How do you solve the system 3x - 4y - 2z = 13, 2x + y + 2z = 5, 6x +2y - z = 16?

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Answer 1 · 2018-01-26T14:44:37

$x = 3, y = - 1, z = 0$ $color(blue)(x = 3, y = -1, z = 0)$

$3 x - 4 y - 2 z = 13$ $3x - 4y - 2z = 13$ Eqn (1)

$2 x + y + 2 z = 5$ $2x + y + 2z = 5$ Eqn (2)

$6 x + 2 y - z = 16$ $6x + 2y - z = 16$ Eqn (3)

Add Eqn (1), (2) to eliminate z

$5 x - 3 y = 18$ $5x - 3y = 18$ Eqn (4)

Multiply Eqn (3) by 2 and add to Eqn (2)

$14 x + 5 y = 37$ $14x + 5y = 37$ Eqn (5)

Find x & y by solving equations (4) & (5)

Multiply Eqn (4) by 5

$25 x - 15 y = 90$ $25x - 15y = 90$ Eqn (6)

Multiply Eqn (5) by 3

$42 x + 15 y = 111$ $42x + 15y = 111$ Eqn (7)

Adding Eqns (6) & (7)

$67 x = 201$ $67x = 201$ or $x = 3$ $x = color(purple)(3)$

Substituting value of x in Eqn (4)

$(5 \cdot 3) - 3 y = 18$ $(5 * 3) - 3y = 18$

$3 y = 15 - 18 = - 3$ $3y = 15 - 18 = -3$ , $y = - 1$ $y = color(purple)( - 1)$

Substituting values of x & y in Eqn (3)

$(6 \cdot 3) + (2 \cdot (- 1)) - z = 16$ $(6*3) + (2 * (-1)) - z = 16$

$z = 18 - 2 - 16 = 0$ $z = 18 - 2 - 16 = 0$

$z = 0$ $z = color (purple)(0)$

$x = 3, y = - 1, z = - 0$ $color(blue)(x = 3, y = -1, z = -0)$