How do you solve $1/2x + 1/3y = 5 and 1/4x + y = 10$ using substitution?

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How do you solve $1/2x + 1/3y = 5 and 1/4x + y = 10$ using substitution?

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Answer 1 · 2018-05-26T08:22:58

$x=4,y=9$

Explanation:

From the second equation we get
$y=10-1/4x$
plugging this in the first one
$1/2x+1/3(10-1/4x)=5$
$1/2x+10/3-1/12x=5$
Multplying by $12$
$6x+40-x=60$
$5x=20$
$x=4$
so
$y=9$

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How do you solve $1/2x + 1/3y = 5 and 1/4x + y = 10$ using substitution?

1 Answer

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How do you solve 1/2x + 1/3y = 5 and 1/4x + y = 10 1/2x + 1/3y = 5 and 1/4x + y = 10 using substitution?

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How do you solve $1/2x + 1/3y = 5 and 1/4x + y = 10$ using substitution?