How do you solve the linear system using substitution $x = 16 - 4 y$ $x = 16 - 4y$ and $3 x + 4 y = 8$ $3x + 4y = 8$ ?

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How do you solve the linear system using substitution $x = 16 - 4 y$ $x = 16 - 4y$ and $3 x + 4 y = 8$ $3x + 4y = 8$ ?

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Answer 1 · 2015-06-05T08:49:52

$x = 16 - 4 y$ $x = color(blue)(16 - 4y$ ......equation $(1)$ $(1)$

$3 x + 4 y = 8$ $3x + 4y = 8$ .........equation $(2)$ $(2)$

substituting equation $(1)$ $(1)$ in equation $(2)$ $(2)$
$3 (16 - 4 y) + 4 y = 8$ $3color(blue)((16-4y)) + 4y = 8$

$48 - 12 y + 4 y = 8$ $48 - 12y + 4y = 8$
$48 - 8 y = 8$ $48 - 8y= 8$
$- 8 y = 8 - 48 = - 40$ $- 8y= 8-48 = -40$
$y = - \frac{40}{- 8}$ $y=-40/-8$
$y = 5$ $color(blue)(y=5$

substituting $y$ $color(blue)(y$ in equation $(1)$ $(1)$ to find $x$ $x$
$x = 16 - 4 y$ $x =16 - 4y$
$x = 16 - 4.5 = 16 - 20$ $x =16 - 4.5 = 16-20$
$x = - 4$ $color(blue)(x = -4$

the solution for the system of equations is
$x = - 4 and y = 5$ $color(blue)(x= -4 and y = 5$

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How do you solve the linear system using substitution $x = 16 - 4 y$ $x = 16 - 4y$ and $3 x + 4 y = 8$ $3x + 4y = 8$ ?

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How do you solve the linear system using substitution $x = 16 - 4 y$ $x = 16 - 4y$ and $3 x + 4 y = 8$ $3x + 4y = 8$ ?