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If x, y, z natural numbers and 5x + 4y +3z = 60 then 12<x + y + z<20

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demadcrazy | Student, Grade 10 | eNoter

Posted June 3, 2009 at 11:31 PM via web

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If x, y, z natural numbers and 5x + 4y +3z = 60 then 12<x + y + z<20

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giorgiana1976 | College Teacher | Valedictorian

Posted June 4, 2009 at 2:04 AM (Answer #1)

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Because x,y,z are natural numbers, we have:

3x<5x

3y<4y

3z<4z

3x+3y+3z<5x+4y+3z=60.

We'll take 3x+3y+3z<60 and we'll divide it by 3,

x+y+z<20

In the same way:

5x=<5x, 4y<5y, 3z<5z

5x+4y+3z=60<5x+5y+5z

Well divide by 5

12<x+y+z

But x+y+z<20

12<x+y+z<20

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fahad-alfahad | eNotes Newbie

Posted June 4, 2009 at 1:42 AM (Answer #2)

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5x + 4y + 3z = 60

2x+y+3x+3y+3z=60  ====> 2x+y+3(x+y+z)=60  divided by 3

(2x+y)/3 + (x+y+z) = 20  ====>  (x+y+z) < 20 .....(1)

because (2x+y)/3 > 0

5x + 4y + 3z = 60

5x+5y+5x-y-2z=60 ====> 5(x+y+z)-(y+2z)=60...divided by 5

(x+y+z)-(y+2z)/5=12 =======> (x+y+z)>12 ....(2)

so  12<(x+y+z)<60

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