The reduced row-echelon form of the augmented matrix for a system of linear equations with variables x1, ... , x5 is given below. Determine the solutions for the system and enter them below.

1 0 0 -3 -3 = 4
0 1 0 -2 -5 =-1
0 0 0 0 0 = 0
If the system has infinitely many solutions, your answer may use expressions involving the parameters r, s, and t.

Please help, it's very urgent.

To determine the solutions for the system of linear equations, we look at the reduced row-echelon form of the augmented matrix given. From the last row of the matrix, we can observe that the equation 0 = 0 holds true. This implies that the system has dependent equations, meaning there are infinitely many solutions.

To find these solutions, we need to express the variables in terms of the parameters r, s, and t. We can start by writing the equations in terms of the variables x1, x2, x3, x4, and x5:

x1 - 3x4 - 3x5 = 4 (Equation 1)
x2 - 2x4 - 5x5 = -1 (Equation 2)

Now let's solve for x1 and x2 in terms of the remaining variables:

From Equation 1, we have:
x1 = 3x4 + 3x5 + 4

From Equation 2, we have:
x2 = 2x4 + 5x5 - 1

Since x3 is a free variable (no pivot in its column), we can express it in terms of a parameter. Let's use r as the parameter:

x3 = r

Finally, we express x4 and x5 in terms of the remaining free variables s and t:

x4 = s
x5 = t

Therefore, the solutions for the system of linear equations are:

x1 = 3s + 3t + 4
x2 = 2s + 5t - 1
x3 = r
x4 = s
x5 = t

If you have any specific values for r, s, and t, you can substitute them into these equations to obtain the corresponding solutions.