Find the slope of two linear functions knowing only the intercept of each function and the properties at the intercept of the functions

Question

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?

Answer 1

From the first two equations it follows that $$\frac{K}{L}=-3b-2c. $$

Substituting this into the last two and solving for $b$ and $c$ yields

$$b=c=0.$$

Since $b+c=0$ is in the denominator of the first two equations it follows that the system does not have a solution.

Answer 2

With the given equations the system has not solutions, indeed

$$\frac{K}L=-(2c+3b)=\frac{1K}{10L} \iff \frac{K}{L}=0 \implies K=0\quad a=b=0$$

Assuming that (II) is $L={{2b-3c} \over {b+c}}$ we have

$$b={-0.5K \over L} \quad c= {{0.7K} \over L} \implies b+c=\frac{K}{5L} \quad 2c+3b=-\frac{1K}{10L} \quad 2b-3c=\frac{21K}{10L} $$

then

$$K= {{2c+3b}\over {b+c}}=-\frac12$$

$$L={{2b-3c} \over {b+c}}=\frac{21}2$$

and from here find $b$ and $c$.

If (II) is $L={{2c-3b} \over {b+c}}$ similar method can be used.

Answer 3

The first part of this answer addresses the revised question (the "Basic Problem").

We know that \begin{align} f_1(x) &= a - kx \tag1\\ f_2(x) &= b + lx \tag2\\ z_1 &= \frac{x_*}{f_1(x_*)} (-k), \tag3\\ z_2 &= \frac{x_*}{f_2(x_*)} l, \tag4 \end{align} for some unknown value $x_*,$ Also, we know that the two functions intersect at $x = x_*$, that is, $$ f_1(x_*) = f_2(x_*).$$

Let $y_* = f_1(x_*).$ We multiply both sides of $(3)$ by $y_*$ to get $$z_1 y_* = -kx_*.$$ But from $(1)$ we know that $-kx_* = f_1(x_*) - a = y_* - a,$ so $$z_1 y_* = y_* - a$$ and therefore $$a = (1 - z_1) y_*. \tag5$$

We multiply both sides of $(4)$ by $y_*$ to get $$z_2 y_* = lx_*.$$ From $(2)$ we know that $lx_* = f_2(x_*) - b = y_* - b,$ so $$z_2 y_* = y_* - b$$ and therefore $$b = (1 - z_2) y_*. \tag6$$ Note that $(5)$ and $(6)$ together withe the given inequalities $a > b > 0$ imply that $z_1 \neq 1$ and $z_2 \neq 1.$

Multiplying both sides of $(5)$ by $1 - z_2$ and both sides of $(6)$ by $1 - z_1,$ we find that $$ a(1 - z_2) = (1 - z_1)(1 - z_2) y_* = b(1 - z_1), \tag7$$ that is, $$a(1 - z_2) = b(1 - z_1), \tag8$$ so if you know any three of the four values $a, b, z_1,$ and $z_2,$ you can deduce the fourth value from $(8).$ You claim to know all four of these values; do they satisfy $(8)$? If not, your problem has no solution.

If your known values do satisfy $a(1 - z_2) = b(1 - z_1),$ then there is only one unknown in $(7),$ which is solved by dividing all the equal expressions in $(7)$ by $(1 - z_1)(1 - z_2)$ (which we know is non-zero), so $$ y_* = \frac{a}{1 - z_1} = \frac{b}{1 - z_2}.$$

Now let's set $x_*$ to any arbitrary value other than zero. We can then solve for $k$ in $(1)$: $$k = \frac{a - y_*}{x_*}.$$ Using this fact to evaluate the right-hand side of $(3),$ we have $$\frac{x_*}{f_1(x_*)} (-k) = \frac{x_*}{y_*} (-k) = \frac{x_*}{y_*}\left(-\frac{a - y_*}{x_*}\right) = \frac{y_* - a}{y_*}, $$ and since we previously found that $\frac{y_* - a}{y_*} = z_1,$ Equation $(3)$ is satisfied. Similarly, we can solve for $l$ in $(2)$: $$l = \frac{y_* - b}{x_*}.$$ Using this fact to evaluate the right-hand side of $(4),$ we have $$\frac{x_*}{f_1(x_*)} l = \frac{x_*}{y_*} l = \frac{x_*}{y_*}\left(\frac{y_* - b}{x_*}\right) = \frac{y_* - b}{y_*}, $$ and since we previously found that $\frac{y_* - b}{y_*} = z_2,$ Equation $(4)$ is satisfied.

In short, we can set $x_*$ to (almost) any value we like and still find values of $k$ and $l$ such that all the given equations are satisfied. Therefore $x_*$ is not determined by the given equations.

For that reason, neither $k$ nor $l$ is determined either. On the other hand, from $(3)$ we know that $k = -\frac{y_*}{x_*} z_1$ and from $(4)$ we know that $l = \frac{y_*}{x_*} z_2,$ so $$ \frac kl = - \frac{z_1}{z_2}.$$ So there is something you can conclude about $k$ and $l,$ even if you cannot fully solve for both values.

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Answer for the original problem:

Working just with equations (I), (III), and (IV), and assuming that $K \neq 0$ and $L \neq 0$ (since otherwise parts of the equations are undefined),

\begin{align} K &= \frac{2\left(\frac{0.7K}L\right) + 3\left(-\frac{0.5K}L\right)} {-\frac{0.5K}L + \frac{0.7K}L} && \text{using (III) and (IV) to substitute $b$ and $c$} \\ &= \frac{2\left({0.7K}\right) + 3\left(-{0.5K}\right)} {-{0.5K} + {0.7K}} && \text{cancel $L$} \\ &= \frac{2\left({0.7}\right) + 3\left(-{0.5}\right)} {-{0.5} + {0.7}} && \text{cancel $K$} \\ &= -\frac12 \\ \end{align} Therefore $bL = 0.25$ (from (III)) and $cL = -0.35$ (from (IV)), and so $(b + c)L = -0.1$ and $L = -\frac{0.1}{b+c},$ contradicting (II).