How do solve the following linear system?: $y = 8 x - 3, x - 3 y = - 1$ $y = 8 x - 3 , x-3y=-1$ ?

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How do solve the following linear system?: $y = 8 x - 3, x - 3 y = - 1$ $y = 8 x - 3 , x-3y=-1$ ?

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Answer 1 · 2015-12-14T22:10:35

$x = \frac{10}{23} and y = \frac{11}{23}$ $x=10/23 and y=11/23$

Explanation:

Substitute the first equation into the second - that is, in the second equation, wherever you see a $y$ $y$ you replace it with $8 x - 3$ $8x-3$ .
Then the second equation becomes :
$x - 3 (8 x - 3) = - 1$ $x-3(8x-3)=-1$ .

This is now 1 equation in 1 unknown and we may solve for $x$ $x$ to obtain
$x - 24 x + 9 = - 1$ $x-24x+9=-1$
$therefore x=-10/-23=10/23$ .

Now substitute this value back into equation 1 to find $y$ :

$therefore y=8(10/23)-3$
$=11/23$

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How do solve the following linear system?: $y = 8 x - 3, x - 3 y = - 1$ $y = 8 x - 3 , x-3y=-1$ ?

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Explanation:

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How do solve the following linear system?: y = 8 x - 3 , x-3y=-1 y=8x−3,x−3y=−1 y = 8 x - 3 , x-3y=-1 ?

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Explanation:

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How do solve the following linear system?: $y = 8 x - 3, x - 3 y = - 1$ $y = 8 x - 3 , x-3y=-1$ ?