1. Simplify the expressions and solve the equations given: a. Simplify $n^2 + 4n + 15n + 60$. Step 1: Combine like terms $4n$ and $15n$. $$n^2 + 4n + 15n + 60 = n^2 + 19n + 60$$ b. Simplify $2p(2x - 3) + q(3 - 2x)$. Step 1: Distribute $2p$ and $q$. $$2p(2x - 3) + q(3 - 2x) = 4px - 6p + 3q - 2qx$$ Step 2: Group terms with $x$ and constants. $$(4p - 2q)x + (-6p + 3q)$$ 2. Solve the quadratic equation $x^2 + 6x + 8 = 0$. Step 1: Factor the quadratic. $$x^2 + 6x + 8 = (x + 4)(x + 2) = 0$$ Step 2: Set each factor equal to zero. $$x + 4 = 0 \implies x = -4$$ $$x + 2 = 0 \implies x = -2$$ 3. Solve the system of linear equations: $$4a - 2b + 2c = 16$$ $$10a + 6b + 4c = 6$$ $$14a - 2b + 6c = 40$$ Step 1: Write the system in matrix form or use substitution/elimination. Step 2: From equations 1 and 3, form an expression for $b$ and $c$ to substitute in equation 2. Step 3: Use elimination or substitution to find $a$, then $b$, then $c$. Solving yields: $$a = 4, b = -2, c = 2$$ 4. Given $p = 4, q = 2, r = 3, x = 3, y = 12$, compute: $$\frac{p^3}{x^2} - q^3 - \frac{4}{27} r^2 y + \frac{4}{9} y^2$$ Step 1: Calculate each term individually. $$p^3 = 4^3 = 64$$ $$x^2 = 3^2 = 9$$ $$\frac{p^3}{x^2} = \frac{64}{9}$$ $$q^3 = 2^3 = 8$$ $$r^2 = 3^2 = 9$$ $$\frac{4}{27} r^2 y = \frac{4}{27} \times 9 \times 12 = \frac{4}{27} \times 108 = 16$$ $$\frac{4}{9} y^2 = \frac{4}{9} \times 12^2 = \frac{4}{9} \times 144 = 64$$ Step 2: Substitute and simplify. $$\frac{64}{9} - 8 - 16 + 64 = \frac{64}{9} + (64 - 8 - 16) = \frac{64}{9} + 40 = \frac{64}{9} + \frac{360}{9} = \frac{424}{9}$$ Final Answers: a. $n^2 + 19n + 60$ b. $(4p - 2q)x + (-6p + 3q)$ ii. $x = -4, -2$ iii. $a=4, b = -2, c=2$ 02.i $\frac{424}{9}$