Question 595500
if the average of x, y, and 35 is 20 less than the average of y, z, and 100,
{{{(x+y+35)/3}}} = {{{(y+z+100)/3}}} - 20
multiply eq by 3, results
x + y + 35 = y + z + 100 - 60
x + y + 35 = y + z + 40
x + y = y + z + 40 - 35
x + y = y + z + 5
Subtract y from both sides
x = z + 5
Subtract z from both sides
x - z = 5