Black-Scholes and solving for both $r$ and $\sigma$ ; Do I have a unqiue solution?

Question

Below is a problem that I am working on. I believe that my incomplete solution is correct as far as it goes. I would like to know if my solution is incorrect. I plan to solve the system of two equations by using a computer program such as SciLab. What I am wondering is, does this system of equations have a unique solution? I believe it does but I cannot offer any real proof.

Problem:

Recall the following equations govern the price of a call option according to the Black-Scholes model. \begin{align*} c &= S_0 N(d_1) - Ke^{-rT}N(d_2) \\ d_1 &= \frac{ \ln{ \frac{S_0}{K} } + ( r + \frac{ {\sigma}^2}{2}) T } { \sigma \sqrt{T } } \\ d_2 &= d_1 - \sigma \sqrt{T} \\ \end{align*} Here is an explanation of the $6$ variables of the model: \begin{align*} c &-\text{The price of the call option} \\ S_0 &- \text{ The initial price of the stock } \\ K &- \text{ The strike price of the option } \\ T &- \text{ The time to expiration of the option.} \\ r &- \text{ The interest rate } \\ \sigma &- \text{ A measure of how volatily the price of the under lying stock is.} \end{align*} Suppose we have a stock with an initial price of $100$. There are two call options with strike prices $100$ and $105$ They both expire in exactly one year. The price of these options are $10$ and $8$ respectively. Find $r$ and $\sigma$.

Answer:

We have: \begin{align*} S_0 &= 100 \\ T &= 1 \\ c_1 &= 10 \\ K_1 &= 100 \\ c_2 &= 8 \\ K_2 &= 105 \\ \end{align*} Now we find $d_{11}$ By $d_{11}$, I mean $d_1$ for the first call option. Similar, by $d_{12}$, I mean $d_2$ for the first call option. \begin{align*} d_{11} &= \frac{ \ln{\left( \frac{100}{100} \right) } + ( r + \frac{ {\sigma}^2}{2}) 1 } { \sigma \sqrt{1 } } \\ d_{11} &= \frac{ \ln{\left( 1 \right) } + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{11} &= \frac{ ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{11} &= \frac{ 2r + \sigma^2 } {2 \sigma } \\ d_{12} &= d_{11} - \sigma \sqrt{T} = \frac{ 2r + \sigma^2 } {2 \sigma } - \sigma \sqrt{1} \\ d_{12} &= \frac{ 2r + \sigma^2 } {2 \sigma } - \sigma = \frac{ 2r + \sigma^2 } {2 \sigma } - \frac{ 2\sigma^2}{2\sigma } \\ d_{12} &= \frac{ 2r - \sigma^2 } { 2 \sigma } \end{align*} Let $c_1$ be the price of first call option and $c_2$ be the price of the second call option. \begin{align*} c_1 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ 10 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ c_1 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ 1 &= 10 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 10e^{-r} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \end{align*} Now we have one equation with two unknowns. We want two equations. \begin{align*} d_{21} &= \frac{ \ln{ \left( \frac{100}{105} \right) } + ( r + \frac{ {\sigma}^2}{2}) 1 } { \sigma \sqrt{1 } } \\ d_{21} &= \frac{ \ln{ \left( \frac{21}{20} \right) } + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{21} &= \frac{ 0.0487902 + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{21} &= \frac{ 2(0.0487902) + 2r + {\sigma}^2 } { 2 \sigma } \\ d_{21} &= \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \\ d_{22} &= d_{21} - \sigma \sqrt{T} = \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } - \sigma \sqrt{1} \\ d_{22} &= \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \\ c_2 &= 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r\left( 1 \right) }N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) \\ 8 &= 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r}N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) \\ \end{align*} Here is the system of two equations that we need to solve: \begin{align*} 10 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 10e^{-r} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) - 1 &= 0 \\ 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r}N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 8 &= 0 \\ \end{align*}

I wrote the following script in SciLab:

function [z]=f(x,y)
z = x*x - 8
endfunction

function [z]=g(x,y)
z = 2*x + y - 8
endfunction

function [z] = fj(x)
        z = [2*x(1),0; 2, 1]
endfunction

function [z]=system1(x)
z = [f(x(1),x(2));g(x(1),x(2))]
endfunction

function [z]=eqn1(rate,sigma)
    tmp1 = 10 * cdfnor("PQ", (2*rate + sigma^2)/(2*sigma), 0, 1)
    tmp2 = 10*%e^(-rate)*cdfnor("PQ", (2*rate - sigma^2)/(2*sigma), 0, 1)
    z = tmp1 - tmp2 - 1
endfunction

function [z]=eqn2(rate,sigma)
    constant1 = 0.0975804
    tmp1 = 100 * cdfnor("PQ", (2*rate + sigma^2 + constant1)/(2*sigma), 0, 1)
    tmp2 = 105*%e^(-rate)*cdfnor("PQ", (2*rate - sigma^2 + constant1)/(2*sigma), 0, 1)
    z = tmp1 - tmp2 - 8
endfunction

function [z]=system2(x)
z = [eqn1(x(1),x(2));eqn2(x(1),x(2))]
endfunction

function [z] = fjacob(x)
    eps = 0.01;
    tmp11 = (eqn1(x(1)+eps,x(2)) - eqn1(x(1),x(2)))/eps
    tmp12 = (eqn1(x(1),x(2)+eps) - eqn1(x(1),x(2)) )/eps
    tmp21 = (eqn2(x(1)+eps,x(2)) - eqn2(x(1), x(2)) )/eps
    tmp22 = (eqn2( x(1), x(2) + eps) - eqn2(x(1),x(2)))/eps
    z = [tmp11,tmp12;tmp21,tmp22]
endfunction

Using this script, I get: $$ r = -0.152815$$ $$ \sigma = 0.3968625$$

A negative interest rate does not make any sense. Therefore, I conclude something is wrong.