Hello Cathy,
The big picture of this question is that we have three variables
x = you.
y = Julian
z = Betty
Now given the parameters of the problem we substitute in to each of the variable equations
x = 20 + y EQ 1
z = 4 + x EQ 2
x + y + z = 280 EQ 3
now we have three equations of three unknowns and are gonna solve for x y and z
substitute equation one into equation 2, and get
z=4 +20 + y
z = 24 + y
substitute equation one for x into EQ 3
20 + y + y + z = 280
2y + z = 260
substitute z = 24 + y into 2y + z = 260 and get
2y + 24+ y=260
3y=236
y=78.67
Substitute this equation one and solve for x
X = 20 + 78.67 = 98.67
Now take the values for x and y and substitute into equation 3 and solve for Z
98.67 + 78.67 + z = 280
z = 102.66
those are the answers for x, y, and z
Seems different that they all have fractional pieces