Question Number 187379 by Humble last updated on 16/Feb/23
$${if}\:{x}\:{and}\:{y}\:{are}\:+{ve}\:{integers} \\ $$$${x}+{xy}+{y}=\mathrm{54} \\ $$$${x}+{y}=? \\ $$
Answered by horsebrand11 last updated on 16/Feb/23
$$\:\begin{cases}{{x}+{xy}+{y}+\mathrm{1}=\mathrm{55}}\\{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)=\mathrm{55}}\end{cases} \\ $$$$\:\begin{cases}{{x}+\mathrm{1}=\mathrm{5}\Rightarrow{x}=\mathrm{4}}\\{{y}+\mathrm{1}=\mathrm{11}\Rightarrow{y}=\mathrm{10}}\end{cases}\:\Rightarrow{x}+{y}=\mathrm{14} \\ $$$$\:{or}\:\begin{cases}{{x}+\mathrm{1}=\mathrm{11}\Rightarrow{x}=\mathrm{10}}\\{{y}+\mathrm{1}=\mathrm{5}\Rightarrow{y}=\mathrm{4}}\end{cases}\Rightarrow{x}+{y}=\mathrm{14} \\ $$
Commented by Humble last updated on 17/Feb/23
$${thank}\:{so}\:{much} \\ $$
Answered by a.lgnaoui last updated on 16/Feb/23
$${x}+{y}=\mathrm{54}−{xy}\:\:\:\:\left(\mathrm{1}\right) \\ $$$${x}\left(\mathrm{1}+{y}\right)+\left({y}+\mathrm{1}\right)=\mathrm{55}\:\:\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\left({y}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)=\mathrm{55} \\ $$$$\Rightarrow\mathrm{1}\bullet\begin{cases}{{y}+\mathrm{1}=\mathrm{5}}\\{{x}+\mathrm{1}=\mathrm{11}}\end{cases} \\ $$$${or}\:\mathrm{2}\bullet\begin{cases}{{y}+\mathrm{1}=\mathrm{11}}\\{{x}+\mathrm{1}=\mathrm{5}}\end{cases} \\ $$$$\:\:\left({x},{y}\right)=\left\{\left(\mathrm{10},\mathrm{4}\right)\:\:;\left(\mathrm{4},\mathrm{10}\right)\right. \\ $$$$ \\ $$
Commented by Humble last updated on 17/Feb/23
$${Beautiful}.\:{thank}\:{you}\:,\:{sir} \\ $$