1. **Problem:** Given vectors $a = \left(-7, -15\right)$ and $b = \left(10, 16\right)$, find $8\left(a + \frac{1}{2}b\right)$. 2. **Formula and rules:** To solve this, we use vector addition and scalar multiplication. The sum of two vectors $a$ and $b$ is $a + b = \left(a_x + b_x, a_y + b_y\right)$. Scalar multiplication of a vector $v = (v_x, v_y)$ by a scalar $k$ is $k v = (k v_x, k v_y)$. 3. **Calculate $\frac{1}{2}b$:** $$\frac{1}{2}b = \frac{1}{2} \times (10, 16) = \left(\frac{1}{2} \times 10, \frac{1}{2} \times 16\right) = (5, 8)$$ 4. **Add vectors $a$ and $\frac{1}{2}b$:** $$a + \frac{1}{2}b = (-7, -15) + (5, 8) = (-7 + 5, -15 + 8) = (-2, -7)$$ 5. **Multiply the result by 8:** $$8 \times (-2, -7) = (8 \times -2, 8 \times -7) = (-16, -56)$$ **Final answer:** $$8\left(a + \frac{1}{2}b\right) = (-16, -56)$$