After some initial confusing on my side I decided to edit the post (and the title). The reason for this is (as many people pointed out) that my original question was (I) confusing and (II) the equations had no solution. So in order to still solve this problem I had to change the post a lot. In the first part you can find the new (basic) problem and in the second part the original question (which has no solution - thank's for all the answers). 

Basic problem:
In short: I want to get the slop of two linear functions and I only know the intercept of each function as well as two additional properties of the two functions in their intersect. 

I have two linear equations:

$f\_1(x)= a-kx$

$f\_2(x)= b+lx$

with $a,b,k,l>0$ and $a>b$.

I know the value for $a$ and $b$ and I want to find the value for $k$ and $l$. I also know that $f\_1$ and $f\_2$ intersect each other at the point $f(x\*);x\*$. In this point I have two additional functional relationships:

$z\_1= {{x\* \over f\_1(x\*)} (-k)}$ 

$z\_2= {{x\* \over f\_2(x\*)} l}$ 

I know the values for $z\_1$ and $z\_2$. This means I want to determine the slope of the two linear functions in this intercept depending on the given values for $z\_1, z\_2, a, b$. 

My experiment: 

Let's donate $f(x)=y$. In the intersect of the two functions both $y$ and $x$ must be the same so we know that:

$x\* = {{a-b}\over {l+k}}$ and $y\*={{a \cdot l+b \cdot b} \over {l+k}}$

Next use the equations for $z\_1$ and $z\_2$ we get the solution for $y$: 

$y\*={{a \cdot z\_2 -b \cdot z\_1} \over {z\_2 -z\_1 }} $ which is exactly what I want. But this is as far as I get. I can't find the value for $x\*$ or $k,l$. 

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Original Problem:

Let's say I have this 4 equations:  

(I) $K= {{2c+3b}\over {b+c}}$ 

(II) $L={{-1} \over {b+c}}$

(III) $b={-0.5K \over L}$ 

(IV) $c= {{0.7K} \over L}$

I want to know $K, L, b$ and $c$. Using substitution I can get the value for $K$ which is:

$K={{2 \cdot 0.7-(-0.5) \cdot 3 \over{}0.7-(-0.5) }}=2.416$

But when I try to solve $b,c$ and $L$ I only get a true statement. But I don't think this is right. What is missing?