MHB Solving equation with 2 variables. (time sensitive)

[Mywork5000]

i am a 50 year old man, lovingly arguing with my 12 year old niece who has an algebra word problem. we're stumped (I'M stumped) -- and would love to get some help. CAN YOU PLEASE HELP ME SOLVE THIS "2 VARIABLE" PROBLEM ASAP (need it by 6am pacific time on 10/7/14). thanks! PROBLEM: Ashlee makes beaded jewelry. She charges 6dollars for a bracelet and 12dollars for a necklace. She earned $168. Let "a" equal the number of bracelets and "b" equal the number of necklaces.
Problem #1: write an equation to model the situation.
Problem #2: Solve for "b".

re #1, my answer is: 6(a) + 12(b) = 168. (correct?)
re #2, I'm stumped. the way I'm TRYING to solve is to isolate "b" (aka, the number of necklaces she sold) on one side of the equation. I'm going like this:
step 1: 6(a) + 12(b) = 168
step 2: divide all groups by 12, giving me
step 3: .5(a) + b = 14. then
step 4: b = 14 - .5(a)
at this point, i hit a wall.
my niece says it can't be answered. i say, they wouldn't put it in the book if it couldn't be answered. THOUGHTS? THANKS!

 

[MarkFL]

Your work is correct...I get:

$$b=\frac{28-a}{2}$$

which is equivalent to your solution.

Now, since we are not given another equation, such as the total number of items Ashlee sold, we cannot get a unique solution, however, we know b must be a whole number, so we then conclude that the only possible solutions are:

$$(a,b)=(0,14),\,(2,13),\,(4,12),\,(6,11),\,(8,10),\,(10,9),\,(12,8),\,(14,7),\,(16,6),\,(18,5),\,(20,4),\,(22,3),\,(24,2),\,(26,1),\,(28,0)$$

But, I suspect they are not asking for the possible solutions, just for $b$ in terms of $a$, which you correctly found.

 

[Mywork5000]

mark. thank you for your stab at this. glad to hear that I'm not the only one coming to that conclusion! much appreciated.

Your work is correct...I get:

$$b=\frac{28-a}{2}$$

which is equivalent to your solution.

Now, since we are not given another equation, such as the total number of items Ashlee sold, we cannot get a unique solution, however, we know b must be a whole number, so we then conclude that the only possible solutions are:

$$(a,b)=(0,14),\,(2,13),\,(4,12),\,(6,11),\,(8,10),\,(10,9),\,(12,8),\,(14,7),\,(16,6),\,(18,5),\,(20,4),\,(22,3),\,(24,2),\,(26,1),\,(28,0)$$

But, I suspect they are not asking for the possible solutions, just for $b$ in terms of $a$, which you correctly found.