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what are the solutions to this system of equation

The system of equations from the image is:

\[
1) \quad x + 2y - z = 5
\]
\[
2) \quad y + 2z = 6
\]
\[
3) \quad 4x + 2z = 5
\]

### Step 1: Solve Equation 2 for \( y \)

From the second equation:
\[
y + 2z = 6
\]
\[
y = 6 - 2z
\]

### Step 2: Substitute \( y = 6 - 2z \) into Equation 1

Now substitute \( y = 6 - 2z \) into the first equation:
\[
x + 2(6 - 2z) - z = 5
\]
Simplify:
\[
x + 12 - 4z - z = 5
\]
\[
x - 5z = 5 - 12
\]
\[
x - 5z = -7
\]
\[
x = 5z - 7 \quad \text{(Equation 4)}
\]

### Step 3: Substitute \( x = 5z - 7 \) into Equation 3

Now substitute \( x = 5z - 7 \) into the third equation:
\[
4(5z - 7) + 2z = 5
\]
Simplify:
\[
20z - 28 + 2z = 5
\]
\[
22z - 28 = 5
\]
\[
22z = 33
\]
\[
z = \frac{33}{22} = \frac{3}{2}
\]

### Step 4: Find \( x \) and \( y \)

Now that we have \( z = \frac{3}{2} \), substitute it back into Equation 4 to find \( x \):
\[
x = 5\left(\frac{3}{2}\right) - 7
\]
\[
x = \frac{15}{2} - 7 = \frac{15}{2} - \frac{14}{2} = \frac{1}{2}
\]

Finally, substitute \( z = \frac{3}{2} \) into \( y = 6 - 2z \) to find \( y \):
\[
y = 6 - 2\left(\frac{3}{2}\right)
\]
\[
y = 6 - 3 = 3
\]

### Solution:
\[
x = \frac{1}{2}, \quad y = 3, \quad z = \frac{3}{2}
\]

Would you like more details on any of the steps or further explanation?

Here are 5 related questions to explore further:
1. What are the different methods to solve systems of equations (e.g., substitution, elimination)?
2. How can matrices be used to solve this system of equations?
3. What is Cramer's Rule, and how can it be applied here?
4. How do changes to the coefficients in a system of equations affect the solution?
5. What are inconsistent and dependent systems of equations?

**Tip:** When solving systems of equations, it's a good idea to check the solution by substituting it back into the original equations.

What are the solutions to this system of equations?

Learn how to solve the system of equations x + 2y - z = 5, y + 2z = 6, and 4x + 2z = 5 using the substitution method. This step-by-step solution explains key algebraic concepts and formulas. Suitable for students in Grades 8-10.

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