SOLUTION: distibute 1000 so that one part has 3 times as large as the second one and the third has the sum of the other two

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: distibute 1000 so that one part has 3 times as large as the second one and the third has the sum of the other two      Log On


   



Question 102343: distibute 1000 so that one part has 3 times as large as the second one and the third has the sum of the other two
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
First Part + Second Part + Third Part = 1000
First Part = 3 x Second Part
Third Part = First Part + Second Part
Let's assign a variable to each part.
X = First Part
Y = Second Part
Z = Third Part
X+%2B+Y+%2B+Z+=+1000 Your original equation.
X+=+3%2AY First Part = 3 x Second Part
Z+=+X+%2B+Y+=+%283%2AY%29+%2B+Y+=+4YThird Part = First Part + Second Part
X+%2B+Y+%2B+Z+=+1000
3%2AY+%2B+Y+%2B+4%2AY+=+1000 Substitute for what you know, in terms of Y. \
8%2AY+=+1000 Simplify.
cross%288%29Y%2Fcross%288%29+=+1000%2F8 Use multiplicative inverse of 8.
Y+=+125
X+=+3%2AY+=+3%2A%28125%29+=+375
Z+=+X+%2B+Y+=+125+%2B+375+=+500
So then,
First Part = 125
Second Part = 375
Third Part = 500