Solve using the substitution method, 5x+5y+7z=10, 4x-2y-5z=33, 2x+4y+6z=0

We are given the system of three equations:

\[
\begin{align*}
1. & \quad 5x + 5y + 7z = 10 \\
2. & \quad 4x - 2y - 5z = 33 \\
3. & \quad 2x + 4y + 6z = 0
\end{align*}
\]

### Step 1: Solve one equation for one variable
Let's solve the third equation, \( 2x + 4y + 6z = 0 \), for \( x \):

\[
2x + 4y + 6z = 0 \quad \text{divide by 2:} \quad x + 2y + 3z = 0
\]

Thus, we have:

\[
x = -2y - 3z
\]

### Step 2: Substitute \( x = -2y - 3z \) into the first and second equations

Substitute into the first equation \( 5x + 5y + 7z = 10 \):

\[
5(-2y - 3z) + 5y + 7z = 10
\]
Simplify:

\[
-10y - 15z + 5y + 7z = 10
\]
\[
-5y - 8z = 10
\]
\[
5y + 8z = -10 \quad \text{(Equation 4)}
\]

Now, substitute \( x = -2y - 3z \) into the second equation \( 4x - 2y - 5z = 33 \):

\[
4(-2y - 3z) - 2y - 5z = 33
\]
Simplify:

\[
-8y - 12z - 2y - 5z = 33
\]
\[
-10y - 17z = 33 \quad \text{(Equation 5)}
\]

### Step 3: Solve the system of two equations in terms of \( y \) and \( z \)

We now have the system:

\[
\begin{align*}
4. & \quad 5y + 8z = -10 \\
5. & \quad -10y - 17z = 33
\end{align*}
\]

To eliminate one variable, let's multiply Equation 4 by 2:

\[
2(5y + 8z) = 2(-10)
\]
\[
10y + 16z = -20 \quad \text{(Equation 6)}
\]

Now, add Equation 6 to Equation 5:

\[
(10y + 16z) + (-10y - 17z) = -20 + 33
\]
Simplify:

\[
-1z = 13 \quad \Rightarrow \quad z = -13
\]

### Step 4: Substitute \( z = -13 \) into one of the original equations
Substitute \( z = -13 \) into Equation 4:

\[
5y + 8(-13) = -10
\]
Simplify:

\[
5y - 104 = -10
\]
\[
5y = 94 \quad \Rightarrow \quad y = \frac{94}{5} = 18.8
\]

### Step 5: Solve for \( x \)
Substitute \( y = 18.8 \) and \( z = -13 \) into the expression for \( x = -2y - 3z \):

\[
x = -2(18.8) - 3(-13)
\]
\[
x = -37.6 + 39 = 1.4
\]

### Final Solution:
The solution to the system is:

\[
x = 1.4, \quad y = 18.8, \quad z = -13
\]

Would you like more details or have any questions?

Here are five related questions to consider:
1. How would you check if the solution is correct?
2. Can substitution be applied to non-linear systems?
3. How do other methods (like elimination) compare to substitution?
4. What happens if a system has no solution or infinite solutions?
5. How would you solve this system graphically?

**Tip:** Always simplify equations before starting substitution to make calculations easier.

Learn how to solve a 3-variable system of equations using the substitution method. The problem involves solving for x, y, and z from the equations 5x + 5y + 7z = 10, 4x - 2y - 5z = 33, and 2x + 4y + 6z = 0. Detailed step-by-step explanation for students in Grades 9-11.

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