How do you solve by substitution $3 x + 2 y = 2$ $3x + 2y = 2$ and $- 2 x + y = 8$ $-2x + y = 8$ ?

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How do you solve by substitution $3 x + 2 y = 2$ $3x + 2y = 2$ and $- 2 x + y = 8$ $-2x + y = 8$ ?

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Answer 1 · 2015-06-09T15:14:54

The solution for the system of equations is
$x = - 2, y = 4$ $color(red)(x =-2 , y =4$

Explanation:

$3 x + 2 y = 2$ $3x + 2y = 2$ ..........equation $(1)$ $(1)$

$- 2 x + y = 8$ $-2x + y = 8$
$y = 8 + 2 x$ $y =color(red)( 8 +2x$ ............equation $(2)$ $(2)$

Substituting equation $(2)$ $(2)$ in equation $(1)$ $(1)$

$3 x + 2 y = 2$ $3x + 2y = 2$
$3 x + 2 (8 + 2 x) = 2$ $3x + 2color(red)((8+2x)) = 2$
$3 x + 16 + 4 x = 2$ $3x + 16 +4x = 2$
$7 x = - 14$ $7x= -14$
$x = - 2$ $color(red)(x=-2$

Substituting $x$ $x$ in equation $(2)$ $(2)$ to find $y$ $y$

$y = 8 + 2 x$ $y = 8 +2x$
$y = 8 + 2 . (- 2)$ $y = 8 +2.(-2)$
$y = 8 - 4$ $y = 8 -4$
$y = 4$ $color(red)(y = 4$

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How do you solve by substitution $3 x + 2 y = 2$ $3x + 2y = 2$ and $- 2 x + y = 8$ $-2x + y = 8$ ?

1 Answer

Explanation:

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How do you solve by substitution 3x + 2y = 23x+2y=23x + 2y = 2 and -2x + y = 8−2x+y=8-2x + y = 8?

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How do you solve by substitution $3 x + 2 y = 2$ $3x + 2y = 2$ and $- 2 x + y = 8$ $-2x + y = 8$ ?