1. **Simplify the expressions involving polynomials** Given: $$P(x) = 2x^2 + x - 6$$ $$H(x) = 6x^2 - 7x - 5$$ (a) Simplify $2P(x) + 3H(x)$: $$2P(x) = 2(2x^2 + x - 6) = 4x^2 + 2x - 12$$ $$3H(x) = 3(6x^2 - 7x - 5) = 18x^2 - 21x - 15$$ Add them: $$2P(x) + 3H(x) = (4x^2 + 2x - 12) + (18x^2 - 21x - 15) = (4x^2 + 18x^2) + (2x - 21x) + (-12 - 15) = 22x^2 - 19x - 27$$ (b) Simplify $P(x) - H(x)$: $$P(x) - H(x) = (2x^2 + x - 6) - (6x^2 - 7x - 5) = 2x^2 + x - 6 - 6x^2 + 7x + 5 = (2x^2 - 6x^2) + (x + 7x) + (-6 + 5) = -4x^2 + 8x - 1$$ 2. **Solve the simultaneous linear equations:** $$\begin{cases} 3a - b = 9 \\ a + 4b = -14 \end{cases}$$ From the first equation: $$b = 3a - 9$$ Substitute into the second: $$a + 4(3a - 9) = -14$$ $$a + 12a - 36 = -14$$ $$13a = 22$$ $$a = \frac{22}{13}$$ Find $b$: $$b = 3\times \frac{22}{13} - 9 = \frac{66}{13} - \frac{117}{13} = -\frac{51}{13}$$ 3. **Triangle with sides AB = 6, BC = 5, AC = 7** This is a triangle with side lengths 6, 5, and 7. We can find the area using Heron's formula: Calculate semi-perimeter: $$s = \frac{6 + 5 + 7}{2} = 9$$ Area: $$\sqrt{s(s - AB)(s - BC)(s - AC)} = \sqrt{9(9 - 6)(9 - 5)(9 - 7)} = \sqrt{9 \times 3 \times 4 \times 2} = \sqrt{216} = 6\sqrt{6}$$ **Final answers:** (a) $2P(x) + 3H(x) = 22x^2 - 19x - 27$ (b) $P(x) - H(x) = -4x^2 + 8x - 1$ Simultaneous solution: $$a = \frac{22}{13}, \quad b = -\frac{51}{13}$$ Triangle area: $$6\sqrt{6}$$