Quick algebra question

Dick and Jane bought some party supplies. Dick bought 6 packages of balloons and 2 packages of favors for $12.60. Jane bought 2 packages of balloons and 1 package of favors for $4.00. Find the price of a package of balloons.
a.) $2.30
b.) $2.10
c.) $2.00
d.) $1.30

Let b = price of balloons and f = price of favors.
6b + 2f = 12.60
2b + f = 4.00
solve the system of equations:
f = 4.00 - 2b
6b + 2(4.00 - 2b) = 12.60
6b + 8 - 4b = 12.60
2b = 4.60
b = $2.30
Answer is (a).

Note, interestingly, f = -$0.60 < 0. This question is poorly designed.

We have to use simultaneous equations here.
We'll denote balloons by B and favors by F.

So,

6B + 2F = 12.6

Dividing by 2, we obtain

3B + F = 6.3
F = 6.3 - 3B

Then we use the other equation, that is

2B + F = 4
F = 4 - 2B

Now we can use the substitution method:

We substitute F = 6.3 - 3B into F = 4 - 2B

So, we get

6.3 - 3B = 4 - 2B
-3B + 2B = 4 - 6.3
-B = -2.3
B = 2.3

So, price of a package of balloon is $2.30.

Answer is A

Let b = cost of the balloons and f = cost of favors

6b + 2f = 1260
2b + f = 400

I get the balloons cost $2.30, but the favors cost -60¢, so that can't be right.