How do you solve the following system?: $\frac{1}{2} x = y - 1, - x + 3 y = - 7$ $1/2x=y-1 , -x+3y=-7$

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Answer 1 · 2018-04-24T05:07:32

$x = - 20$ $x=-20$

$y = - 9$ $y=-9$

The given equations are

$\frac{1}{2} x = y - 1$ $1/2x = y-1$
$- x + 3 y = - 7$ $-x+3y = -7$

From $\frac{1}{2} x = y - 1$ $1/2x = y-1$

$x = 2 (y - 1)$ $x=2(y-1)$

Substituting this $x$ $x$ value in equation $- x + 3 y = - 7$ $-x+3y = -7$

$\Rightarrow - (2 (y - 1) + 3 y = - 7$ $=> - (2(y-1) + 3y = -7$

$\Rightarrow - 2 y + 2 + 3 y = - 7$ $=> -2y +2 +3y = -7$

$\Rightarrow y + 2 = - 7$ $=> y+2 = -7$

$\Rightarrow y = - 7 - 2$ $=> y= -7-2$

$\Rightarrow y = - 9$ $=> y = -9$

Substitute this $y$ $y$ value in any of the given equations to obtain the $x$ $x$ value

$- x + 3 y = - 7$ $-x +3y = -7$

$\Rightarrow - x + 3 (- 9) = - 7$ $=> -x + 3(-9) = -7$

$\Rightarrow - x - 27 = - 7$ $=> -x -27 = -7$

$\Rightarrow x = - 20$ $=> x = -20$

How do you solve the following system?: 1/2x=y-1 , -x+3y=-7 12x=y−1,−x+3y=−7 1/2x=y-1 , -x+3y=-7