How do you solve the system of equations $-x - 8y = - 16$ and $6x + 4y = 8$ ?

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Answer 1 · 2018-03-31T13:48:10

$(x,y)-=(0,2)$

$-x-8y=-16$

$6x+4y=8$

If
$a_(11)x+a_(12)y=b_(11)$

$a_(21)x+a_(22)y=b_(22)$

Cramer's rule can be applied

$(x,y)=((b_(11)a_(22)-b_(22)a_(12))/(a_(11)a_(22)-a_(21)a_(12)),(a_(11)b_(22)-a_(22)b_(12))/(a_(11)a_(22)-a_(21)a_(12)))$

Thus, comparing

$x=((-16xx4-8xx(-8)))/((-1xx4-6xx-8))= ((-64+64))/((-4+48))=0/44=0$

$y=((-1xx8-6xx(-16)))/((-1xx4-6xx-8))=((-8+96))/((-4+48))=88/44=2$

$(x,y)-=(0,2)$

How do you solve the system of equations -x - 8y = - 16-x - 8y = - 16 and 6x + 4y = 86x + 4y = 8?