Solving system of 4 equations and 4 variables.

Question

I am looking to solve this system of equations but cannot think of the solution. Wolfram alpha just give the answer and not the steps involved. Can anyone help?

$A+C=0$

$AC+B+D=3$

$AD+BC=6$

$BD=10$

Answer 1

After the easy substitutions $c = -a$ and $d = 3 - ac - b$ from the first two equations, you are left with $$ \eqalign{a^3 - 2 a b + 3 a - 6 &= 0\cr a^2 b - b^2 + 3 b - 10 &= 0\cr}$$ From the first of these, $$b = \frac{a^2}{2} + \frac{3}{2} - \frac{3}{a}$$ and the final equation becomes $$ \frac{a^4}{4} + \frac{3 a^2}{2} - \frac{31}{4} - \frac{9}{a^2} = 0 $$ Yes, after multiplying by $a^2$ you have an equation of degree $6$ in $a$, or of degree $3$ in $a^2$, but don't despair: it's easy to guess a small integer solution. The Rational Root Theorem may help.

Answer 2

With some effort,I was able to arrive at the equation

a^6 + 6a^4 - 31a^2 - 36 = 0

Using rational root theorem, I was able to get a = 2 and a = -2 as roots.

Using this I got:

A = +2 or -2

B = 2 or 5

C = -2 or 2

D = 5 or 2