How do solve the following linear system?: $\frac{1}{3} x + 3 y = 9, 8 x + 3 y = - 9$ $1/3x + 3y = 9 , 8x + 3y= -9$ ?

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How do solve the following linear system?: $\frac{1}{3} x + 3 y = 9, 8 x + 3 y = - 9$ $1/3x + 3y = 9 , 8x + 3y= -9$ ?

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Answer 1 · 2015-12-26T19:11:58

You can solve it either using substitution or elimination, but the following proof will use substitution.

Explanation:

$\frac{1}{3} x$ $1/3x$ + 3y = 9
3y = 9 - $\frac{1}{3} x$ $1/3x$
y = 3 - $\frac{1}{9} x$ $1/9x$
8x + 3(3 - $\frac{1}{9} x$ $1/9x$ ) = -9
8x + 9 - $\frac{1}{3} x$ $1/3x$ = -9
$\frac{23}{3} x$ $23/3x$ + 9 = -9
$\frac{23}{3} x$ $23/3x$ = -18
23 = -54x
$- \frac{54}{23}$ $-54/23$ = x
8( $- \frac{54}{23}$ $-54/23$ ) + 3y = -9
$- \frac{432}{23}$ $-432/23$ + 3y = -9
$- \frac{432}{23}$ $-432/23$ + 9 = -3y
$- \frac{225}{23}$ $-225/23$ = -3y
$\frac{75}{23}$ $75/23$ = y

The solution set is ( $- \frac{54}{23}$ $-54/23$ , $\frac{75}{23}$ $75/23$ )

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How do solve the following linear system?: $\frac{1}{3} x + 3 y = 9, 8 x + 3 y = - 9$ $1/3x + 3y = 9 , 8x + 3y= -9$ ?

1 Answer

Explanation:

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How do solve the following linear system?: $\frac{1}{3} x + 3 y = 9, 8 x + 3 y = - 9$ $1/3x + 3y = 9 , 8x + 3y= -9$ ?