# Solving 3 variables - simultaneous equations

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!

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 ## Homework 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. ## 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 ## Answers and Replies Related Precalculus Mathematics Homework Help News on Phys.org Ray Vickson Science Advisor Homework Helper Dearly Missed 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! ## Homework Statement 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

## Homework 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.

## 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

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: