Question Number 26274 by NECx last updated on 23/Dec/17
$${could}\:{there}\:{be}\:{an}\:{analytical}\:{or} \\ $$$${numerical}\:{meghod}\:{for}\:{solving} \\ $$$${this}\:{non}-{linear}\:{simultaneous} \\ $$$${equation} \\ $$$${x}+{y}=\mathrm{5} \\ $$$${x}^{{x}} +{y}^{{y}} =\mathrm{31} \\ $$$$ \\ $$$${please}\:{help}\:{if}\:{possible} \\ $$
Commented by mrW1 last updated on 23/Dec/17
$${by}\:{try}\:\&\:{error}: \\ $$$${x}=\mathrm{3},\:{y}=\mathrm{2}\:{or} \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{3} \\ $$
Commented by prakash jain last updated on 24/Dec/17
$${f}\left({x}\right)={x}^{{x}} +\left(\mathrm{5}−{x}\right)^{\mathrm{5}−{x}} −\mathrm{31} \\ $$$$\mathrm{Graphing}\:\mathrm{technique}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{approx}\:\mathrm{value}. \\ $$