Let's solve the system of equations step by step. The system of linear equations is:

1. $2x_1 + x_2 - 3x_3 = 4$
2. $4x_1 + 2x_3 = 10$
3. $-6x_1 + 3x_2 - 15x_3 = -18$

Step 1: Solve the second equation for $x_1$

The second equation is:

$4x_1 + 2x_3 = 10$

Divide through by 2 to simplify:

$2x_1 + x_3 = 5$

Now solve for $x_1$:

$x_1 = \frac{5 - x_3}{2}$

Step 2: Substitute $x_1$ into the first and third equations

First equation:

The first equation is:

$2x_1 + x_2 - 3x_3 = 4$

Substitute $x_1 = \frac{5 - x_3}{2}$ into this equation:

$2\left(\frac{5 - x_3}{2}\right) + x_2 - 3x_3 = 4$

Simplifying:

$(5 - x_3) + x_2 - 3x_3 = 4$

$5 - 4x_3 + x_2 = 4$

Now solve for $x_2$:

$x_2 = 4x_3 - 1$

Third equation:

The third equation is:

$-6x_1 + 3x_2 - 15x_3 = -18$

Substitute $x_1 = \frac{5 - x_3}{2}$ and $x_2 = 4x_3 - 1$:

$-6\left(\frac{5 - x_3}{2}\right) + 3(4x_3 - 1) - 15x_3 = -18$

Simplifying:

$-3(5 - x_3) + 12x_3 - 3 - 15x_3 = -18$

$-15 + 3x_3 + 12x_3 - 3 - 15x_3 = -18$

$-18 = -18$

This is always true, which means the third equation is dependent on the others. So, there are infinitely many solutions.

Step 3: Express $x_1, x_2, x_3$ in terms of a parameter

Let $x_3 = t$ (where $t$ is a free parameter). Then:

$x_1 = \frac{5 - t}{2}$ $x_2 = 4t - 1$ $x_3 = t$

Thus, the solution is:

$(x_1, x_2, x_3) = \left( \frac{5 - t}{2}, 4t - 1, t \right)$

Final Answer:

$(x_1, x_2, x_3) = \left( \frac{5 - t}{2}, 4t - 1, t \right)$

Would you like more details or have any questions?

Related Questions:

1. How can you check if a system of equations has no solution?
2. What are dependent and independent equations in a system?
3. How do you interpret parametric solutions geometrically?
4. Can you solve a system of equations using matrices?
5. How does the method of substitution compare to Gaussian elimination?

Tip: For large systems, matrix methods like Gaussian elimination or using inverses simplify the solving process.