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1 year ago

Can someone help me on number 9 please?

Mathematics

2 answers:

Vikki [24]1 year ago

8 0

$-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
$

Minchanka [31]1 year ago

3 0

$\left\{\begin{array}{ccc}-x+3y=0\\2x+6y=12&|divide\ both\ sides\ by\ (-2)\end{array}\right\\\\+\left\{\begin{array}{ccc}-x+3y=0\\-x-3y=-6\end{array}\right\ add\ sides\ of\ equations\\----------\\.\ \ \ \ \ -2x=-6\ \ \ \ |divide\ both\ sides\ by\ (-2)\\.\ \ \ \ \ \ \ \boxed{x=3}\\\\-3+3y=0\ \ \ \ \ \ |add\ 3\ to\ both\ sides\\3y=3\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\boxed{y=1}\\\\Answer:x=3\ and\ y=1\to(3;\ 1).$

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