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2. Tentukan nilai [tex]\( u \)[/tex], [tex]\( y \)[/tex], dan [tex]\( z \)[/tex] dari persamaan berikut:
[tex]\[
\begin{array}{l}
2x + 3y + z = 2 \\
u - 2x - 3z = 1 \\
-3u - 5y + z = 0 \\
\end{array}
\][/tex]

Sagot :

To determine the values of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] from the given system of equations:

1. Starting with the system of equations:
[tex]\[ \begin{aligned} 2x + 3y + z &= 2 \quad \text{(1)} \\ u - 2u - 3z &= 1 \quad \text{(2)} \\ -3u - 5y + z &= 0 \quad \text{(3)} \end{aligned} \][/tex]

2. Simplify the second equation:
[tex]\[ u - 2u - 3z = 1 \implies -u - 3z = 1 \implies -u = 1 + 3z \implies u = -1 - 3z \][/tex]

3. Substitute [tex]\( u = -1 - 3z \)[/tex] into the third equation:
[tex]\[ -3(-1 - 3z) - 5y + z = 0 \rightarrow 3 + 9z - 5y + z = 0 \rightarrow 10z - 5y + 3 = 0 \rightarrow 5y = 10z + 3 \rightarrow y = 2z + \frac{3}{5} \][/tex]

4. Next, substitute [tex]\( y = 2z + \frac{3}{5} \)[/tex] into the first equation:
[tex]\[ 2x + 3\left(2z + \frac{3}{5}\right) + z = 2 \rightarrow 2x + 6z + \frac{9}{5} + z = 2 \rightarrow 2x + 7z + \frac{9}{5} = 2 \rightarrow 2x + 7z = 2 - \frac{9}{5} \][/tex]
Simplify the right-hand side:
[tex]\[ 2 - \frac{9}{5} = \frac{10}{5} - \frac{9}{5} = \frac{1}{5} \][/tex]
Thus:
[tex]\[ 2x + 7z = \frac{1}{5} \rightarrow 2x = \frac{1}{5} - 7z \rightarrow x = \frac{1}{10} - \frac{7z}{2} \][/tex]

5. Simplify [tex]\( x \)[/tex]:
[tex]\[ x = \frac{1}{10} - \frac{7z}{2} = \frac{1 - 35z}{10} \][/tex]

6. Rewrite [tex]\( x \)[/tex] in a simpler form:
[tex]\[ x = \frac{1}{10} - \frac{35z}{10} \rightarrow x = \frac{1 - 35z}{10} \][/tex]

Finally, we have:

[tex]\[ \begin{aligned} x &= \frac{1}{10} - \frac{7}{2}z \rightarrow x = \frac{7z}{6} + \frac{19}{15} \\ y &= 2z + \frac{3}{5} \rightarrow y = -2z/3 - 1/15 \\ z &= - \frac{u}{3} - \frac{1}{3} \end{aligned} \][/tex]

The values of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] in terms of [tex]\( u \)[/tex] are:
[tex]\[ \boxed{x = \frac{7u}{6} + \frac{19}{15}, \quad y = -\frac{2u}{3} - \frac{1}{15}, \quad z = -\frac{u}{3} - \frac{1}{3}} \][/tex]
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