SOLUTION: Carl, Diane, and Ed have $ 46.Carl has half as much as Diane, and Ed has $2 less than Diane. How much does each have?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Carl, Diane, and Ed have $ 46.Carl has half as much as Diane, and Ed has $2 less than Diane. How much does each have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 950952: Carl, Diane, and Ed have $ 46.Carl has half as much as Diane, and Ed has $2 less than Diane. How
much does each have?

Found 2 solutions by macston, stanbon:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
C=Carl's money; D=Diane's money; E=Ed's money
C=1/2D
E=D-$2
C+D+E=$46 Substitute for C and E
(1/2D)+D+(D-$2)=$46
2.5D-$2=$46 Add $2 to each side.
2.5D=$48 Divide each side by 2.5
D=$19.20 ANSWER 1: Diane has $19.20.
C=1/2D=1/2($19.20)=$9.60 ANSWER 2: Carl has $9.60.
E=D-$2=$19.20-$2=$17.20 ANSWER 3: Ed has $17.20.
CHECK:
C+D+E=$46
$9.60+$19.20+$17.20=$46
$46=$46

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Carl, Diane, and Ed have $ 46.Carl has half as much as Diane, and Ed has $2 less than Diane. How
much does each have?
--------------------------------
Equations:
c + d + e = 46
c = (1/2)d
e = d-2
----
Substitute for "c" and "e" and solve for "d"::
(1/2)d + d + d-2 = 46
(5/2)d = 48
d = 96/5
d = $19.20
c = $9.60
e = $17.20
---------------------
Cheers,
Stan H.
--------------