The sum of three numbers is 98. The sum of the first and second is two more than the third number and the second is four times the first. Find the numbers?

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let the numbers be ;x ,y z

x+y+z=98-----------------i

x+y+2=z------------------ii

4x=y----------------------iii

or 4x-y=0

then solve

x+y+z=98

x+y-z=-2

Add equations i and ii to obtain iii

2x+2y=96------------------iv

using iii and iv

2x+2y=96

(4x-y=0)2=8x-2y=0

Add the two equations

+

8x-2y=0

10x=96

x =9.6

y=2.4

z=86

CORRECTION PLEASE CHECK

x+y-2=z------------------ii

x+y-z=2

2x+2y=100------------------iv

2x+2y=100

10x=100

x =10

y=40

z=48

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