Monthly Archives: March 2023

Find real solutions for the following equations: $$a+bcd = 2$$ $$b+cda=2$$ $$c+dab=2$$ $$d + abc=2$$ Solution: Because $a+bcd=2$, $b+cda=2$, we have $a+bcd=b+cda$. Factorizing it, we have $$(a-b)(cd-1)=0$$ Therefore either $a=b$ or $cd=1$. Case 1: If $a=b$, we have $$a+ac^2=2$$ $$c+da^2=2$$ … Continue reading →