marcuslum2
marcuslum2
24-08-2014
Mathematics
contestada
Can someone help me on number 9 please?
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konrad509
konrad509
24-08-2014
[tex]-x+3y=0\\ 2x+6y=12\\\\ -x+3y=0\\ x+3y=6\\ ------\\ 6y=6\\ y=1\\\\ x+3\cdot1=6\\ x+3=6\\ x=3 [/tex]
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Аноним
Аноним
24-08-2014
[tex]\left\{\begin{array}{ccc}-x+3y=0\\2x+6y=12&|divide\ both\ sides\ by\ (-2)\end{array}\right\\\\+\left\{\begin{array}{ccc}-x+3y=0\\-x-3y=-6\end{array}\right\ add\ sides\ of\ equations\\----------\\.\ \ \ \ \ -2x=-6\ \ \ \ |divide\ both\ sides\ by\ (-2)\\.\ \ \ \ \ \ \ \boxed{x=3}\\\\-3+3y=0\ \ \ \ \ \ |add\ 3\ to\ both\ sides\\3y=3\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\boxed{y=1}\\\\Answer:x=3\ and\ y=1\to(3;\ 1).[/tex]
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