How do you solve the system $w + z = 14$ $w + z = 14$ and $w - 2 z = 2$ $w - 2z = 2$ by substitution?

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How do you solve the system $w + z = 14$ $w + z = 14$ and $w - 2 z = 2$ $w - 2z = 2$ by substitution?

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Answer 1 · 2015-05-31T15:41:33

$w + z = 14$ $w+z = 14$
$w = 14 - z$ $w= color(red)(14-z$ ......equation $(1)$ $(1)$
$w - 2 z = 2$ $w - 2z = 2$ .....equation $(2)$ $(2)$

Substituting equation $(1)$ $(1)$ in $(2)$ $(2)$
$(14 - z) - 2 z = 2$ $color(red)(( 14-z))-2z =2$
$14 - 3 z = 2$ $14 - 3z = 2$
$- 3 z = - 12$ $- 3z =- 12$
$z = \frac{12}{3} = 4$ $z = 12/3 = 4$

Substituting $z$ $z$ in equation $(1)$ $(1)$ to find $w$ $w$
$w = 14 - z$ $w= 14-z$
$w = 14 - 4 = 10$ $w= 14-4 = 10$

The solution for the system of equations is
$z = 4, w = 10$ $color(blue)(z = 4, w=10$

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How do you solve the system $w + z = 14$ $w + z = 14$ and $w - 2 z = 2$ $w - 2z = 2$ by substitution?

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How do you solve the system $w + z = 14$ $w + z = 14$ and $w - 2 z = 2$ $w - 2z = 2$ by substitution?