Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x1, x2, and x3 in terms of the parameter t.)
2x1	 + 	x2	 − 	3x3	 = 	4
4x1			 + 	2x3	 = 	10
−6x1	 + 	3x2	 − 	15x3	 = 	−18
(x1, x2, x3) =

We are given a system of three linear equations:

1. \( 2x_1 + x_2 - 3x_3 = 4 \)
2. \( 4x_1 + 2x_3 = 10 \)
3. \( -6x_1 + 3x_2 - 15x_3 = -18 \)

We will solve this system step by step.

### Step 1: Simplify the second equation
The second equation is:
\[
4x_1 + 2x_3 = 10
\]
We can divide the entire equation by 2:
\[
2x_1 + x_3 = 5 \quad \text{(Equation 4)}
\]

### Step 2: Solve the first equation for \( x_2 \)
From the first equation:
\[
2x_1 + x_2 - 3x_3 = 4
\]
We solve for \( x_2 \):
\[
x_2 = 4 - 2x_1 + 3x_3 \quad \text{(Equation 5)}
\]

### Step 3: Simplify the third equation
The third equation is:
\[
-6x_1 + 3x_2 - 15x_3 = -18
\]
We divide the entire equation by 3:
\[
-2x_1 + x_2 - 5x_3 = -6 \quad \text{(Equation 6)}
\]

### Step 4: Substitute \( x_2 \) from Equation 5 into Equation 6
Substitute \( x_2 = 4 - 2x_1 + 3x_3 \) into Equation 6:
\[
-2x_1 + (4 - 2x_1 + 3x_3) - 5x_3 = -6
\]
Simplifying:
\[
-2x_1 + 4 - 2x_1 + 3x_3 - 5x_3 = -6
\]
Combine like terms:
\[
-4x_1 - 2x_3 + 4 = -6
\]
Subtract 4 from both sides:
\[
-4x_1 - 2x_3 = -10
\]
Divide by -2:
\[
2x_1 + x_3 = 5 \quad \text{(Equation 7)}
\]

### Step 5: Compare Equation 7 with Equation 4
Equation 7 is the same as Equation 4:
\[
2x_1 + x_3 = 5
\]
This indicates that there are infinitely many solutions because the equations are dependent. We can express the solution in terms of a parameter \( t \).

### Step 6: Express \( x_1 \), \( x_2 \), and \( x_3 \) in terms of \( t \)
Let \( x_3 = t \) (parameter \( t \)). Then from Equation 4:
\[
2x_1 + t = 5
\]
Solving for \( x_1 \):
\[
x_1 = \frac{5 - t}{2}
\]

Substitute \( x_1 \) and \( x_3 \) into Equation 5 to solve for \( x_2 \):
\[
x_2 = 4 - 2\left(\frac{5 - t}{2}\right) + 3t
\]
Simplifying:
\[
x_2 = 4 - (5 - t) + 3t
\]
\[
x_2 = 4 - 5 + t + 3t
\]
\[
x_2 = -1 + 4t
\]

### Final Answer:
\[
(x_1, x_2, x_3) = \left(\frac{5 - t}{2}, 4t - 1, t\right)
\]

This system has infinitely many solutions, with \( x_1 \), \( x_2 \), and \( x_3 \) expressed in terms of the parameter \( t \).

Would you like more details on any of the steps? Here are some related questions:

1. What does it mean for a system of equations to have infinitely many solutions?
2. How can we verify if a system of equations has no solution?
3. What are some methods to solve systems of linear equations?
4. How does the elimination method compare to substitution for solving such systems?
5. What is the role of parameters in describing solution sets for systems with infinite solutions?

**Tip**: When a system has dependent equations, always try to simplify and identify relations between the variables using parameters!

Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x1, x2, and x3 in terms of the parameter t.)
2x1 + x2 − 3x3 = 4
4x1 + 2x3 = 10
−6x1 + 3x2 − 15x3 = −18
(x1, x2, x3) =

Learn how to solve a system of three linear equations with parametric solutions. This problem involves concepts from linear algebra and solving systems of equations. Suitable for advanced high school algebra or college-level algebra students.

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