Question 39080
Let the second one be x
Let the first one be y
Let the third one be z
FIRST EQUATION:
x+20=4(y-20)
x=4y-80-20
x=4y-100 (SUBSITUTION 1)


SECOND EQUATION:
z+60=2(y-20+x+20-60)
z+60=2y-40+2x+40-60
z=2y+2x-120 (SUBSITUTION 2)


THIRD EQUATION:
z-40=2y+80


ALLOW THE SUBSITUTION:
Subsitute for z:
2y+2x-120=2y+80
2x=200
x=100

FROM 1st EQUATION:
x=4y-100
{{{y=x/4+25}}}
{{{y=100/4+25}}}
y=50


z=2(50)+2(100)-120
z=180 ( I have no idea how to get 120)
Hence, the money in three purses were $50, $100, and $180.
Paul.