SOLUTION: 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 f

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.

Let a,b,c, and d respectively represent the first four numbers.

That tells us that

 a + b + c + d = 9900

That tells us that

b = a + a

That tells us that

c = (a + b) + 300

That tells us that

d = (a + b + c) + 300

So we have the system:



Simplify by removing parentheses and
multiplying the second one through by 7
to clear of fractions.



Simplify by combining terms in the 2nd one:



Simplify by rearranging terms:



Add the 1st and 4th equations term by term:







So substitute that in the first three equations: 




 
Add the 1st and 3rd equation:






Substitute that in the first two
equations:





Rearrange:



To eliminate the a's multiply the first equation by 8



Add term by term:




a = 1050, b = 1200, c = 2550, d = 5100 

Edwin