Solve the linear system represented by the Augmentad matrix. \( \left[\begin{array}{ccccc}1 & -7 & -5 & 8 & 10 \\ 0 & 1 & -3 & 3 & -5 \\ 0 & 0 & 0 & 1 & 4\end{array}\right] \)

To solve the linear system represented by the augmented matrix, we can convert it into a system of equations. The provided matrix corresponds to the following equations: 1. \( x_1 - 7x_2 - 5x_3 + 8x_4 = 10 \) 2. \( x_2 - 3x_3 + 3x_4 = -5 \) 3. \( x_4 = 4 \) Now let's solve these equations step by step: From the third equation, we have: \( x_4 = 4 \) Now, substitute \( x_4 \) into the second equation: \( x_2 - 3x_3 + 3(4) = -5 \) This simplifies to: \( x_2 - 3x_3 + 12 = -5 \) So, \( x_2 - 3x_3 = -17 \) Now, we can express \( x_2 \) in terms of \( x_3 \): \( x_2 = -17 + 3x_3 \) (Equation 1) Next, substitute \( x_4 = 4 \) into the first equation: \( x_1 - 7x_2 - 5x_3 + 8(4) = 10 \) This simplifies to: \( x_1 - 7x_2 - 5x_3 + 32 = 10 \) So, \( x_1 - 7x_2 - 5x_3 = -22 \) Substituting Equation 1 into this equation, we get: \( x_1 - 7(-17 + 3x_3) - 5x_3 = -22 \) Distributing gives: \( x_1 + 119 - 21x_3 - 5x_3 = -22 \) Combining the like terms yields: \( x_1 - 26x_3 + 119 = -22 \) Thus, \( x_1 = -22 - 119 + 26x_3 \) Which simplifies to: \( x_1 = 26x_3 - 141 \) (Equation 2) Now we summarize our findings: 1. \( x_1 = 26x_3 - 141 \) 2. \( x_2 = -17 + 3x_3 \) 3. \( x_4 = 4 \) \( x_3 \) can be any real number since it is a free variable. Thus, the solution to the linear system can be expressed in parametric form, where \( x_3 = t \) (for some \( t \in \mathbb{R} \)): 1. \( x_1 = 26t - 141 \) 2. \( x_2 = -17 + 3t \) 3. \( x_3 = t \) 4. \( x_4 = 4 \) This represents the infinite solutions for the original linear system.