# How do you solve the system of equations algebraically -x-3y+z=54, 4x+2y-3z=-32, 2y+8z=78?

Aug 6, 2017

Solution : $x = 11 , y = - 17 , z = 14$

#### Explanation:

$- x - 3 y + z = 54 \left(1\right) , 4 x + 2 y - 3 z = - 32 \left(2\right)$

$2 y + 8 z = 78 \left(3\right) : y = \frac{78 - 8 z}{2} = 39 - 4 z$ Putting $y = 39 - 4 z$

In equation (1) and (2) we get $- x - 3 \left(39 - 4 z\right) + z = 54$ or

$- x + 13 z = 171 \left(4\right) \mathmr{and} 4 x + 2 \left(39 - 4 z\right) - 3 z = - 32 \mathmr{and} 4 x - 11 z = - 110 \left(5\right)$

$\left(4\right) \cdot 4 : - 4 x + 52 z = 684 \left(6\right)$ Adding (5) and (6) we get

$41 z = 574 \therefore z = 1 4 \therefore 2 y + 8 \cdot 14 = 78 \therefore 2 y = 78 - 112$

or 2y = -34 :. y = -17 ; -x - 3* -17 + 14 =54 or x = 51+14-54=11

Solution : $x = 11 , y = - 17 , z = 14$ [Ans]