How do you solve by substitution $y - 2 x = 8$ $y - 2x = 8$ and $16 + 4 x = 2 y$ $16 + 4x = 2y$ ?

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How do you solve by substitution $y - 2 x = 8$ $y - 2x = 8$ and $16 + 4 x = 2 y$ $16 + 4x = 2y$ ?

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Answer 1 · 2015-06-06T22:44:30

Add $2 x$ $2x$ to both sides of the first equation to get:

$y = 8 + 2 x$ $y = 8 + 2x$

Divide both sides of the second equation by $2$ $2$ to get:

$y = 8 + 2 x$ $y = 8 + 2x$

The two equations determine the same conditions on $x$ $x$ and $y$ $y$ so this system of equations has an infinite number of solutions, namely all points on the line:

$y = 2 x + 8$ $y = 2x + 8$

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How do you solve by substitution $y - 2 x = 8$ $y - 2x = 8$ and $16 + 4 x = 2 y$ $16 + 4x = 2y$ ?

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Humanities

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How do you solve by substitution y−2x=8y - 2x = 8 and 16+4x=2y16 + 4x = 2y?

1 Answer

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How do you solve by substitution $y - 2 x = 8$ $y - 2x = 8$ and $16 + 4 x = 2 y$ $16 + 4x = 2y$ ?