# Thread: Price modelling - Simultaneous Linear Equations.

1. ## Price modelling - Simultaneous Linear Equations.

I decided to put my new algebra skills into practice recently. But it didn't work!

I obtained a price list the other day for some products we are looking to buy. Their price decreases the more you buy.

between 50-99 qty their price is £8.76
between 100-199 qty their price is £6.57

There will become a point where it's the same price to buy x amount in the lower price bracket than it would to buy the same money's worth in the higher price bracket. I.e, you'd get more for the same price.

I tried using linear equations for both prices and solving them simultaneously to find a point of intersection where both equations were equal. But the lines only intersect at 0.

I know it is possible to do this. I'm just stumbling somewhere. Bit of help would be good. I am thinking I may need to add a value to the constant of the linear equations, but am unsure where to start.

2. There are multiple points where you get more for the same price.
The prices lie between 99*€8.76=€867.24 and 100*€6.57 = €657
So, you can get either 99 or x*6.57=867.24 => x = 132 products for €867.24
but you can also get 100 or x*8.76=657 => x = 75 products for €657
So, from 77 products and onwards, you should be buying more, as to end up in the higher bracket.
I used € because can't find how to get pounds on here.
Also, I realized I did some redundant math, but I decided to leave it there if you might need it/are interested in it.

3. Thanks.

I appreciate your input, however, I think you may have missed my point (probably my poor explanation).

I know from experimenting with this that it I can buy 100 at £6.57 or 75 £8.76.

I was hoping there was a way to model this, linearly, and solve linearly. More out of curiosity, and dare I say it, fun?

I know both the prices can be modelled linearly, I just need an equation to do it. In 'my' theory, their lines should intersect at 75?

Am I trying to do the impossible here?

4. I don't think you can get intersecting lines.
You could think of this as solving the equation 100*6.57 = x*8.76
$f(x)=8.76*x$

$D_f = 50 \leq x \leq 99$

$g(x)=6.57*x$

$D_g = 100 \leq x \leq 199$

$f(x_1) = g(x_2)$

$8.76*x_1=6.57*x_2$

$x_1 = 0.75*x_2$

Here you can plug in $x_2 = 100$
Therefore you can conclude that
$x_1 = 0.75*x_2$ can be (validly) computed for

$75 \leq x_1 \leq 99$

$100 \leq x_2 \leq 132$

I must add that I have no idea if this last bit is real mathematics. It's how I imagined you could solve it using equations.