Question

solve each system of equations. If the system has no solution, say that it is inconsistent

x + 4y - 3z = -8

3x - y + 3z = 12

x + y + 6z = 1

Short answer

The solution of the system can be given as$\left\{\right(x,y,z)=(-3,-\frac{8}{3},\frac{1}{9}\left)\right\}$

Step-by-step solution

Given information

We are given a system of equations

$$\begin{gathered} x+4y-3z=-8\left(1\right) \\ 3x-y+3z=12\left(2\right) \\ x+y+6z=1\left(3\right) \end{gathered}$$

Add equation 1 and 2

We get,

$x+4y-3z=-8+3x-y+3z=124x+3y=4$

Hence we get, $4x+3y=4$ (4)

Now Multiply equation 1 by 2 and add it to equation 3

We get,

$$\begin{gathered} 2(x+4y-3z=-8) \\ 2x+8y-6z=-16 \end{gathered}$$

Also

$2x+8x-6z=-16+x+y+6z=13x+9y=-15$

Hence we get

$$\begin{gathered} 3x+9y=-15 \\ x+3y=-5\left(5\right) \end{gathered}$$

Subtract equation 4 and 5

We get,

$4x+3y=4-x-3y=53x=9$

hence x=3

Find the values of y and z

We get,

$$\begin{gathered} x+3y=-5 \\ 3+3y=-5 \\ 3y=-8 \\ y=-\frac{8}{3} \end{gathered}$$

Now also we have

$$\begin{gathered} x+4y-3z=-8 \\ -3+4(-\frac{8}{3})-3z=-8 \\ z=\frac{1}{9} \end{gathered}$$

The solution set is$\left\{\right(x,y,z)=(-3,-\frac{8}{3},\frac{1}{9}\left)\right\}$