A very simple problem tends to become very hard. Perhaps I am overthinking it.
If the square of Winslow's age is added to Abby's age, the sum is 209.
If the square of Abby's age is added to Winslow's age, the sum if 183.
What is the sum of their ages?
I start off as usual by setting equations:
$$w^2+a=209$$ $$w+a^2=183$$
where $w$ and $a$ are Winslow's and Abby's ages, respectively.
Now, I try to solve:
$$w=183-a^2$$$$\implies (183-a^2)^2+a=209$$$$\implies a^4-366a^2+a+33280=0$$
I have no way to continue solving, as I can't set $a^2$ to some variable due to there being an $a$ in the equation. I try to solve it again by substituting for $w$ but still get the very complicated equation.
Desperate, I try adding the equations together:
$$\implies w^2+w+a^2+a=392$$$$\implies w(w+1)+a(a+1)=392$$
Here, I become stuck again, as I cannot factor this thing further into the form $(a+b)(c+d)$.
How do I solve this problem? Have I missed a point?
Thanks!
Max0815
Best Answer
Instead of adding the equations, you could subtract them:
$209-183=w^2+a-a^2-w=(w+a)(w-a)+(a-w)=(w-a)(w+a-1).$
Since $209-183=26=13 \cdot 2=1\cdot 26$, and $w+a-1 > w-a$, we have that either $w+a-1=26$ or that $w+a-1=13$. But since $w$ and $a$ must have different parity (if they had the same parity, $w^2+a$ and $a^2+w$ would be even, which is not the case), their sum must be odd, meaning that we can rule out the $w+a=14$ case. Therefore, $w+a=27$.
(I have edited this comment, I dont know if that is something that I am supposed to mention, but now I have).