How do you solve the following system?: $\frac{1}{2} x = y - 1, \frac{1}{3} x = y - 5$ $1/2x=y-1 , 1/3x=y-5$

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How do you solve the following system?: $\frac{1}{2} x = y - 1, \frac{1}{3} x = y - 5$ $1/2x=y-1 , 1/3x=y-5$

1 Answer

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Answer 1 · 2016-07-25T17:06:40

$x = 24$ $x=24$
$y = 13$ $y=13$

Explanation:

$\frac{1}{2} x = y - 1$ $1/2x=y-1$ , $\frac{1}{3} x = y - 5$ $1/3x=y-5$
or
$x = 2 (y - 1)$ $x=2(y-1)$ , $x = 3 (y - 5)$ $x=3(y-5)$
or
$x = 2 y - 2$ $x=2y-2$ , $x = 3 y - 15$ $x=3y-15$
or
$2 y - 2 = 3 y - 15$ $2y-2=3y-15$
or
$3 y - 2 y = 15 - 2$ $3y-2y=15-2$
or
$y = 13$ $y=13$ ---------Ans $1$ $1$

Since $x = 2 y - 2$ $x=2y-2$
or
$x = 2 (13) - 2$ $x=2(13)-2$
or
$x = 26 - 2$ $x=26-2$
or
$x = 24$ $x=24$ --------Ans $2$ $2$

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How do you solve the following system?: $\frac{1}{2} x = y - 1, \frac{1}{3} x = y - 5$ $1/2x=y-1 , 1/3x=y-5$

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How do you solve the following system?: $\frac{1}{2} x = y - 1, \frac{1}{3} x = y - 5$ $1/2x=y-1 , 1/3x=y-5$