1. **Problem 1:** Find $x$ given that $|P| = -10$, where $$P = \begin{bmatrix} x+3 & x+2 \\ x+1 & x-1 \end{bmatrix}.$$ The determinant of $P$ is $$|P| = (x+3)(x-1) - (x+2)(x+1).$$ Calculate the determinant: $$= (x^2 - x + 3x - 3) - (x^2 + x + 2x + 2)$$ $$= (x^2 + 2x - 3) - (x^2 + 3x + 2)$$ $$= x^2 + 2x - 3 - x^2 - 3x - 2 = -x - 5.$$ 2. We are given $$|P| = -10,$$ so $$-x - 5 = -10.$$ Solve for $x$: $$-x = -10 + 5 = -5,$$ $$x = 5.$$ **Answer for Problem 1:** $x=5$ 3. **Problem 2:** Given $$2^n = 128,$$ find the value of $$ (2^{n-1})(5^{n-2}).$$ Since $128 = 2^7,$ we have $$n=7.$$ Calculate the expression: $$ (2^{7-1})(5^{7-2}) = 2^6 \cdot 5^5.$$ Calculate powers: $$2^6 = 64,$$ $$5^5 = 3125.$$ Multiply: $$64 \times 3125 = 200000 = 2 \times 10^5.$$ **Answer for Problem 2:** $2(10^5)$ 4. **Problem 3:** Find $p, q$ satisfying the matrix equation: $$\begin{bmatrix} 2p & 8 \\ 3 & -5q \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 24 \\ -17 \end{bmatrix}.$$ Multiply the matrix and vector: $$\begin{cases} 2p \cdot 1 + 8 \cdot 2 = 24 \\ 3 \cdot 1 + (-5q) \cdot 2 = -17 \end{cases}$$ Simplify equations: $$2p + 16 = 24,$$ $$3 - 10q = -17.$$ Solve first: $$2p = 24 -16 = 8,$$ $$p = 4.$$ Solve second: $$-10q = -17 - 3 = -20,$$ $$q = 2.$$ **Answer for Problem 3:** $(p,q) = (4,2)$ 5. **Problem 4:** Given $$\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}} = x + y \sqrt{15},$$ Find $x$ (and possibly $y$). Simplify the left side: $$= \frac{\sqrt{3}}{\sqrt{5}} + \frac{\sqrt{5}}{\sqrt{5}} = \sqrt{\frac{3}{5}} + 1.$$ Rewrite $$\sqrt{\frac{3}{5}} = \frac{\sqrt{15}}{5}.$$ So the expression is: $$1 + \frac{\sqrt{15}}{5} = x + y \sqrt{15}.$$ By comparing terms: $$x = 1,$$ $$y = \frac{1}{5}.$$ **Answer for Problem 4:** $x=1$ Final summary: - Problem 1 answer: $5$ - Problem 2 answer: $2(10^5)$ - Problem 3 answer: $(4,2)$ - Problem 4 answer: $x=1$