d) \( \left\{\begin{array}{c}3 x+2 y+z=1 \\ 5 x+3 y+4 z=2 \\ x+y-z=1\end{array}\right. \)

To solve the system of equations, we can use methods such as substitution or elimination. First, let's rearrange the equations according to a standard form and identify a simple equation to start from. 1. \( 3x + 2y + z = 1 \) (Equation 1) 2. \( 5x + 3y + 4z = 2 \) (Equation 2) 3. \( x + y - z = 1 \) (Equation 3) We can express \( z \) from the third equation: \( z = x + y - 1 \). Now, substitute \( z \) into the first two equations. Substituting \( z \) into Equation 1: \( 3x + 2y + (x + y - 1) = 1 \) \( 4x + 3y - 1 = 1 \) \( 4x + 3y = 2 \) (Equation 4) Substituting \( z \) into Equation 2: \( 5x + 3y + 4(x + y - 1) = 2 \) \( 5x + 3y + 4x + 4y - 4 = 2 \) \( 9x + 7y - 4 = 2 \) \( 9x + 7y = 6 \) (Equation 5) Now we have a new system of two equations: 1. \( 4x + 3y = 2 \) 2. \( 9x + 7y = 6 \) We can solve these two equations simultaneously. Let's multiply Equation 4 by 7 and Equation 5 by 3 to eliminate \( y \): 1. \( 28x + 21y = 14 \) 2. \( 27x + 21y = 18 \) Now, subtract Equation 2 from Equation 1: \( (28x + 21y) - (27x + 21y) = 14 - 18 \) \( x = -4 \) Substituting \( x = -4 \) back into Equation 4: \( 4(-4) + 3y = 2 \) \( -16 + 3y = 2 \) \( 3y = 18 \) \( y = 6 \) Now substituting \( x \) and \( y \) back into Equation 3 to find \( z \): \( -4 + 6 - z = 1 \) \( 2 - z = 1 \) \( z = 1 \) The solution to the system of equations is: \( x = -4 \), \( y = 6 \), and \( z = 1 \).