solving 3 variables - simultaneous equations

[orangesun]

hi, my sister came to me with this problem today, and i was stumped to believe this is 'year 8/9' work, but nevertheless, could you please provide me with a path, been trying to solve this for her for ages!

1. The problem statement, all variables and given/known data

p = $1.60 each, p = 40c each and a = 70c each.
At the shop she gave a $10 note and received a dozen pieces of fruit and $1 change.
how many A does she have

2. Relevant equations
I have a strange feeling I might need to use higher level maths to solve this (Gaussian) but theyre still at a young level.


3. The attempt at a solution
I set up the equations at least

1.60p + 0.40b + 0.7a = 9 (times 10 to get rid of annoying decimal)

p + b + a = 12
16p + 4b + 7a = 90

I then eliminated to get:
12b + 9a = 102
12p + 3a = 42
9p + 3b = 6

I don't know where to go from here, I hope you can help me! Thanks

[Ray Vickson]

hi, my sister came to me with this problem today, and i was stumped to believe this is 'year 8/9' work, but nevertheless, could you please provide me with a path, been trying to solve this for her for ages!

1. The problem statement, all variables and given/known data

p = $1.60 each, p = 40c each and a = 70c each.
At the shop she gave a $10 note and received a dozen pieces of fruit and $1 change.
how many A does she have

2. Relevant equations
I have a strange feeling I might need to use higher level maths to solve this (Gaussian) but theyre still at a young level.


3. The attempt at a solution
I set up the equations at least

1.60p + 0.40b + 0.7a = 9 (times 10 to get rid of annoying decimal)

p + b + a = 12
16p + 4b + 7a = 90

I then eliminated to get:
12b + 9a = 102
12p + 3a = 42
9p + 3b = 6

I don't know where to go from here, I hope you can help me! Thanks

From your two equations p + b + a = 12 and 16p + 4b + 7a = 90 you do not have enough to conditions to determine a unique solution. If you know (as is reasonable) that only *whole* numbers of fruits can be bought, so that p, b and a are whole numbers >= 0, then you can enumerate all the possibilities. For example, if we solve for a and b in terms of p we have: a = 14-4p, and b=3p-2. Since b >= 0 we must have 3p >= 2, which means that the integer p is >= 1. Now a must be >= 0, so 4p <= 14; that means that the integer p must be <= 3. So, now you know you must have p = 1, 2 or 3, and for each p-value you can figure out a and b.

RGV

[BloodyFrozen]

hi, my sister came to me with this problem today, and i was stumped to believe this is 'year 8/9' work, but nevertheless, could you please provide me with a path, been trying to solve this for her for ages!

This actually is around year 8 work (advanced), but they might not necessarily realize it. It involves elimination, and/or ref/rref. A student could easily recognize this "method"

Anyways, back to your question. I'd recommend RGV's method. It takes a little "deductive reasoning". Good Luck:tongue2: