Question

решите задание♡♡♡
♡♡♡​

Answer 1

$\left\{\begin{array}{ccc}3x+2y=5 \ | \ \cdot 3\\2x-3y=7 \ | \ \cdot2\end{array}\right\\\\\\+\left\{\begin{array}{ccc}9x+6y=15 \ \\4x-6y=14 \ \end{array}\right\\---------\\13x=29\\\\x=\dfrac{29}{13} \\\\2y=5-3x=5-3\cdot\dfrac{29}{13}=5-\dfrac{87}{13}=5-6\dfrac{9}{13}=-1\dfrac{9}{13} \\\\y=-1\dfrac{9}{13}:2=-\dfrac{22}{13}\cdot\dfrac{1}{2}=-\dfrac{11}{13}\\\\\\13(x-y)=13\cdot\Big[ \dfrac{29}{13}-(-\dfrac{11}{13})\Big]=13\cdot\Big(\dfrac{29}{13}+\dfrac{11}{13}\Big)=$

$=13\cdot \dfrac{40}{13}=\boxed{40}$