# How do you solve the system of equations algebraically x+2y+3z=5, 3x+2y-2z=-13, 5x+3y-z=-11?

Dec 30, 2016

The Soln. is x=1,y=-4, &, z=4.

#### Explanation:

Let us give nos. (1),(2) & (3) to the given eqns.

$3 \left(3\right) + \left(1\right) \Rightarrow 15 x + 9 y - 3 z + x + 2 y + 3 z = - 33 + 5$

$\Rightarrow 16 x + 11 y = - 28. \ldots \ldots \ldots \ldots \ldots \ldots . . \left(4\right)$

$\left(2\right) - 2 \left(3\right) \Rightarrow 3 x + 2 y - 2 z - 10 x - 6 y + 2 z = - 13 + 22$

$\Rightarrow - 7 x - 4 y = 9. \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(5\right)$

$4 \left(4\right) + 11 \left(5\right) \Rightarrow 64 x + 44 y - 77 x - 44 y = - 112 + 99$

$\Rightarrow - 13 x = - 13 \Rightarrow x = 1. \ldots \ldots \ldots \ldots . \left(6\right)$

(6) & (5) rArr -7-4y=9 rArr -4y=16

$y = - 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(7\right)$

(6),(7) & (3) rArr 5-12-z=-11

$z = 4. \ldots \ldots \ldots \ldots \ldots . \left(8\right)$

Hence, the Soln. is x=1,y=-4, &, z=4.