1. Problem 32 (first part): Given two parallel lines $l \parallel m$ crossed by a transversal $n$, with angles: $\angle 1 = (2x + y)^\circ$, $\angle 4 = (x + 2y)^\circ$, $\angle 6 = (3y + 20)^\circ$, Find $\angle 7$ and $\angle 8$. Step 1: Since $l \parallel m$, alternate interior and corresponding angles are equal. Step 2: Angles $1$ and $4$ are consecutive interior angles: their sum is $180^\circ$. $$ (2x + y) + (x + 2y) = 180 $$ Simplify: $$ 3x + 3y = 180 \implies x + y = 60 $$ Step 3: Angles $4$ and $6$ are corresponding angles and therefore equal: $$ x + 2y = 3y + 20 $$ Rearranged: $$ x - y = 20 $$ Step 4: Solve the system: $$ x + y = 60 $$ $$ x - y = 20 $$ Adding: $$ 2x = 80 \implies x = 40 $$ Substitute $x$: $$ 40 + y = 60 \implies y = 20 $$ Step 5: Calculate $\angle 7$ and $\angle 8$. Since angles $6$ and $7$ form a linear pair: $$ \angle 7 = 180 - \angle 6 $$ Evaluate $\angle 6$: $$ 3y + 20 = 3(20) + 20 = 60 + 20 = 80 $$ So: $$ \angle 7 = 180 - 80 = 100^\circ $$ Angles $7$ and $8$ are vertically opposite and thus equal: $$ \angle 8 = 100^\circ $$ --- 2. Problem 32 (second part): Given $AB \parallel CD$, and angles at $P$ are $120^\circ$, $150^\circ$ and $x^\circ$ adjacent to line $CD$. Step 1: Note that at point $P$, the angles around the point sum to $360^\circ$. Step 2: Sum of known angles plus $x$: $$ 120 + 150 + x = 180 $$ This equation doesn't balance, so likely $x$ is supplementary to $150^\circ$ since $AB \parallel CD$ implies alternate interior angles. Step 3: Since $x$ and $150^\circ$ are supplementary: $$ x + 150 = 180 \implies x = 30 $$ --- 3. Problem 33: Given $$ x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1},\quad y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} $$ Find: $$ x^2 + y^2 + 2xy $$ Step 1: Recognize that $x^2 + y^2 + 2xy = (x + y)^2$. Step 2: Find $x + y$: $$ x + y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} $$ Step 3: Bring to common denominator: $$ x + y = \frac{(\sqrt{5}+1)^2 + (\sqrt{5} -1)^2}{(\sqrt{5} -1)(\sqrt{5} +1)} $$ Step 4: Calculate numerator: $$(\sqrt{5} + 1)^2 = (\sqrt{5})^2 + 2\cdot \sqrt{5} \cdot 1 + 1^2 = 5 + 2\sqrt{5} + 1 = 6 + 2\sqrt{5}$$ $$(\sqrt{5} -1)^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5}$$ Sum numerator: $$ (6 + 2\sqrt{5}) + (6 - 2\sqrt{5}) = 12 $$ Step 5: Calculate denominator: $$(\sqrt{5} -1)(\sqrt{5} +1) = (\sqrt{5})^2 - 1^2 = 5 - 1 = 4$$ Step 6: Thus, $$ x + y = \frac{12}{4} = 3 $$ Step 7: Finally, $$ x^2 + y^2 + 2xy = (x + y)^2 = 3^2 = 9 $$ --- 4. Problem 34: Factorize polynomial $$ x^3 - 2x^2 - 5x + 6 $$ Step 1: Use Rational Root Theorem, test possible roots $\pm1, \pm2, \pm3, \pm6$. Step 2: Test $x = 1$: $$ 1 - 2 - 5 + 6 = 0 $$ So, $x = 1$ is a root. Step 3: Divide polynomial by $(x -1)$ using synthetic division: Coefficients: 1, -2, -5, 6 Carry down 1 Multiply and add: $1 \times 1=1$, sum $-2 +1 = -1$ $1 \times -1 = -1$, sum $-5 + (-1) = -6$ $1 \times -6 = -6$, sum $6 + (-6) = 0$ Result: $x^2 - x - 6$ Step 4: Factor quadratic: $$ x^2 - x - 6 = (x - 3)(x + 2) $$ Step 5: Final factorization: $$ (x - 1)(x - 3)(x + 2) $$ ---