Question 614418
Cost of 1st handbag=x
Cost of 2nd handbag=y
Cost of 3rd handbag=z
{{{(x+y+z)/3}}}=46.70
x+y+z=46.70x3=140.10
y=3x
z=y-15.30
We have three equations and three unknowns
substitute 3x for y we get z = 3x-15.30
also x + 3x + z =140.10 or 4x + z = 140.10 or z=140.10-4x
z=3x-15.30 and z=140.10-4x so
3x-15.30=140.10-4x
combine like terms
7x=155.40
divide each side by 7
x=22.20
y=3x22.20=66.60
z=66.60-15.30=51.30
difference between the first and third is 51.30-22.20=29.10