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

Apr 17, 2017

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

Explanation:

Given -

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

Take the first two equations and eliminate $y$

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

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

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

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

Take equations (4) and (7)

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

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

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

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

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

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

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