We are given a system of linear equations:
[tex]\[
\begin{cases}
-16x + 16y + 6z = 40 \\
8x - 5y - 4z = 0 \\
-4x + 4y + 2z = 8
\end{cases}
\][/tex]
along with [tex]\( x = 0 \)[/tex]. We need to find the values of [tex]\( y \)[/tex] and [tex]\( z \)[/tex].
1. Substitute [tex]\( x = 0 \)[/tex] into the equations:
[tex]\[
\begin{cases}
-16(0) + 16y + 6z = 40 \\
8(0) - 5y - 4z = 0 \\
-4(0) + 4y + 2z = 8
\end{cases}
\][/tex]
Simplifying these, we get:
[tex]\[
\begin{cases}
16y + 6z = 40 \\
-5y - 4z = 0 \\
4y + 2z = 8
\end{cases}
\][/tex]
2. Solve the simplified system of equations:
First, let's use the third equation (it has simpler coefficients):
[tex]\[
4y + 2z = 8
\][/tex]
Divide everything by 2:
[tex]\[
2y + z = 4 \quad \text{(Equation 1)}
\][/tex]
Next, let's simplify the first equation:
[tex]\[
16y + 6z = 40
\][/tex]
Divide everything by 2 as well:
[tex]\[
8y + 3z = 20 \quad \text{(Equation 2)}
\][/tex]
We now have:
[tex]\[
\begin{cases}
2y + z = 4 \\
8y + 3z = 20
\end{cases}
\][/tex]
3. Solve for [tex]\( y \)[/tex] and [tex]\( z \)[/tex] using substitution or elimination:
To use elimination, we can multiply Equation 1 by 3 to align [tex]\( z \)[/tex]-coefficients:
[tex]\[
3(2y + z) = 3(4) \\
6y + 3z = 12 \quad \text{(Equation 3)}
\][/tex]
Now, subtract Equation 3 from Equation 2:
[tex]\[
(8y + 3z) - (6y + 3z) = 20 - 12 \\
8y + 3z - 6y - 3z = 8 \\
2y = 8 \\
y = 4
\][/tex]
Substituting [tex]\( y = 4 \)[/tex] back into Equation 1:
[tex]\[
2(4) + z = 4 \\
8 + z = 4 \\
z = 4 - 8 \\
z = -4
\][/tex]
Thus, the solutions are:
[tex]\[
y = 4, \quad z = -4
\][/tex]
Therefore, the values are [tex]\( \boxed{4} \)[/tex] for [tex]\( y \)[/tex] and [tex]\( \boxed{-4} \)[/tex] for [tex]\( z \)[/tex].
