Let x be the mass of the liquid itself, and y the mass of the empty bottle. Then
x + y = 1 1/2 kg
x = y + 3/4 kg
Substitute x into the first equation and solve for y :
(y + 3/4 kg) + y = 1 1/2 kg
(y + 3/4 kg) + y = 3/2 kg
2y + 3/4 kg = 3/2 kg
2y = 3/2 kg - 3/4 kg
2y = 6/4 kg - 3/4 kg
2y = (6 - 3)/4 kg
2y = 3/4 kg
y = 3/8 kg
Now solve for x :
x = 3/8 kg + 3/4 kg
x = 3/8 kg + 6/8 kg
x = (3 + 6)/8 kg
x = 9/8 kg
That is, the empty bottle weighs 3/8 kg, and the liquid that fills the bottle 3/4 of the way weighs 9/8 kg.
If the bottle is completely filled, then it contains an amount of liquid weighing z such that 3/4 of z weighs 9/8 kg. Solve for z :
3/4 z = 9/8 kg
z = (9/8)/(3/4) kg
z = (9/8)•(4/3) kg
z = (9•4)/(8•3) kg
z = 36/24 kg
z = 3/2 kg
So, the total weight of the completely filled bottle is
y + z = 3/8 kg + 3/2 kg
y + z = 3/8 kg + 12/8 kg
y + z = (3 + 12)/8 kg
y + z = 15/8 kg