How do you solve the system $x - 5 y = 2$ $x-5y=2$ and $9 x + 8 = 32 y$ $9x+8=32y$ ?

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How do you solve the system $x - 5 y = 2$ $x-5y=2$ and $9 x + 8 = 32 y$ $9x+8=32y$ ?

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Answer 1 · 2015-05-16T21:28:38

Substitute $x = 5 y + 2$ $x=5y+2$ (first equation) into the second to get:
$9 (5 y + 2) + 8 = 32 y$ $9(5y+2)+8=32y$
$45 y + 18 + 8 = 32 y$ $45y+18+8=32y$
$13 y = - 26$ $13y=-26$
$y = - 2$ $y=-2$
Use this back into the first equation:
$x = (5 \cdot - 2) + 2 = - 8$ $x=(5*-2)+2=-8$

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How do you solve the system $x - 5 y = 2$ $x-5y=2$ and $9 x + 8 = 32 y$ $9x+8=32y$ ?

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How do you solve the system x-5y=2x−5y=2x-5y=2 and 9x+8=32y9x+8=32y9x+8=32y?

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How do you solve the system $x - 5 y = 2$ $x-5y=2$ and $9 x + 8 = 32 y$ $9x+8=32y$ ?