1. **State the problem:** We are given vectors $\overline{A} = (4, -6)$, $\overline{B} = (1, 4)$, and $\overline{P} = (-1, -15)$. We want to verify or find $\overline{P}$ using the equation: $$\overline{P} = \frac{1}{2} \overline{A} - 4 \overline{B}$$ 2. **Recall the formula and rules:** Vector scalar multiplication means multiplying each component by the scalar. Vector subtraction means subtracting corresponding components. 3. **Calculate each term:** $$\frac{1}{2} \overline{A} = \frac{1}{2} (4, -6) = \left(\frac{1}{2} \times 4, \frac{1}{2} \times -6\right) = (2, -3)$$ $$4 \overline{B} = 4 (1, 4) = (4 \times 1, 4 \times 4) = (4, 16)$$ 4. **Substitute and simplify:** $$\overline{P} = (2, -3) - (4, 16) = (2 - 4, -3 - 16) = (-2, -19)$$ 5. **Compare with given $\overline{P}$:** The calculated $\overline{P}$ is $(-2, -19)$, but the given $\overline{P}$ is $(-1, -15)$. 6. **Conclusion:** The given $\overline{P}$ does not satisfy the equation $\overline{P} = \frac{1}{2} \overline{A} - 4 \overline{B}$. The correct vector $\overline{P}$ from the equation is $(-2, -19)$.