Algebra problem help!?
Heres my question:
Jack bought four things at a store.
The first item cost as much as the second plus half of the third.
The second costs as much as the fourth minus the cost of the third.
The third cost one third of the first.
The fourth cost as much as the second and the third together.
He paid $80 in total. How much did Jack pay for each of the items?
Thanks!
Update:Also, please show your work, thanks!!!
Comments
Let "f" be the cost of the first item
"s" is cost of the second item
t is the cost of the third item
o is the cost of the fourth item, alright?
f+s+t+o = 80
f = s + 1/2(t)
s = o - t
t = 1/3( f )
o = s + t
Now you have to use substitution.
[ s + 1/2(t) ] + [ o-t ] + [ 1/3(f) ] + [s+t] = 80
[ s + 1/2(t) ] + [ o-t ] + [ 1/3( s + 1/2(t) ) ] + [s+t] =80
[ (o-t) + 1/2(t) ] + [ o-t ] + [ 1/3( (o+t) + 1/2(t) ) ] + [(o-t)+t] =80
[o-1/2(t)] + [o-t] + [1/3(o) + 5/6(t)] + [o] =80
Simplify. Arghh. I am working frantically. I can't explain what I did. Here are the correct answers:
You end up with:
1) 24
2) 20
3) 8
4) 28
Let's call the cost of the four items Jack bought a, b, c and d. Then a + b + c + d = $80.
From the information given, a = b + c/2, b = d - c, c = a/3, and d = c + b.
From that we can infer this:
c = a/3 -----> 3c = a
a = b + c/2
3c = b + c/2 (We substituted 3c for a.)
6c = 2b + c (We multiplied the equation above by 2 to clear fractions.)
5c = 2b ----> b = 5c/2.
From all this information we get this:
a + b + c + d = (3c) + (5c/2) + c + (5c/2 + c) = 80.
Recall that d = b + c , but b = 5c/2, as we just found above. Therefore, d = (5c/2 + c).
Now we have the sum in terms of c only, for which we can solve and plug into the other expressions to find other values:
(3c) + (5c/2) + c + (c + 5c/2) = 5c + 2 (5c/2) = 10c = 80
10c = 80
c = 80/10
c = 8.
Now we can calculate a:
c = a/3
3c = a
3 (8) = a
24 = a.
With a we can now find b, since we also know c:
a = b + c/2
24 = b + (8)/2
24 = b + 4
24 - 4 = b
20 = b.
Now that we know b and c, we can use that information to find d:
d = b + c
d = 20 + 8
d = 28.
Now let's plug all our values into the original equation to make sure they sum to $80:
a + b + c + d = $80
$24 + $20 + $8 + $28 = $80
$80 = $80.
The equation balances, so we seem to have the correct answers.
a = $24
b = $20
c = $8
d = $28.
well if its an odd # prob look in the back of your text book!..haha lol...heres the answer!
1.) 24
2.) 20
3.) 8
4.) 28
=$80