Question 212175
This problem is from the second century. Four numbers have a sum of 9900.
 The second exceeds the first by one-seventh of the first. 
The third exceeds the sum of the first two by 300. 
The fourth exceeds the sum of the first three by 300. Find the four numbers. 
:
Write an equation for each statement:
:
Four numbers have a sum of 9900.
w + x + y + z = 9900
: 
The second exceeds the first by one-seventh of the first.
x = {{{8/7}}}w
Multiply both sides by 7
7x = 8w
: 
The third exceeds the sum of the first two by 300.
y = w + x + 300
: 
The fourth exceeds the sum of the first three by 300.
z = w + x + y + 300
Rearrange and subtract from the 1st equation:
w + x + y + z = 9900
w + x + y - z = -300
----------------------eliminates w, x, y
2z = 10200
z = 5100
:
Using the 1st equation replace z with 5100 
w + x + y + 5100 = 9900
w + x + y = 9900 - 5100
w + x + y = 4800
:
Rearrange the 3rd equation to: w + x - y = -300 subtract from the above eq;
w + x + y = 4800
w + x - y = -300
------------------eliminate w and x
2y = 5100
y = {{{5100/2}}}
y = 2550
:
Using the 1st equation replace y and z
w + x + 2550 + 51000 = 9900 
w + x = 9900 - 2550 - 5100
w + x = 2250
w = (2250 - x); use for substitution
:
Using 7x = 8w, replace w, find x
7x = 8(2250-x)
7x = 18000 - 8x
7x + 8x = 18000
x = {{{18000/15}}}
x = 1200
:
w = 2250 - x; replace x
w = 2250 - 1200
w = 1050
;
The 4 numbers
w = 1050
x = 1200
y = 2550
z = 5100
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tot:9900