\( \left. \begin{array} { l } { \vec { u } = ( - 7,8 ) } \\ { \vec { w } = ( - 5 , - 1 ) } \\ { \frac { 1 } { 2 } \vec { u } - 2 \vec { w } = ( \square , \square ) } \end{array} \right. \)

Given \[ \vec{u} = (-7,8) \quad \text{and} \quad \vec{w} = (-5,-1), \] we want to calculate \[ \frac{1}{2}\vec{u} - 2\vec{w}. \] 1. First, compute \(\frac{1}{2}\vec{u}\): \[ \frac{1}{2}\vec{u} = \left(\frac{1}{2} \cdot (-7), \frac{1}{2} \cdot 8\right) = \left(-\frac{7}{2}, 4\right). \] 2. Next, compute \(-2\vec{w}\): \[ -2\vec{w} = -2 \cdot (-5,-1) = \left(-2 \cdot (-5), -2 \cdot (-1)\right) = (10,2). \] 3. Now, add the results: \[ \frac{1}{2}\vec{u} - 2\vec{w} = \left(-\frac{7}{2}, 4\right) + (10,2) = \left(-\frac{7}{2} + 10,\, 4 + 2\right). \] Convert \(10\) to a fraction with denominator 2: \[ 10 = \frac{20}{2}. \] Therefore, the \(x\)-coordinate is: \[ -\frac{7}{2} + \frac{20}{2} = \frac{13}{2}. \] 4. The \(y\)-coordinate is: \[ 4 + 2 = 6. \] Thus, the final answer is: \[ \frac{1}{2}\vec{u} - 2\vec{w} = \left(\frac{13}{2}, 6\right). \]