Hi Chaos,
This will come out to a system of 3 equations with 3 unknowns.
Those can be a bit unwieldy, but if you take your time and follow
the logic trails, they aren't that bad.
First, pick variables to represent the unknowns
Let x = Artesia's age
Let y = Amuro's age
Let z = Char's age
Make equations from the information given:
Artesia is as old as the combined age of her two brothers:
x = y + z {equation 1}
Char is two years older than Amuro:
z = y + 2 {equation 2}
The combined age of the three last year was ¾ their combined ages at present:
Note: Last year their ages were.. x-1, y-1, z-1
(x-1)+(y-1)+(z-1) = (3/4)(x+y+z)
Let's simplify this down
x + y + z - 3 = (3/4)(x+y+z)
Clear the fraction by multiplying through by 4
4x + 4y + 4z - 12 = 3(x+y+z)
4x + 4y + 4z - 12 = 3x + 3y + 3z
Subtract (3x+3y+3z) from both sides
x + y + z - 12 = 0
x + y + z = 12 {equation 3}
Let's look at what we have so far:
x = y + z {equation 1}
z = y + 2 {equation 2}
x + y + z = 12 {equation 3}
We want to find Char's age now ... that is "z"
From equation 1:
y = x - z
From equation 2:
y = z - 2
so
x - z = z - 2
x = 2z - 2
So, we have x in terms of z and y in terms of z..
Substitute 2z-2 in place of x and z-2 in place of y in equation 3
(2z-2)+(z-2)+z = 12
4z - 4 = 12
4z = 16
z = 4
Char is 4 years old...
y=z-2 = 2 years old for Amuro
x = 2z-2 = 6 years old for Artesia
If you plug those 3 values into the 3 equations you will find they add up correctly.
Inactive Tutor
10/13/17