Question Number 195806 by Mingma last updated on 11/Aug/23
Answered by som(math1967) last updated on 11/Aug/23
Commented by som(math1967) last updated on 11/Aug/23
$$\mathrm{13}^{\mathrm{2}} −{a}^{\mathrm{2}} ={x}^{\mathrm{2}} −{b}^{\mathrm{2}} \:…\mathrm{1} \\ $$$$\mathrm{14}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} −{b}^{\mathrm{2}} \:….\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\:\:−\left(\mathrm{2}\right) \\ $$$$\mathrm{13}^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{100}−\mathrm{27} \\ $$$$\:{x}=\sqrt{\mathrm{73}} \\ $$
Commented by Mingma last updated on 13/Aug/23
Perfect ��