IMO Solving System Of Equation Problem (First Attempt)

Solve the following system of equations in real $a,\,b,\,c,\,d$:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

This doesn't look like it is a hard system to attack, as we have four variables with four equations, a very standard system instead.

But, from the given info, we have no idea if $a,\,b,\,c,\,d$ are all positive real, or if they are integers. Things are beginning to get very unclear. Says we want to solve the system with the elimination route, we see that we could get it down to:

$a+b=9$

$ab+c+d=29$ implies $c=29-ab-d$

$ad+bc=39$ implies $ad+b(29-ab-d)=39$, or $d=\dfrac{39+ab^2-29b}{a-b}$

Lastly,

$cd=18$ gives

$(29-ab-d)\left(\dfrac{39+ab^2-29b}{a-b}\right)=18$

$\left(29-ab-\left(\dfrac{39+ab^2-29b}{a-b}\right)\right)\left(\dfrac{39+ab^2-29b}{a-b}\right)=18$

Simplifying we get:

$(29(a-b)-ab(a-b)-(39+ab^2-29b))(39+ab^2-29b)=18(a-b)^2$

$(29a-29b-a^2b+ab^2-39-ab^2+29b)(39+ab^2-29b)=18(a-b)^2$

$(29a-a^2b-39)(39+ab^2-29b)=18(a-b)^2$

Now we substitute $b=9-a$ into the equation above to obtain:

$(29(9-a)-a^2(9-a)-39)(39+a(9-a)^2-29(9-a))=18(a-(9-a))^2$

$(261-29a-9a^2+a^3-39)(39+81a-18a^2+a^3-261+29a)=18(2a-9)^2$

$(a^3-9a^2-29a+222)(a^3-18a^2+110a-222)=18(2a-9)^2$

Wow, this has gone way too far, we have gotten a sextic polynomial (a polynomial of degree six), and it's already a very exhausting task to solve for cubic equation, now, we have to deal with a sextic equation?

But, if we don't solve it for $a$ but other variable, says $b$, we would still end up with the polynomial of degree six in that variable. All in all, this elimination method just won't work in this instance.

Have I ignited your passion to attempt at this beautiful problem by now?

Yes, we can learn the problem solving skills from others when we watch how other approached the problem that we failed to solve, and but that boils down to if we tried hard enough on it because learning from mistakes requires one important thing:

You have to be willing to put yourself in situations where you can make interesting mistakes.

I will post in another blog post to see what else we could do to attempt at this daunting yet good problem using another approach.