e) \( \left\{\begin{array}{c}3 x+y+4 z=2 \\ 2 x+6 y+2 z=17 \\ 4 x+2 y-2 z=35\end{array}\right. \)

To solve this system of equations, we can use the method of elimination or substitution. First, let's organize the equations clearly: 1) \( 3x + y + 4z = 2 \) 2) \( 2x + 6y + 2z = 17 \) 3) \( 4x + 2y - 2z = 35 \) Let's take a closer look at the second equation. We can simplify it by dividing all terms by 2: \( x + 3y + z = 8.5 \). Now we have a simpler system. It's helpful to express one variable in terms of the others. Let's solve for \( z \) in terms of \( y \) and \( x \): From the second simplified equation: \( z = 8.5 - x - 3y \). Next, we can substitute \( z \) back into the first and third equations and solve for the remaining variables \( x \) and \( y \): Substituting into the first equation: \( 3x + y + 4(8.5 - x - 3y) = 2 \). This expands to: \( 3x + y + 34 - 4x - 12y = 2 \). Simplifying gives us: \( -x - 11y = -32 \) or \( x + 11y = 32 \) (Equation 4). Next, substitute \( z \) into the third equation: \( 4x + 2y - 2(8.5 - x - 3y) = 35 \). This expands to: \( 4x + 2y - 17 + 2x + 6y = 35 \). Combining terms, we have: \( 6x + 8y - 17 = 35 \) or \( 6x + 8y = 52 \) (Equation 5). Now we can solve Equations 4 and 5: From Equation 4: \( x + 11y = 32 \) From Equation 5: \( 6x + 8y = 52 \) To eliminate \( x \), we can multiply Equation 4 by 6: \( 6x + 66y = 192 \) (Equation 6). Now subtract Equation 5 from Equation 6: \[ (6x + 66y) - (6x + 8y) = 192 - 52 \] Which simplifies to: \( 58y = 140 \) or \( y = \frac{140}{58} \approx 2.41 \). Now substitute \( y = 2.41 \) back into Equation 4 to find \( x \): \( x + 11(2.41) = 32 \) \( x + 26.51 = 32 \) \( x \approx 5.49 \). Finally, substitute \( x \) and \( y \) back into the expression for \( z \): \( z = 8.5 - 5.49 - 3(2.41) \approx -3.87 \). So the solution to the system is approximately: \( x \approx 5.49, y \approx 2.41, z \approx -3.87 \).