How do you solve the system of equations $4 x + 2 y + z = 7, 2 x + 2 y + 4 z = - 4, x + 3 y - 2 z = - 8$ $4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8$ ?

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How do you solve the system of equations $4 x + 2 y + z = 7, 2 x + 2 y + 4 z = - 4, x + 3 y - 2 z = - 8$ $4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8$ ?

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Answer 1 · 2017-04-17T07:29:49

$x = \frac{47}{11}$ $x=47/11$
$y = \frac{- 51}{11}$ $y=(-51)/11$
$z = \frac{- 9}{11}$ $z=(-9)/11$

Explanation:

Given -

$4 x + 2 y + z = 7$ $4x+2y+z=7$ ----------------(1)
$2 x + 2 y + 4 z = - 4$ $2x+2y+4z=-4$ -------------(2)
$x + 3 y - 2 z = - 8$ $x+3y-2z=-8$ ----------------(3)

Take the first two equations and eliminate $y$ $y$

$4 x + 2 y + z = 7$ $4x+2y+z=7$ ----------------(1)
$2 x + 2 y + 4 z = - 4$ $2x+2y+4z=-4$ -------------(2) -------- $(1) - (2)$ $(1)-(2)$
$2 x - 3 z = 11$ $2x-3z=11$ -------------(4)

Take equations (1) and (3)
We have to elimiate $y$ $y$ term

$4 x + 2 y + z = 7$ $4x+2y+z=7$ ----------------(1) ------- $\times 3$ $xx3$
$x + 3 y - 2 z = - 8$ $x+3y-2z=-8$ ----------------(3) ------- $\times 2$ $xx2$

$12 x + 6 y + 3 z = 21$ $12x+6y+3z=21$ ------------(5)
$2 x + 6 y - 4 z = - 16$ $2x+6y-4z=-16$ -------------(6) ----- $(5) - (6)$ $(5)-(6)$
$10 x + 7 z = 37$ $10x+7z=37$ ----------------(7)

Take equations (4) and (7)

$2 x - 3 z = 11$ $2x-3z=11$ -------------(4) -------- $\times 5$ $xx5$
$10 x + 7 z = 37$ $10x+7z=37$ ----------------(7)

$10 x - 15 z = 55$ $10x-15z=55$ ---------(8)
$10 x + 7 z = 37$ $10x+7z=37$ -----------(7) ------ $(8) - (7)$ $(8)-(7)$
$- 22 z = 18$ $-22z=18$
$z = - \frac{18}{22}$ $z=-18/22$

Plugin > $z = - \frac{18}{22}$ $z=-18/22$ in equation (7)

$10 x + 7 (- \frac{18}{22}) = 37$ $10x+7(-18/22)=37$
$10 x - \frac{126}{22} = 37$ $10x-126/22=37$
$10 x = 37 + \frac{126}{22} = \frac{814 + 126}{22} = \frac{940}{22} = \frac{470}{11}$ $10x=37+126/22=(814+126)/22=940/22=470/11$
$x = \frac{470}{11} \times \frac{1}{10} = \frac{470}{110} = \frac{47}{11}$ $x=470/11xx1/10=470/110=47/11$
$x = \frac{47}{11}$ $x=47/11$

Plugin > $z = - \frac{18}{22}$ $z=-18/22$ and > $x = \frac{47}{11}$ $x=47/11$ in equation (1)

$4 x + 2 y + z = 7$ $4x+2y+z=7$ ----------------(1)
$4 (\frac{47}{11}) + 2 y - \frac{18}{22} = 7$ $4(47/11)+2y-18/22=7$
$4 (\frac{47}{11}) + 2 y - \frac{9}{11} = 7$ $4(47/11)+2y-9/11=7$
$\frac{188}{11} + 2 y - \frac{9}{11} = 7$ $188/11+2y-9/11=7$
$\frac{188 - 9}{11} + 2 y = 7$ $(188-9)/11+2y=7$
$\frac{179}{11} + 2 y = 7$ $179/11+2y=7$
$2 y = 7 - \frac{179}{11} = \frac{77 - 179}{11} = \frac{- 102}{11}$ $2y=7-179/11=(77-179)/11=(-102)/11$
$y = \frac{- 102}{11} \times \frac{1}{2} = \frac{- 102}{22} =$ $y=(-102)/11xx1/2=(-102)/22=$
$y = \frac{- 51}{11}$ $y=(-51)/11$

$x = \frac{47}{11}$ $x=47/11$
$y = \frac{- 51}{11}$ $y=(-51)/11$
$z = - \frac{18}{22} = \frac{- 9}{11}$ $z=-18/22=(-9)/11$

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How do you solve the system of equations $4 x + 2 y + z = 7, 2 x + 2 y + 4 z = - 4, x + 3 y - 2 z = - 8$ $4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8$ ?

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How do you solve the system of equations 4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-84x+2y+z=7,2x+2y+4z=−4,x+3y−2z=−84x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8?

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How do you solve the system of equations $4 x + 2 y + z = 7, 2 x + 2 y + 4 z = - 4, x + 3 y - 2 z = - 8$ $4x+2y+z=7, 2x+2y+4z=-4, x+3y-2z=-8$ ?