Answer:
14, 21, and 63
Explanation:
we can set the three unknown numbers as x, y, and z
so
x + y + z =98
now we are given that the first (x) is seven less than the second (y)
x = y - 7
the third number (z) is three times the second (y)
z = 3y
since we have an equation for z and x in terms of y, we can plug it into the first equation and solve for y.
(y-7) + y + (3y) =98
y - 7 + y + 3y =98
y + y + 3y =105
5y = 105
y=21
now that we know y, we can plug its value into either the second or the third equation to find x or z (I chose to find z first)
z = 3y
z = 3(21)
z = 63
now we can plug in y to find x
x = y - 7
x = 21 - 7
x = 14
so the three numbers should be 14, 21, and 63