages of three brothers are three consecutive even integers. If the sum of the 1st, four times the 2nd, and twice the 3rd is 86, what is the age of the oldest brother?
the numbers are x-2,x,x+2
x-2 + 4x + 2(x+2) = 86
7x +2 = 86
x = 12
So, the brothers are 10,12,14
Let's represent the ages of the three brothers as x, x+2, and x+4, where x is the age of the youngest brother.
According to the problem, the sum of the 1st (x), four times the 2nd (4(x+2)), and twice the 3rd (2(x+4)) is 86.
x + 4(x+2) + 2(x+4) = 86
Now we can solve for x:
x + 4x + 8 + 2x + 8 = 86
7x + 16 = 86
7x = 86 - 16
7x = 70
x = 70/7
x = 10
So the youngest brother is 10 years old.
The age of the middle brother is x+2 = 10+2 = 12 years old.
And the age of the oldest brother is x+4 = 10+4 = 14 years old.
Therefore, the age of the oldest brother is 14 years old.
To solve this problem, let's first represent the ages of the three brothers using variables. Let's assume that the age of the youngest brother is x. Since they are consecutive even integers, the ages of the three brothers will be x, x + 2, and x + 4.
Now, let's form an equation based on the given information. According to the problem, the sum of the 1st brother's age, four times the 2nd brother's age, and twice the 3rd brother's age is 86. Translating this into an equation, we get:
x + 4(x + 2) + 2(x + 4) = 86
Simplifying this equation, we have:
x + 4x + 8 + 2x + 8 = 86
7x + 16 = 86
7x = 86 - 16
7x = 70
x = 70 / 7
x = 10
Therefore, the age of the youngest brother (x) is 10. To find the age of the oldest brother, we add 4 to x:
Oldest brother's age = x + 4 = 10 + 4 = 14
Hence, the age of the oldest brother is 14.