Question 119484
The four oldest people in Golden City have lived a total of 384 years put together. The difference in ages for the youngest and the second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest. What are their ages? 
The ages are ____, _____, _____, and _____ .

<pre><b>
Suppose their ages, youngest to oldest are x, y, z, and w,

That is x < y < z < w

>>...have lived a total of 384 years put together...<<

So

x + y + z + w = 384

>>...The difference in ages for the youngest and the second 
oldest is 14...<<

so

z - x = 14

>>...The second youngest is 3 years older than the youngest.,,<<

so

y = x + 3


>>...The oldest is 20 years older than the average of the second oldest and youngest...<<

The average of the second oldest, z, and the youngest x, is {{{(x+z)/2}}}

So the oldest, w, is {{{(x+z)/2 + 20}}} or

w = {{{(x+z)/2 + 20}}}

So we have the system of four equations in four unknowns:


x + y + z + w = 384
z - x = 14
y = x + 3
w = {{{(x+z)/2 + 20}}} 

Clear the last one of fractions by multiplying through by 2:

x + y + z + w = 384
z - x = 14
y = x + 3
2w = x+z+40

Solve the 2nd for z, getting z = 14+x
Substitute 14+x for z in the 4th, getting 2w = x+(14+x)+40, or
2w = x+14+x+40 or 2w = 2x+54.  Then divide thru by 2, getting
w = x+27. 

Now substitute x+3 for y, 14+x for z and x+27 for w, all
in the 1st:

              x + y + z + w = 384
x + (x+3) + (14+x) + (x+27) = 384
            x+x+3+14+x+x+27 = 384
                      4x+44 = 384
                         4x = 340
                          x = 85
  
y = x+3, so y = 85 + 3 = 88
z = 14+x, so z = 85+14 = 99
w = x+27 = 85 + 27 = 112 
                        
Their ages are 85, 88, 99, and 112.

Edwin</pre>