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

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Previously we tried the elimination route and we failed.

What if we try to get another equation with not $a,\,b,\,c,$ or $d$ the subject, but perhaps try to make $c+d$ the subject of the equation?

$(a+b)(c+d)=ac+ad+bc+bd$

$9(c+d)=ac+39+bd$

It's tempted to replace $c+d$ by $29-ab$ to the equation above, but hey, we want to make $c+d$ the subject of the equation, or perhaps an equation in terms of only $c$ and/or $d$.

So, NOPE, we should not replace $c+d$ by $29-ab$! That also means we should get rid of $a$ and $b$:

$9(c+d)=ac+39+bd$

$9(c+d)=(9-b)c+39+bd$

$9(c+d)=9c+39-bc+bd$

$9(c+d)=9c+39-b(c-d)$

We need to find an expression of $b$ that is purely in terms of $c$ and/or $d$:

$a+b=9$ implies $a=9-b$

$ab+c+d=29$ implies $(9-b)b+c+d=29$

We have a quadratic equation in $b$, which isn't what we're after.

$ad+bc=39$ implies

$(9-b)d+bc=39$

$9d-bd+bc=39$

$9d-b(d-c)=39$

$9d-39=b(d-c)$

$b=\dfrac{9d-39}{d-c}=\dfrac{39-9d}{c-d}$

I think it's quite obvious that we will get back $9(c+d)=9c+39-b(c-d)=9(c+d)$ upon substituting $b$ by $\dfrac{39-9d}{c-d}$.

It seems as though there is nothing much left to do with this approach and we are at the end of the cul-de-sac.

I think this seemingly like a very straightforward problem has turned out to be a very vexing (in a good way) problem. I will post back to show my solution some time later today or by tomorrow.

## No comments:

## Post a Comment