Let x, y, & z be the 3 numbers. Based on the information given
x+y+z = 110
z = 4y
y= x+10 substituting into the first equation
x + x + 10 +4(x+10) = 110
or 6x = 60
then x = 10
y = 20
z = 80
Let x, y, & z be the 3 numbers. Based on the information given
x+y+z = 110
z = 4y
y= x+10 substituting into the first equation
x + x + 10 +4(x+10) = 110
or 6x = 60
then x = 10
y = 20
z = 80
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