How do you solve the system $- 3 x + y = 2$ $−3x + y = 2$ and $4 x + 2 y = 4$ $4x + 2y = 4$ ?

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How do you solve the system $- 3 x + y = 2$ $−3x + y = 2$ and $4 x + 2 y = 4$ $4x + 2y = 4$ ?

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Answer 1 · 2015-05-29T09:04:41

$- 3 x + y = 3 \Rightarrow y = 3 + 3 x$ $-3x + y =3 => y = 3 +3x$ ;
$4 x + 2 y = 4 \Rightarrow 4 x + 2 \cdot (3 + 3 x) = 4 \Rightarrow 4 x + 6 + 6 x = 4$ $4x + 2y = 4 => 4x + 2*(3+3x)= 4 => 4x+6+6x=4$
$\Rightarrow 10 x + 6 = 4 \Rightarrow 10 x = - 2 \Rightarrow x = - \frac{2}{10} = - \frac{1}{5}$ $=> 10x +6 =4 => 10x =-2 => x=-2/10=-1/5$
$y = 3 + 3 x \Rightarrow y = 3 + 3 \cdot (- \frac{1}{5}) \Rightarrow y = 3 - \frac{3}{5} \Rightarrow y = \frac{15}{5} - \frac{3}{5}$ $y=3+3x => y=3+3*(-1/5) => y= 3-3/5 => y= 15/5 - 3/5$
$y = \frac{12}{5}$ $y= 12/5$

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How do you solve the system $- 3 x + y = 2$ $−3x + y = 2$ and $4 x + 2 y = 4$ $4x + 2y = 4$ ?

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How do you solve the system $- 3 x + y = 2$ $−3x + y = 2$ and $4 x + 2 y = 4$ $4x + 2y = 4$ ?