Solve the following system of equations for all three variables.
4, x, minus, 8, y, plus, 5, z, equals, minus, 8
4x−8y+5z=
−8
minus, 9, x, plus, 8, y, minus, 6, z, equals, minus, 4
−9x+8y−6z=
−4
10, x, minus, 8, y, plus, 7, z, equals, 0
10x−8y+7z=
0

We have the following system of linear equations:

\[
\begin{aligned}
1. \quad 4x - 8y + 5z &= -8 \\
2. \quad -9x + 8y - 6z &= -4 \\
3. \quad 10x - 8y + 7z &= 0
\end{aligned}
\]

Let's solve this system using the elimination or substitution method.

### Step 1: Add Equation 1 and Equation 2 to eliminate \( y \)
\[
(4x - 8y + 5z) + (-9x + 8y - 6z) = -8 + (-4)
\]
Simplifying:
\[
(4x - 9x) + (-8y + 8y) + (5z - 6z) = -12
\]
\[
-5x - z = -12 \quad \text{(Equation 4)}
\]

### Step 2: Add Equation 1 and Equation 3 to eliminate \( y \)
\[
(4x - 8y + 5z) + (10x - 8y + 7z) = -8 + 0
\]
Simplifying:
\[
(4x + 10x) + (-8y - 8y) + (5z + 7z) = -8
\]
\[
14x + 12z = -8 \quad \text{(Equation 5)}
\]

### Step 3: Solve Equations 4 and 5 as a system of two variables (\( x \) and \( z \))
We have:
1. \( -5x - z = -12 \quad \text{(Equation 4)} \)
2. \( 14x + 12z = -8 \quad \text{(Equation 5)} \)

#### Solve Equation 4 for \( z \):
\[
z = -12 + 5x
\]

#### Substitute into Equation 5:
\[
14x + 12(-12 + 5x) = -8
\]
\[
14x + 12(-12) + 12(5x) = -8
\]
\[
14x - 144 + 60x = -8
\]
\[
74x - 144 = -8
\]
\[
74x = 136
\]
\[
x = \frac{136}{74} = \frac{68}{37}
\]

### Step 4: Substitute \( x = \frac{68}{37} \) into \( z = -12 + 5x \)
\[
z = -12 + 5 \times \frac{68}{37} = -12 + \frac{340}{37}
\]
Converting -12 to have a common denominator:
\[
z = \frac{-444}{37} + \frac{340}{37} = \frac{-104}{37}
\]
Thus, \( z = \frac{-104}{37} \).

### Step 5: Solve for \( y \)
Substitute \( x = \frac{68}{37} \) and \( z = \frac{-104}{37} \) into one of the original equations, say Equation 1:
\[
4x - 8y + 5z = -8
\]
Substitute the values of \( x \) and \( z \):
\[
4\left(\frac{68}{37}\right) - 8y + 5\left(\frac{-104}{37}\right) = -8
\]
Simplify:
\[
\frac{272}{37} - 8y + \frac{-520}{37} = -8
\]
Combine the fractions:
\[
\frac{272 - 520}{37} - 8y = -8
\]
\[
\frac{-248}{37} - 8y = -8
\]
Convert -8 to have the same denominator:
\[
\frac{-248}{37} - 8y = \frac{-296}{37}
\]
Now solve for \( y \):
\[
-8y = \frac{-296}{37} + \frac{248}{37}
\]
\[
-8y = \frac{-48}{37}
\]
\[
y = \frac{6}{37}
\]

### Final Solution:
\[
x = \frac{68}{37}, \quad y = \frac{6}{37}, \quad z = \frac{-104}{37}
\]

Let me know if you'd like more details on any step!

### Related Questions:
1. What are the conditions for a system of linear equations to have a unique solution?
2. How can you solve a system of equations using matrices?
3. What is Cramer's Rule and how does it apply to systems of equations?
4. How can elimination and substitution methods be combined for solving systems of equations?
5. How does one handle systems of nonlinear equations?

**Tip:** Always check your final solution by substituting it back into the original equations to verify correctness.

Solve the following system of equations for all three variables:
4x − 8y + 5z = −8
−9x + 8y − 6z = −4
10x − 8y + 7z = 0

Discover how to solve the system of linear equations 4x − 8y + 5z = −8, −9x + 8y − 6z = −4, and 10x − 8y + 7z = 0. This solution uses both the elimination and substitution methods, making it perfect for high school students studying algebra. Includes step-by-step breakdowns of each method.

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