How do you solve by substitution $x + 7 y = 33$ $x + 7y = 33$ and $3 x - 2 y = 7$ $3x - 2y = 7$ ?

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Answer 1 · 2015-06-07T12:00:13

$x + 7 y = 33$ $x + 7y = 33$

$x = 33 - 7 y$ $x = color(red)(33 - 7y$ .......equation $(1)$ $(1)$
$3 x - 2 y = 7$ $3x - 2y = 7$ ........equation $(2)$ $(2)$

Substituting equation $(1)$ $(1)$ in $(2)$ $(2)$ to find $y$ $y$

$3 (33 - 7 y) - 2 y = 7$ $3color(red)((33 - 7y)) - 2y = 7$

$99 - 21 y - 2 y = 7$ $99 - 21y- 2y = 7$
$- 23 y = 7 - 99$ $-23y= 7-99$
$- 23 y = - 92$ $-23y= -92$
$y = 4$ $color(blue)(y= 4$

Substituting $y$ $y$ in equation $(1)$ $(1)$ to find $x$ $x$
$x = 33 - 7 y$ $x =33 - 7y$

$x = 33 - 7.4 = 33 - 28$ $x =33 - 7.4 = 33 - 28$
$x = 5$ $color(blue)(x =5$

The solution for the system of equations is:
$x = 5, y = 4$ $color(blue)( x = 5 , y=4$

How do you solve by substitution x + 7y = 33x+7y=33x + 7y = 33 and 3x - 2y = 73x−2y=73x - 2y = 7?