Question

I need help on the third question

Answer 1

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

Answer 2

For number 3? I think it's 33 because 33x3= 99 and 99+20= 119.