How do you solve the following linear system: $- x + 4 y = - 8, - 4 x + 2 y = - 8$ $-x + 4y = -8 , -4x+2y=-8$ ?

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Answer 1 · 2018-06-25T15:34:50

$x = \frac{8}{7}$ $x=8/7$ and $y = - \frac{12}{7}$ $y=-12/7$

$2 \cdot (- 4 x + 2 y) - (- x + 4 y) = 2 \cdot (- 8) - (- 8)$ $2*(-4x+2y)-(-x+4y)=2*(-8)-(-8)$

$- 8 x + 4 y + x - 4 y = - 8$ $-8x+4y+x-4y=-8$

$- 7 x = - 8$ $-7x=-8$ , so $x = \frac{- 8}{- 7} = \frac{8}{7}$ $x=(-8)/(-7)=8/7$

Hence,

$- 4 \cdot \frac{8}{7} + 2 y = - 8$ $-4*8/7+2y=-8$

$- \frac{32}{7} + 2 y = - 8$ $-32/7+2y=-8$

$2 y = - \frac{24}{7}$ $2y=-24/7$ or $y = - \frac{12}{7}$ $y=-12/7$

How do you solve the following linear system: −x+4y=−8,−4x+2y=−8-x + 4y = -8 , -4x+2y=-8 ?