1. Problem A: Simplify the expression $$(3^2 - 8) + \frac{2^2}{2} \cdot 3 - 3$$ using order of operations. 2. Calculate powers and parentheses first: $$3^2 = 9 \implies (9 - 8) = 1$$ $$2^2 = 4$$ 3. Substitute back into the expression: $$1 + \frac{4}{2} \cdot 3 - 3$$ 4. Division and multiplication from left to right: $$\frac{4}{2} = 2$$ $$2 \cdot 3 = 6$$ 5. Now expression is: $$1 + 6 - 3$$ 6. Addition and subtraction from left to right: $$1 + 6 = 7$$ $$7 - 3 = 4$$ **Final answer for A:** 4 --- 1. Problem B: Simplify the algebraic expression $$4x + 2 [ 3 (x - 1) + 2]$$ 2. Start inside the brackets, distribute multiplication: $$3(x - 1) = 3x - 3$$ 3. Substitute back: $$4x + 2 [ (3x - 3) + 2 ] = 4x + 2 (3x - 1)$$ 4. Distribute the 2: $$4x + 6x - 2$$ 5. Combine like terms: $$4x + 6x = 10x$$ **Final answer for B:** $$10x - 2$$ --- 1. Problem 3A: Solve the linear equation $$2x - 4 + 2(2x + 1) = 10$$ 2. Distribute the 2: $$2x - 4 + 4x + 2 = 10$$ 3. Combine like terms: $$2x + 4x = 6x$$ $$-4 + 2 = -2$$ 4. Simplify the equation: $$6x - 2 = 10$$ 5. Add 2 to both sides: $$6x = 12$$ 6. Divide both sides by 6: $$x = 2$$ **Final answer for 3A:** $$x = 2$$ --- 1. Problem 3B: Solve the quadratic equation $$x^2 + 4x + 3 = 0$$ 2. Factor the quadratic: Find two numbers that multiply to 3 and add to 4, which are 3 and 1. 3. Write factored form: $$(x + 3)(x + 1) = 0$$ 4. Solve each factor: $$x + 3 = 0 \Rightarrow x = -3$$ $$x + 1 = 0 \Rightarrow x = -1$$ **Final answers for 3B:** $$x = -3, -1$$