j+s is no more than 120
j+s≤120
if jacket is 20 more than 3 times skirt
j=20+3s
find max price for each
so we have
j+s≤120 and
j=20+3s
sub 20+3s for j
20+3s+s≤120
20+4s≤120
minus 20 oth sides
4s≤100
divide both sides by 4
s≤25
sub back
if s is 25
j=20+3s
j=20+3(25)
j=20+75
j=95
at most
jacket=$95
skirt=$25
