1. ## Equation system

Hi, i would like some help on how to solve the following equation system:

$\displaystyle \begin{array}{lcr}3x+10y+0.5k = 100 \\ x+y+k = 100 \end{array}$

I know that you normally reduce down a system like this to only two variables

For equations on the form $\displaystyle ax+by=c$ there is a formula:
$\displaystyle (cu-bn,cv+an), n \in \mathbb{Z}$

But how do you solve this when you have to deal with floating point numbers?

2. $\displaystyle 3x + 11y + 0.5k = 100$ $\displaystyle / *2$

$\displaystyle 6x + 22y + k = 200$

3. okay,

So we have $\displaystyle \begin{array}{lcr}6x + 20y + k = 200 \\ x + y + k = 100 \end{array}$

If we subtract the second from the first, we end up with $\displaystyle 5x+19y=100$

But how are we supposed to get the solution for k from this equation?

4. Originally Posted by Jones
Hi, i would like some help on how to solve the following equation system:

$\displaystyle \begin{array}{lcr}3x+10y+0.5k = 100 \\ x+y+k = 100 \end{array}$

I know that you normally reduce down a system like this to only two variables

For equations on the form $\displaystyle ax+by=c$ there is a formula:
$\displaystyle (cu-bn,cv+an), n \in \mathbb{Z}$

But how do you solve this when you have to deal with floating point numbers?
Your formula above only gives integer values of x and y. It is not just that a b and c must be integers.

From 5x+ 19y= 100 you can get x= 20- (19/5)y.

Putting that back into x+y+k= 100 gives 20- (19/5)y+ y+ k= 100 or -(14/5)y+ k= 80. Then k= (14/5)y+ 80.

For any value of y, x= 20- (19/5)y and z= (14/5)y+ 80 so(20-(19/5)y, y, (14/5)y+ 80) is a solution.

If you are only interested in integer values, it is easy to see that x and z will be integers if and only if y is a multiple of 5. If you let y= 5n, that becomes (20-19n, 5n, 14n+80).

5. Thanks!

So clearly $\displaystyle n=1$ is a solution that would work...