Odpowiedź:
[tex]\huge\boxed{a)\ \left\{\begin{array}{ccc}x=11\\y=-1\end{array}\right}\\\boxed{b)\ \left\{\begin{array}{ccc}x=2\\y=-2\end{array}\right}\\\boxed{c)\ \left\{\begin{array}{ccc}x=3\\y=-7\end{array}\right}[/tex]
Szczegółowe wyjaśnienie:
a)
[tex]\left\{\begin{array}{ccc}\dfrac{x+y}{2}=5&|\cdot2\\\dfrac{x-y}{3}=4&|\cdot3\end{array}\right\\\left\{\begin{array}{ccc}x+y=10&|-y\\x-y=12\end{array}\right\\\left\{\begin{array}{ccc}x=10-y&(1)\\x-y=12&(2)\end{array}\right[/tex]
podstawiamy (1) do (2):
[tex](10-y)-y=12\\10-y-y=12\qquad|-10\\-2y=2\qquad|:(-2)\\\boxed{y=-1}[/tex]
podstawiamy do (1):
[tex]x=10-(-1)\\x=10+1\\\boxed{x=11}[/tex]
b)
[tex]\left\{\begin{array}{ccc}\dfrac{2x-y}{3}=\dfrac{3x+y}{2}&|\text{mnozymy na krzyz}\\x=2(x+y+1)\end{array}\right\\\left\{\begin{array}{ccc}2(2x-y)=3(3x+y)\\x=2x+2y+2&|-2x\end{array}\right\\\left\{\begin{array}{ccc}4x-2y=9x+3y&|+2y-9x\\-x=2y+2&|\cdot(-1)\end{array}\right\\\left\{\begin{array}{ccc}-5x=5y&|:(-5)\\x=-2y-2&|+2y\end{array}\right\\\left\{\begin{array}{ccc}x=-y&(1)\\x+2y=-2&(2)\end{array}\right[/tex]
podstawiamy (1) do (2):
[tex]-y+2y=-2\\\boxed{y=-2}[/tex]
podstawiamy do (1):
[tex]x=-(-2)\\\boxed{x=2}[/tex]
c)
[tex]\left\{\begin{array}{ccc}1-\dfrac{x-1}{2}=2+\dfrac{y+1}{3}&|\cdot6\\x+y+4=0&|-y-4\end{array}\right\\\left\{\begin{array}{ccc}6\cdot1-6\!\!\!\!\diagup^3\dfrac{x-1}{2\!\!\!\!\diagup_1}=6\cdot2+6\!\!\!\!\diagup^2\dfrac{y+1}{3\!\!\!\!\diagup_1}&|\cdot6\\x=-y-4\end{array}\right\\\left\{\begin{array}{ccc}6-3(x-1)=12+2(y+1)\\x=-y-4\end{array}\right\\\left\{\begin{array}{ccc}6-3x+3=12+2y+2\\x=-y-4\end{array}\right\\\left\{\begin{array}{ccc}9-3x=14+2y&|-9-2y\\x=-y-4\end{array}\right[/tex]
[tex]\left\{\begin{array}{ccc}-3x-2y=5&(1)\\x=-y-4&(2)\end{array}\right[/tex]
podstawiamy (2) do (1):
[tex]-3(-y-4)-2y=5\\3y+12-2y=5\qquad|-12\\\boxed{y=-7}[/tex]
podstawiamy do (2):
[tex]x=-(-7)-4\\x=7-4\\\boxed{x=3}[/tex]
