How do you solve $x + 9 y = 80$ $x+9y=80$ and $y = - 64 + 9 x$ $y=-64+9x$ using substitution?

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How do you solve $x + 9 y = 80$ $x+9y=80$ and $y = - 64 + 9 x$ $y=-64+9x$ using substitution?

1 Answer

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Answer 1 · 2016-06-10T05:31:52

$x = 8$ $x=8$ ; $y = 8$ $y=8$

Explanation:

Since $y = - 64 + 9 x$ $y=-64+9x$
let's substitute it in the first equatiom so that
$x + 9 (- 64 + 9 x) = 80$ $x+9(-64+9x)=80$
then let's simplify
$x - 576 + 81 x = 80$ $x-576+81x=80$
$82 x = 656$ $82x=656$
$x = \frac{656}{82}$ $x=656/82$
$x = 8$ $x=8$
$y = - 64 + 9 \cdot 8$ $y=-64+9*8$
$y = 8$ $y=8$

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How do you solve $x + 9 y = 80$ $x+9y=80$ and $y = - 64 + 9 x$ $y=-64+9x$ using substitution?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you solve x+9y=80x+9y=80 and y=−64+9xy=-64+9x using substitution?

1 Answer

Explanation:

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How do you solve $x + 9 y = 80$ $x+9y=80$ and $y = - 64 + 9 x$ $y=-64+9x$ using substitution?