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

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

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Answer 1 · 2015-06-03T08:39:44

$y = 3 - 4 x$ $y=color(red)(3-4x$ ....equation $(1)$ $(1)$
$- 3 x - y = 2$ $-3x-y=2$ .......equation $(2)$ $(2)$

substituting equation $(1)$ $(1)$ in equation $(2)$ $(2)$

$- 3 x - y = 2$ $-3x-y=2$
$- 3 x - (3 - 4 x) = 2$ $-3x-color(red)((3-4x))=2$
$- 3 x - 3 + 4 x = 2$ $-3x color(red)(- 3 +4x)=2$
$x = 2 + 3, x = 5$ $x=2 + 3, color(blue)(x = 5$

substituting $x$ $x$ in equation $(1)$ $(1)$ to find $y$ $y$
$y = 3 - 4 x$ $y=3-4x$
$y = 3 - 20, y = - 17$ $y=3-20 , color(blue)(y = -17$

the solutions are : $x = 5, y = - 17$ $color(blue)( x = 5 , y = -17$

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

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

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