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

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

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Answer 1 · 2018-03-31T13:48:10

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

Explanation:

$- x - 8 y = - 16$ $-x-8y=-16$

$6 x + 4 y = 8$ $6x+4y=8$

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

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

Cramer's rule can be applied

$(x, y) = (\frac{b_{11} a_{22} - b_{22} a_{12}}{a_{11} a_{22} - a_{21} a_{12}}, \frac{a_{11} b_{22} - a_{22} b_{12}}{a_{11} a_{22} - a_{21} a_{12}})$ $(x,y)=((b_(11)a_(22)-b_(22)a_(12))/(a_(11)a_(22)-a_(21)a_(12)),(a_(11)b_(22)-a_(22)b_(12))/(a_(11)a_(22)-a_(21)a_(12)))$

Thus, comparing

$x = \frac{(- 16 \times 4 - 8 \times (- 8))}{(- 1 \times 4 - 6 \times - 8)} = \frac{(- 64 + 64)}{(- 4 + 48)} = \frac{0}{44} = 0$ $x=((-16xx4-8xx(-8)))/((-1xx4-6xx-8))= ((-64+64))/((-4+48))=0/44=0$

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

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

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

1 Answer

Explanation:

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

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