To find the [tex]\( z \)[/tex]-coordinate of the point [tex]\((x, y, z)\)[/tex] that satisfies the system of linear equations:
[tex]\[
\begin{array}{l}
5x - y - z = 10 \\
2x + y + z = 11 \\
y + 5 = z
\end{array}
\][/tex]
we need to solve the system step-by-step.
1. Start with the third equation:
[tex]\[
y + 5 = z
\][/tex]
We can rearrange this to express [tex]\( z \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
z = y + 5
\][/tex]
2. Substitute [tex]\( z = y + 5 \)[/tex] into the first and second equations.
For the first equation:
[tex]\[
5x - y - (y + 5) = 10
\][/tex]
Simplify this:
[tex]\[
5x - y - y - 5 = 10
\][/tex]
[tex]\[
5x - 2y - 5 = 10
\][/tex]
Add 5 to both sides:
[tex]\[
5x - 2y = 15
\][/tex]
Divide everything by 5:
[tex]\[
x - \frac{2}{5}y = 3
\][/tex]
Multiply everything by 5 to clear the fraction:
[tex]\[
5x - 2y = 15 \quad \text{(Equation A)}
\][/tex]
For the second equation:
[tex]\[
2x + y + (y + 5) = 11
\][/tex]
Simplify this:
[tex]\[
2x + y + y + 5 = 11
\][/tex]
[tex]\[
2x + 2y + 5 = 11
\][/tex]
Subtract 5 from both sides:
[tex]\[
2x + 2y = 6
\][/tex]
Divide everything by 2:
[tex]\[
x + y = 3 \quad \text{(Equation B)}
\][/tex]
3. Now we have a simplified system of two equations:
[tex]\[
\begin{array}{l}
5x - 2y = 15 \quad \text{(Equation A)} \\
x + y = 3 \quad \text{(Equation B)}
\end{array}
\][/tex]
4. Solve Equation B for [tex]\( x \)[/tex]:
[tex]\[
x = 3 - y
\][/tex]
5. Substitute [tex]\( x = 3 - y \)[/tex] into Equation A:
[tex]\[
5(3 - y) - 2y = 15
\][/tex]
Distribute 5:
[tex]\[
15 - 5y - 2y = 15
\][/tex]
Combine like terms:
[tex]\[
15 - 7y = 15
\][/tex]
Subtract 15 from both sides:
[tex]\[
-7y = 0
\][/tex]
Divide by -7:
[tex]\[
y = 0
\][/tex]
6. Substitute [tex]\( y = 0 \)[/tex] back into Equation B:
[tex]\[
x + 0 = 3
\][/tex]
[tex]\[
x = 3
\][/tex]
7. Finally, use [tex]\( z = y + 5 \)[/tex] to find [tex]\( z \)[/tex]:
[tex]\[
z = 0 + 5
\][/tex]
[tex]\[
z = 5
\][/tex]
Hence, the [tex]\( z \)[/tex]-coordinate of the solution is [tex]\( \boxed{5} \)[/tex].
