Question

Dau coroana, va rog sa rezolvati d

Answer 1

$\left\{\begin{array}{ccc}\dfrac{3x-y+2}{5}=\dfrac{4x-2y-6}{3} &\\\\\dfrac{4x+y+2}{4}=\dfrac{3x+2y+8}{5} \end{array}\right \Rightarrow \left\{\begin{array}{ccc}3(3x-y+2)=5(4x-2y-6)\\5(4x+y+2)=4(3x+2y+8)\end{array}\right\Rightarrow \\\\\\\Rightarrow \left\{\begin{array}{ccc}9x-3y+6=20x-10y-30\\20x+5y+10=12x+8y+32\end{array}\right \Rightarrow$

$\Rightarrow \left\{\begin{array}{ccc}9x-20x-3y+10y=-30-6\\20x-12x+5y-8y=32-10\end{array}\right \Rightarrow \left\{\begin{array}{ccc}-11x+7y=-36~|\cdot 3\\8x-3y=22~|\cdot7\end{array}\right \Rightarrow \\\\\\\Rightarrow \left\{\begin{array}{ccc}-33x+21y=-108\\56x-21y=154\end{array}\right\\~~~~~-----------\\~~~~~~~~~~~23x~~~~/~~~=46 \Rightarrow x=\dfrac{46}{23} \Rightarrow \boxed{x=2}$

$x=2\Rightarrow 8x-3y=22\Rightarrow 8\cdot2-3y=22\Rightarrow 16-3y=22\Rightarrow \\\\\Rightarrow -3y=22-16\Rightarrow -3y=6\Rightarrow x=-\dfrac{6}{3} \Rightarrow \boxed{x=-2}$