Let's work through the problem you provided.

Part (a):

We are given the parametric equations: $x_1 = t, \quad x_2 = 2 + t, \quad x_3 = 3 - t$ We need to find two linear equations that result in this parametric solution.

To solve this, let's express the parametric equations in terms of linear relationships. From the parametric equations, we observe the following relationships:

1. From $x_1 = t$, we have $t = x_1$.
2. Substituting $t = x_1$ into $x_2 = 2 + t$, we get: $x_2 = 2 + x_1.$ So, the first equation is: $x_2 - x_1 = 2.$
3. Similarly, substitute $t = x_1$ into $x_3 = 3 - t$, yielding: $x_3 = 3 - x_1.$ Thus, the second equation is: $x_3 + x_1 = 3.$

So the system of two linear equations is: $x_2 - x_1 = 2, \quad x_3 + x_1 = 3.$

Part (b):

We are asked to find another parametric solution where the parameter is $s$ and $x_3 = s$.

In part (a), the relation between $x_1$ and $x_3$ is $x_3 = 3 - x_1$. Given that $x_3 = s$, this implies: $s = 3 - x_1 \quad \Rightarrow \quad x_1 = 3 - s.$

Next, we use the equation $x_2 - x_1 = 2$ from part (a). Substituting $x_1 = 3 - s$ into this equation: $x_2 - (3 - s) = 2 \quad \Rightarrow \quad x_2 = 2 + 3 - s = 5 - s.$

Thus, the new parametric equations are: $x_1 = 3 - s, \quad x_2 = 5 - s, \quad x_3 = s.$

Summary:

• (a) The system of two linear equations is: $x_2 - x_1 = 2, \quad x_3 + x_1 = 3.$
• (b) The parametric solution where $x_3 = s$ is: $x_1 = 3 - s, \quad x_2 = 5 - s, \quad x_3 = s.$

Would you like any further details or have any questions?

Here are 5 follow-up questions to consider:

1. How do we convert parametric equations to a system of linear equations?
2. Why does the second parametric solution differ from the first one?
3. How can we check if a given set of equations is linear?
4. What method can we use to solve systems of linear equations algebraically?
5. How does changing the parameter in parametric equations affect the solution set?

Tip: Always check if your parametric solutions satisfy the original system of linear equations.