Vì x - y = -1
⇒ x = y - 1 ⇒ 3x = 3y - 3
Ta có:
$\frac{2}{x}$ + $\frac{3}{y}$ = 2
⇔ $\frac{2y+3x}{xy}$ =2
⇔ 2xy = 2y + 3x
⇔ 2y(y-1) = 2y +3y - 3
⇔ 2y² - 2y = 5y - 3
⇔ 2y² - 2y - 5y + 3 = 0
⇔ 2y² - 7y + 3 = 0
⇔ 2y² - 6y - y + 3 = 0
⇔ 2y(y-3) - (y-3) = 0
⇔ (y-3)(2y-1) = 0
⇔ \(\left[ \begin{array}{l}y-3 = 0\\2y-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}y=3\\y= \frac{1}{2} \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=\frac{1}{2}\end{array} \right.\)
