Question

method with back substitution to solve the following system of linear equations. if there is a solution, write your for
o solution\ or write \is\ for \infinite solutions.\ \\(\

$$\begin{cases} 2x + y - z = 2 \\\\ x + y - 3z = -2 \\\\ 3x + y + z = 6 \\end{cases}$$

\\)

Explanation:

Step1: Eliminate \( y \) from equations

Subtract the second equation from the first: \((2x + y - z) - (x + y - 3z) = 2 - (-2)\)
Simplify: \(2x + y - z - x - y + 3z = 4\) → \(x + 2z = 4\) (Equation 4)

Subtract the second equation from the third: \((3x + y + z) - (x + y - 3z) = 6 - (-2)\)
Simplify: \(3x + y + z - x - y + 3z = 8\) → \(2x + 4z = 8\) (Equation 5)

Step2: Simplify Equation 5

Divide Equation 5 by 2: \(\frac{2x + 4z}{2} = \frac{8}{2}\) → \(x + 2z = 4\) (same as Equation 4, so dependent)

Let \( z = t \) (parameter). From Equation 4: \(x + 2t = 4\) → \(x = 4 - 2t\)

Step3: Substitute \( x \) and \( z \) into second equation

\((4 - 2t) + y - 3t = -2\) → \(4 - 5t + y = -2\) → \(y = 5t - 6\)

Answer:

Infinite solutions (IS), with \( x = 4 - 2t \), \( y = 5t - 6 \), \( z = t \) (where \( t \) is a real number)