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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?

$A+C=0$

$AC+B+D=3$

$AD+BC=6$

$BD=10$

Hosam Hajeer
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Bflat
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  • What’s your attempt, thanks! – user1176409 Jul 18 '23 at 13:37
  • Multiply eq 1 and 2 and substitute from eq 1, 2 and 3. That gets me to a(b-d) =-6. Thereafter, I use d=10/b. With this I am able to get a quadratic ab^2 + 6b -10a = 0. The plan was to solve each equation in terms of a and find the real values. However, the solution is becoming lengthy and complex. – Bflat Jul 18 '23 at 13:46

2 Answers2

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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.

Robert Israel
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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

Bflat
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