a) \( \left\{\begin{array}{c}x+y+w=10 \\ 2 x-2 y+w=4 \\ 3 x-y+w=3\end{array}\right. \)

To solve the system of equations, we can express it in matrix form or use substitution or elimination methods. First, let's write the equations again for clarity: 1. \( x + y + w = 10 \) (Equation 1) 2. \( 2x - 2y + w = 4 \) (Equation 2) 3. \( 3x - y + w = 3 \) (Equation 3) We can start with Equation 1 and express \( w \) in terms of \( x \) and \( y \): \( w = 10 - x - y \) Now, substitute \( w \) in the other two equations: 1. For Equation 2: \( 2x - 2y + (10 - x - y) = 4 \) This simplifies to: \( 2x - 2y + 10 - x - y = 4 \) \( x - 3y + 10 = 4 \) Leading to: \( x - 3y = -6 \quad (Equation 4) \) 2. For Equation 3: \( 3x - y + (10 - x - y) = 3 \) This simplifies to: \( 3x - y + 10 - x - y = 3 \) \( 2x - 2y + 10 = 3 \) Leading to: \( 2x - 2y = -7 \quad (Equation 5) \) Now we have a new system comprised of Equations 4 and 5: 1. \( x - 3y = -6 \) 2. \( 2x - 2y = -7 \) From Equation 4, we can express \( x \): \( x = 3y - 6 \) Now substitute this expression for \( x \) into Equation 5: \( 2(3y - 6) - 2y = -7 \) This simplifies to: \( 6y - 12 - 2y = -7 \) \( 4y - 12 = -7 \) \( 4y = 5 \) \( y = \frac{5}{4} \) Now plug \( y \) back into the expression for \( x \): \( x = 3\left(\frac{5}{4}\right) - 6 \) Calculating this gives: \( x = \frac{15}{4} - \frac{24}{4} = -\frac{9}{4} \) Finally, substitute both \( x \) and \( y \) back into the expression for \( w \): \( w = 10 - (-\frac{9}{4}) - \frac{5}{4} = 10 + \frac{9}{4} - \frac{5}{4} = 10 + \frac{4}{4} = 10 + 1 = 11 \) So the solution to the system is: \( x = -\frac{9}{4}, \, y = \frac{5}{4}, \, w = 11 \).