How do you solve the system $7 + 2 y = 8 x$ $7 + 2y = 8x$ and $3 x - 2 y = 0$ $3x - 2y = 0$ by substitution?

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How do you solve the system $7 + 2 y = 8 x$ $7 + 2y = 8x$ and $3 x - 2 y = 0$ $3x - 2y = 0$ by substitution?

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Answer 1 · 2018-06-22T07:06:11

$x = \frac{7}{5}, y = \frac{21}{10}$ $x=7/5,y=21/10$

Explanation:

from the second equation we get

$x = \frac{2}{3} y$ $x=2/3y$
plugging this in the first one:

$7 + 2 y = \frac{16}{3} y$ $7+2y=16/3y$
$7 = \frac{16}{3} y - 2 y$ $7=16/3y-2y$
$7 = \frac{10}{3} y$ $7=10/3y$
so $y = \frac{21}{10}$ $y=21/10$
$x = \frac{7}{5}$ $x=7/5$

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How do you solve the system $7 + 2 y = 8 x$ $7 + 2y = 8x$ and $3 x - 2 y = 0$ $3x - 2y = 0$ by substitution?

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Explanation:

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How do you solve the system 7 + 2y = 8x7+2y=8x7 + 2y = 8x and 3x - 2y = 03x−2y=03x - 2y = 0 by substitution?

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Explanation:

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How do you solve the system $7 + 2 y = 8 x$ $7 + 2y = 8x$ and $3 x - 2 y = 0$ $3x - 2y = 0$ by substitution?