5x+4y+2z=
–
13
x+3y–5z=
–
4
4x–2y+6z=18

find x y z

To solve for \(x\), \(y\), and \(z\) in the given system of linear equations:

\[
\begin{align*}
5x + 4y + 2z &= -13 \quad \text{(Equation 1)} \\
x + 3y - 5z &= -4 \quad \text{(Equation 2)} \\
4x - 2y + 6z &= 18 \quad \text{(Equation 3)}
\end{align*}
\]

we can use the method of elimination or substitution. Here, we'll use the elimination method to find the solution step-by-step.

### Step 1: Eliminate \(x\) from Equations 1 and 3

First, we will eliminate \(x\) from Equation 1 and Equation 3 by multiplying Equation 2 by 4 and subtracting from Equation 3:

\[
\text{Equation 2} \times 4: \quad 4x + 12y - 20z = -16
\]

Subtract this result from Equation 3:

\[
(4x - 2y + 6z) - (4x + 12y - 20z) = 18 - (-16)
\]

Simplifying, we get:

\[
-14y + 26z = 34 \quad \text{(Equation 4)}
\]

### Step 2: Eliminate \(x\) from Equations 1 and 2

Next, let's eliminate \(x\) from Equations 1 and 2. We can multiply Equation 2 by 5 and subtract from Equation 1:

\[
\text{Equation 2} \times 5: \quad 5x + 15y - 25z = -20
\]

Subtract this from Equation 1:

\[
(5x + 4y + 2z) - (5x + 15y - 25z) = -13 - (-20)
\]

Simplifying, we get:

\[
-11y + 27z = 7 \quad \text{(Equation 5)}
\]

### Step 3: Solve for \(y\) and \(z\)

Now we have two equations with two unknowns (\(y\) and \(z\)):

\[
\begin{align*}
-14y + 26z &= 34 \quad \text{(Equation 4)} \\
-11y + 27z &= 7 \quad \text{(Equation 5)}
\end{align*}
\]

To eliminate \(y\), multiply Equation 4 by 11 and Equation 5 by 14:

\[
\begin{align*}
-154y + 286z &= 374 \quad \text{(Equation 4) multiplied by 11} \\
-154y + 378z &= 98 \quad \text{(Equation 5) multiplied by 14}
\end{align*}
\]

Subtract the first from the second:

\[
(286z - 378z) = 374 - 98
\]

Simplifying, we get:

\[
-92z = 276
\]

So,

\[
z = -3
\]

Now substitute \(z = -3\) back into Equation 4:

\[
-14y + 26(-3) = 34
\]

\[
-14y - 78 = 34
\]

\[
-14y = 112
\]

\[
y = -8
\]

### Step 4: Solve for \(x\)

Now substitute \(y = -8\) and \(z = -3\) back into Equation 2:

\[
x + 3(-8) - 5(-3) = -4
\]

\[
x - 24 + 15 = -4
\]

\[
x - 9 = -4
\]

\[
x = 5
\]

### Solution

The solution to the system is:

\[
x = 5, \quad y = -8, \quad z = -3
\]

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

---

Here are 5 related questions you might find interesting:

1. How do you solve a system of linear equations using the substitution method?
2. What are some other methods to solve systems of equations besides elimination and substitution?
3. How can you determine if a system of linear equations has no solution or infinitely many solutions?
4. What is the geometric interpretation of solving a system of three linear equations in three variables?
5. How can systems of equations be applied to real-world problems?

**Tip:** When solving systems of equations, always check the results by substituting back into the original equations to ensure accuracy.

Learn how to solve a system of linear equations 5x + 4y + 2z = -13, x + 3y - 5z = -4, 4x - 2y + 6z = 18 using the elimination method step-by-step. This problem involves solving for variables x, y, and z using algebraic techniques. Suitable for high school students.

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