Need help solving three nonlinear equations with two unknowns

Question

Could you please help me to solve the system for c and d:

$c^2 + (2d - 5/2)c + (3d)/2 - 1/2 = 0$

$d^2 + (2c - 5/2)d + (3c)/2 - 3/2 = 0$

$-c^2d - (9c^2)/4 - cd^2 + (cd)/2 + 6c + (3d^2)/4 - 7d + 9/4 = 0$

when I used MATLAB with "solve", this is what I got:

Empty sym: 0-by-1

Empty sym: 0-by-1

So MATLAB could not find any answer. So I tried to solve it by myself and I got some answers but when I substituted in the LHS of the equations, I did not get zero. So there are some mistakes in my answers but I don't know where.

Any help in how to find solutions for this system? or can you tell me what is wrong with my answer below.

Adding the idea of my answer:

I multiply Eq.1 by $d$ and Eq.2 by $c$ and adding the new two equations together and got

$3c^2d + (3c^2)/2 + 3cd^2 - 5cd - (3c)/2 + (3d^2)/2 - d/2 = 0$

then I found $cd^2+dc^2$ from this equation and substitute it in Eq.3 in the question and I got

$(11c)/2 - (43d)/6 - (7cd)/6 - (7c^2)/4 + (5d^2)/4 + 9/4 = 0$

and if we substitute $c^2$ from Eq.1 and $d^2$ from Eq.2 from the question in this equation we got

$cd = 39/2 - (17d)/2 - (9c)/2$

hence, from this equation we find $d$ ($d = -(9c - 39)/(2c + 17)$) and also if we substitute $cd$ from this equation in Eq.1 and Eq.2 in the question we got these two equations:

$c^2 = (23c)/2 + (31d)/2 - 77/2$

$d^2 = (15c)/2 + (39d)/2 - 75/2$

then if we substitute the value of $d = -(9c - 39)/(2c + 17)$ in these two equations we got

$c^2 = (23c^2 - 21c - 50)/(2c + 17)$

$(9c - 39)^2/(2c + 17)^2 = (15c)/2 - (39(9c - 39))/(2(2c + 17)) - 75/2$

which are equivalent to these two equations:(by multiplying the first one by $(2c + 17)$ and second one by $(2c + 17)^2$)

$2c^3 - 6c^2 + 21c + 50 = 0$

$81c^2 - 702c + 1521 = 30c^3 + 9c^2 - 1845c + 2091$

then in these two equations if we find $c^3$ from the first one($c^3 = 3c^2 - (21c)/2 - 25$) and substitute it in the second one we got

$81c^2 - 702c + 1521 = 99c^2 - 2160c + 1341$

and if we solve this equation, we have

$c = 81/2 - \sqrt(6601)/2$

$c= 81/2+\sqrt(6601)/2 $

therefore we found some solutions of this system!

If we assume the calculation in these steps are correct(which I did it by Matlab), what are the mathematical mistakes in these steps?(please let me know all the mistakes if you find more then one)

Answer 1

You have $3$ equations in $2$ unknowns, so unless you know for a fact that there is a solution, you should expect that the system might be inconsistent.

And in fact, that's the case here.

Considering only the first two equations, we get the equivalent system $$ \begin{cases} 6c^4-26c^3-18c^2+27c+10=0\\[4pt] 31d = -24c^3+122c^2-35c-43\\ \end{cases} $$ Solving numerically, we get the $4$ solutions \begin{align*} (c,d)&\approx (-1.065872615, 5.224837188)\\[4pt] (c,d)&\approx (-.3345918428, -.5397478364)\\[4pt] (c,d)&\approx (.9838796319, .5743420457)\\[4pt] (c,d)&\approx (4.749918159, -.9260980641)\\[4pt] \end{align*} however none of the above pairs $(c,d)$ satisfy the third equation.

Therefore the given system has no solutions.

As regards the error in your analysis . . .

It's not that you made an algebraic error.

As I noted in the comments, all you found are potential solutions.

Based on your work, it follows that if $(c,d)$ is a solution to the original system, then \begin{align*} (c,d)&={\small{\left(\frac{81+\sqrt{6601}}{2},\frac{-19-\sqrt{6601}}{26}\right)}}\\[4pt] &\;\text{or}\\[4pt] (c,d)&={\small{\left(\frac{81-\sqrt{6601}}{2},\frac{-19+\sqrt{6601}}{26}\right)}}\\[4pt] \end{align*} However it's easily verified that neither of the above pairs is an actual solution to the original system, and from that, it follows that the original system has no solutions.

Let me illustrate your misconception with a much simpler example . . .

Consider the following system $$ \begin{cases} x+y^2-xy=1&\;\;\;(\text{eq}1)\\[4pt] x+y^2+xy=3&\;\;\;(\text{eq}2)\\[4pt] x+y^2+y=5&\;\;\;(\text{eq}3)\\ \end{cases} $$ of $3$ non-linear equations in the $2$ unknowns $x,y$.

Computing $$(\text{eq}1)+(\text{eq}2)-2(\text{eq}3)$$ we get $y=3$.

Replacing $y$ by $3$ in $(\text{eq}3)$, we get $x=-7$.

So we've found the solution $(x,y)=(-7,3)$, right?

Wrong!

As explained, the result $(x,y)=(-7,3)$ is only a necessary condition, i.e., a potential solution.

To see if it's an actual solution, we need to test it to see if it satisfies the original system.

Noting that $(x,y)=(-7,3)$ doesn't satisfy $(\text{eq}1)$, it follows that the original system has no solutions.

Answer 2

Taking advantage that the first equation is linear in $d$, we have, assuming that $c \neq -\frac 34$, $$d=\frac{-2 c^2+5 c+1}{4 c+3}$$ Plugging in the second and third equations, we have $$6 c^4-26 c^3-18 c^2+27 c+10=0\tag1$$ $$4 c^4-31 c^3+50 c^2+74 c-17=0\tag 2$$ Each of these can be solved with nasty radicals. The first one as solutions $$\{c_1= -1.06587\},\{c_2= -0.334592\},\{c_3= 0.98388\},\{c_4= 4.74992\}$$ and the second one has only two real roots $$\{c_5= -1.04221\},\{c_6= 0.204876\}$$ If you consider the norm $$\Phi=(6 c^4-26 c^3-18 c^2+27 c+1)^2+(4 c^4-31 c^3+50 c^2+74 c-17)^2$$ it does not show any real root and its minimum value is $1.26204$ obtained for $c_*=-1.04449$.

We can notice that $c_*$ is "quite close" to $c_1$ and $c_5$. So, maybe, there could be a typo in one of the equations.

Edit

Naming $f_1,f_2,f_3$ the three equations, let us consider the norm $$\Phi=f_1^2+f_2^2+f_3^2$$ and minimize it. Solving in the real domain the eqauions $\frac{\partial \Phi}{\partial c}=\frac{\partial \Phi}{\partial d}=0$ reveals $7$ soutions.

Minimizing $\Phi$ leads to a minimum value equal to $0.34515$ for $c=0.88087$ and $d=0.75068$.

$3D$ and contour plots of $\Phi$ reveal the same.

All of the above seems to confirm that there is no solution to the problem.