Question

what is the solution to the following system?\\\

$$\begin{cases}4x + 3y - z = -6\\\\6x - y + 3z = 12\\\\8x + 2y + 4z = 6\\end{cases}$$

\\\\(\boldsymbol{x = 1, y = -3, z = -1}\\)\\(\boldsymbol{x = 1, y = -3, z = 1}\\)\\(\boldsymbol{x = 1, y =, z = 19}\\)\\(\boldsymbol{x = 1, y = 3, z = -2}\\)

Explanation:

Step1: Test the first option \(x = 1, y=-3, z = -1\) in the first equation \(4x + 3y - z\)

Substitute \(x = 1\), \(y=-3\), \(z = -1\) into \(4x+3y - z\):
\(4(1)+3(-3)-(-1)=4 - 9 + 1=-4
eq - 6\). So this option is wrong.

Step2: Test the second option \(x = 1, y=-3, z = 1\) in the first equation \(4x + 3y - z\)

Substitute \(x = 1\), \(y = - 3\), \(z=1\) into \(4x + 3y-z\):
\(4(1)+3(-3)-1=4-9 - 1=-6\).

Step3: Test the second option in the second equation \(6x-y + 3z\)

Substitute \(x = 1\), \(y=-3\), \(z = 1\) into \(6x-y + 3z\):
\(6(1)-(-3)+3(1)=6 + 3+3 = 12\).

Step4: Test the second option in the third equation \(8x+2y + 4z\)

Substitute \(x = 1\), \(y=-3\), \(z = 1\) into \(8x + 2y+4z\):
\(8(1)+2(-3)+4(1)=8-6 + 4=6\).
All three equations are satisfied. We can stop here as we have found the correct solution.

Answer:

\(x = 1, y=-3, z = 1\) (the second option: \(x = 1, y=-3, z = 1\))