To solve this system of equations, let's organize it and use either substitution or elimination to find values for $x$, $y$, $z$, and $t$.

Given Equations

1. $x + y + z = 0$
2. $x + z + t = 0$
3. $2z + t = 0$
4. $x + 2y - 3z + t = 0$

Step 1: Solve for $t$ in terms of $z$ from Equation 3

From Equation 3, we get: $t = -2z$

Step 2: Substitute $t = -2z$ into Equations 2 and 4

Substitute into Equation 2:

$x + z + (-2z) = 0$ $x - z = 0 \Rightarrow x = z$

Substitute into Equation 4:

$x + 2y - 3z + (-2z) = 0$ Substituting $x = z$ gives: $z + 2y - 3z - 2z = 0$ $2y - 4z = 0 \Rightarrow y = 2z$

Step 3: Substitute $x = z$, $y = 2z$, and $t = -2z$ back into the equations

Verify in Equation 1:

$x + y + z = 0$ $z + 2z + z = 0$

\Rightarrow z = 0$$ ### Step 4: Solve for $$x$$, $$y$$, and $$t$$ using $$z = 0$$ - $$x = z = 0$$ - $$y = 2z = 0$$ - $$t = -2z = 0$$ ### Solution The only solution is: $$x = 0, \; y = 0, \; z = 0, \; t = 0$$ This system has a unique solution at the origin $$(0, 0, 0, 0)$$. Would you like further details on the steps or any clarifications? --- **Related Questions:** 1. How would the solution change if there were no zero constant term? 2. What if an additional equation was added to make the system inconsistent? 3. How would you approach solving this system using matrix methods? 4. What happens if you substitute a non-zero value for $$z$$ in these equations? 5. Can this system have any non-trivial (non-zero) solutions? **Tip:** When dealing with systems of equations, start by simplifying individual equations for single variables, then substitute systematically to reduce the number of unknowns in each equation.