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Selesaikan menggunakan metode substitusi dan eliminasi

The system of equations you provided is:

\[
\begin{aligned}
1) & \quad 3x + 2y + 5z = 2 \\
2) & \quad 5x + 3y - 2z = 4 \\
3) & \quad 2x - 5y - 3z = 14
\end{aligned}
\]

To solve this system using the substitution and elimination methods, follow these steps:

### Step 1: Elimination of one variable
Let's eliminate \(z\) by manipulating equations 1) and 2).

Multiply equation 1 by 2 and equation 2 by 5, then add them together to eliminate \(z\).

\[
\begin{aligned}
2 \times (3x + 2y + 5z) &= 2 \times 2 \quad \text{(Multiply equation 1 by 2)} \\
6x + 4y + 10z &= 4 \quad \text{(Equation 4)}
\end{aligned}
\]

\[
\begin{aligned}
5 \times (5x + 3y - 2z) &= 5 \times 4 \quad \text{(Multiply equation 2 by 5)} \\
25x + 15y - 10z &= 20 \quad \text{(Equation 5)}
\end{aligned}
\]

Now, add equation 4 and equation 5:

\[
(6x + 4y + 10z) + (25x + 15y - 10z) = 4 + 20
\]
\[
31x + 19y = 24 \quad \text{(Equation 6)}
\]

### Step 2: Elimination of the same variable \(z\) using another pair of equations
Now use equation 1) and 3) to eliminate \(z\).

Multiply equation 1 by 3 and equation 3 by 5:

\[
\begin{aligned}
3 \times (3x + 2y + 5z) &= 3 \times 2 \quad \text{(Multiply equation 1 by 3)} \\
9x + 6y + 15z &= 6 \quad \text{(Equation 7)}
\end{aligned}
\]

\[
\begin{aligned}
5 \times (2x - 5y - 3z) &= 5 \times 14 \quad \text{(Multiply equation 3 by 5)} \\
10x - 25y - 15z &= 70 \quad \text{(Equation 8)}
\end{aligned}
\]

Add equation 7 and equation 8:

\[
(9x + 6y + 15z) + (10x - 25y - 15z) = 6 + 70
\]
\[
19x - 19y = 76 \quad \text{(Equation 9)}
\]
\[
x - y = 4 \quad \text{(Equation 10)}
\]

### Step 3: Substitute and solve for \(x\) and \(y\)
From equation 10, \(x = y + 4\).

Substitute \(x = y + 4\) into equation 6:

\[
31(y + 4) + 19y = 24
\]
\[
31y + 124 + 19y = 24
\]
\[
50y = -100
\]
\[
y = -2
\]

Substitute \(y = -2\) back into equation 10:

\[
x = -2 + 4 = 2
\]

### Step 4: Solve for \(z\)
Use equation 1) to solve for \(z\):

\[
3(2) + 2(-2) + 5z = 2
\]
\[
6 - 4 + 5z = 2
\]
\[
2 + 5z = 2
\]
\[
5z = 0
\]
\[
z = 0
\]

### Final Answer:
The solution to the system of equations is:

\[
x = 2, \quad y = -2, \quad z = 0
\]

Would you like more details on any step, or do you have any other questions?

---

**Related Questions:**

1. How would the solution change if the third equation was modified slightly?
2. What if one of the variables had a non-zero initial condition, how would the approach differ?
3. Can substitution and elimination methods be used effectively in larger systems of equations, like 4 or 5 variables?
4. How do you solve a system of equations graphically compared to algebraically?
5. How can matrix operations be applied to solve this system of equations?

**Tip:** When using the elimination method, always choose the variable to eliminate first by considering which will simplify the process most efficiently.

Learn how to solve a system of equations using the substitution and elimination methods with a detailed step-by-step solution. This problem involves three linear equations with variables x, y, and z. Suitable for high school students studying algebra and linear equations.

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