a+2=e
b-2=e
2c=e
d/2=e
dunno wat last part is but since they all equal e
a+2=b-2=2c=d/2
I am guessing that a+b+c+d=45 such that a+2=b-2=2c=d/2
interesting
make in terms of one thing
let's do a
a+2=b-2
a+4=b
a+2=2c
(a+2)/2=c
a+2=d/2
2(a+2)=d
sub
a+(a+4)+[(a+2)/2]+(2(a+2))=45
distribute some
a+a+4+(a+2)/2+2a+4=45
add like terms
4a+8+(a+2)/2=45
minus 8
4a+(a+2)/2=37
mult both sides by 2
8a+a+2=74
add like terms
9a+2=74
minus 2
9a=72
divide 9
a=8
sub back
a+4=b, 8+4=b=12
(a+2)/2=c, (8+2)/2=10/2=c=5
2(a+2)=d, 2(8+2)=2(10)=d=20
a=8
b=12
c=5
d=20