Which point would be a solution to the system of linear inequalities shown below? \[ y>-4 x+6 \quad y>\frac{1}{3} x-7 \] Answer \( \begin{array}{ll}(9,-7) & (-12,-2) \\ (12,1) & (-12,-7)\end{array} \)

We need to determine which of the given points satisfies both inequalities: 1. y > -4x + 6 2. y > (1/3)x - 7 Let's test each option. • For (9, -7):  – First inequality: -7 > -4(9) + 6 = -36 + 6 = -30.   Since -7 > -30 is true, the first inequality is satisfied.  – Second inequality: -7 > (1/3)(9) - 7 = 3 - 7 = -4.   Since -7 > -4 is false, (9, -7) does not satisfy both. • For (-12, -2):  – First inequality: -2 > -4(-12) + 6 = 48 + 6 = 54.   Since -2 > 54 is false, it fails and we don't need to check the second inequality. • For (12, 1):  – First inequality: 1 > -4(12) + 6 = -48 + 6 = -42.   Since 1 > -42 is true, the first inequality is satisfied.  – Second inequality: 1 > (1/3)(12) - 7 = 4 - 7 = -3.   Since 1 > -3 is also true, this point satisfies both. • For (-12, -7):  – First inequality: -7 > -4(-12) + 6 = 48 + 6 = 54.   Since -7 > 54 is false, it fails immediately. The only point that satisfies both inequalities is (12, 1). Thus, the solution is (12, 1).