Solve the System. Give answer as (x,y,z).

2x-6y-3z=-26

-1x-12y+3z=-11

4x-24y-3z=-64

We are given the system of equations:

\[
1. \ 2x - 6y - 3z = -26
\]
\[
2. \ -x - 12y + 3z = -11
\]
\[
3. \ 4x - 24y - 3z = -64
\]

To solve this system, let's proceed step by step.

### Step 1: Simplify the equations, if possible

Equations 1 and 3 are already simplified. For Equation 2, we can multiply the whole equation by \(-1\) to make the coefficient of \(x\) positive:

\[
x + 12y - 3z = 11
\]

### Step 2: Eliminate one variable

We will eliminate \(z\) by adding or subtracting the equations. 

First, let's add Equation 1 and Equation 3:

\[
(2x - 6y - 3z) + (4x - 24y - 3z) = -26 + (-64)
\]
\[
6x - 30y - 6z = -90
\]

Now divide the entire equation by 6:
\[
x - 5y - z = -15 \quad \text{(Equation 4)}
\]

### Step 3: Eliminate \(z\) again with a different pair of equations

Now, let's add Equation 1 and the modified Equation 2 (after multiplying Equation 2 by 1):

\[
(2x - 6y - 3z) + (x + 12y - 3z) = -26 + 11
\]
\[
3x + 6y - 6z = -15
\]

Now divide this equation by 3:
\[
x + 2y - 2z = -5 \quad \text{(Equation 5)}
\]

### Step 4: Solve the new system

Now we have a simpler system of two equations in terms of \(x\) and \(y\) from Equation 4 and Equation 5:

1. \(x - 5y - z = -15\)
2. \(x + 2y - 2z = -5\)

We can now solve these two equations using substitution or elimination. Let's use substitution:

From Equation 1:
\[
x - 5y = z - 15 \quad \text{or} \quad z = x - 5y + 15
\]

Substitue in...

Solve the System. Give answer as (x,y,z).

2x - 6y - 3z = -26

-1x - 12y + 3z = -11

4x - 24y - 3z = -64

Discover how to solve a system of linear equations step-by-step using substitution and elimination methods. This problem involves three equations in three variables and is suitable for Grades 10-12.

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