VicKmath7 wrote: Solve the system in reals: $\frac{4-a}{b}=\frac{5-b}{a}=\frac{10}{a^2+b^2}$. Let $c$ be the common value. First equality implies $c^2\ne 1$ and $a=\frac{4-5c}{1-c^2}$ and $b=\frac{5-4c}{1-c^2}$ Plugging in $\frac{10}{a^2+b^2}=c$, this becomes $(2c-1)(c-2)(5c^2-8c+5)=0$ and so Either $c=\frac 12$ and $\boxed{\text{S1 : }(a,b)=(2,4)}$ Either $c=2$ and $\boxed{\text{S2 : }(a,b)=(2,1)}$

Can anyone share a more elegant solution?

Attachments:

Let $m,n,p,q \in \mathbb{R}$ such that $n \neq 0, q \neq 0$ and $n+q \neq 0$. Then $\frac{m}{n} = \frac{p}{q} = r \Rightarrow \frac{m+p}{n+q} = r$ Proof: Since $n \neq 0, q \neq 0$ we have $m = rn, p=rq$. Since $n+q \neq 0$, $\frac{m+p}{n+q} = \frac{r(n+q)}{n+q} = r. \;\;\square$ Since $a \neq 0, b \neq 0$, we have $$\frac{b(4-a)}{b^2} = \frac{a(5-b)}{a^2} = \frac{10}{a^2+b^2} \Rightarrow \frac{4b-ab+5a-ab}{a^2+b^2} = \frac{10}{a^2+b^2} $$$$\Leftrightarrow 4b+5a-2ab-10 = 0 \Leftrightarrow (a-2)(5-2b) = 0 \Leftrightarrow a = 2 \; \lor \; b = \frac{5}{2}$$Letting $\boxed{a=2}$ we get $\boxed{b=1}$ and $\boxed{b=4}$ as solutions. Letting $b=\frac{5}{2}$ we get no solutions at all.

Thanks from Turkey for this great solutions.