# Thread: Solving two equations with three variables

1. ## Solving two equations with three variables

We have been taught that in order to solve three variables, you need three equations. Here is a two equations with three variables that can be solved: 8X+2Y+Z=48 AND X+Y+Z=12; Solve for X, Y and Z.

Hints: You may use limits an inequalities to get to simultaneus equations

2. Originally Posted by lgsalmon
We have been taught that in order to solve three variables, you need three equations. Here is a two equations with three variables that can be solved: 8X+2Y+Z=48 AND X+Y+Z=12; Solve for X, Y and Z.

Hints: You may use limits an inequalities to get to simultaneus equations
What are the restrictions on $\displaystyle X, Y,$ and $\displaystyle Z$? Otherwise, you will get a free variable when solving for $\displaystyle X, Y$, and $\displaystyle Z$ with no restrictions.

3. ## Restriction

The problem is maths professor gave you $48 to buy fruits for your 12 claamates (including you). Apples are$8 each, Oranges are $2 each and Peaches are$1 each.

You need to buy twelve fruits and use all \$48, no change remain. How many of each in that dozen fruits?

4. So X, Y, and Z must be positive integers. That's important information!

From X+ Y+ Z= 12 and 8X+2Y+Z=48, multiply the first equation by 2 and subtract from the second: 6X- Z= 24. Because X, Y, and Z must be integers, that is what is known as a "Diophantine" equation. Notice that 6(1)- 5= 1. Multiplying by 24, 6(24)- (120)= 24 so X= 24, Z= 120 would work. Unfortunately, that gives Y= 12- X- Z= 12- 24- 120= -132.

However If X and Z are any solution to 6X- Z= 24, so is X+ n, Z+ 6n for n any integer because 6(X+ n)- (Z+6n)= 6X- Z+ 6n- 6n= 6X- Z= 24. If we put 24+ n and 120+ 6n into the equation Y= 12- X- Z we get Y= 12- 24- n+ -120- 6n= -132- 7n. That will be larger than 0 when Y= -132- 7n> 0 or -7n> -132, n< -132/7= -18.8. If we take n= -19 we get X= 24-19= 5, Z= 120- 114= 6 and Y= 12- 5- 6= 1.

If we take n to be any number larger than -19 we get Y negative. If we take n= -20, X= 24- 20= 4, Z= 120- 120= 0, and Y= 12- 4= 8.

If you must have purchased at least on of every fruit, you must have purchased 5 apples, 1 orange and 6 peaches.

If it is not required that you purchased at least one of every fruit, 4 apples and 8 oranges would also work.
Why in the world are apples so much more expensive than peaches?