1. Solve the equation $n - (3 + 2n) = -6$. Start by distributing and combining like terms: $$n - 3 - 2n = -6$$ $$-n - 3 = -6$$ Add 3 to both sides: $$-n = -6 + 3$$ $$-n = -3$$ Multiply both sides by $-1$: $$n = 3$$ 2. Weekend bonus: Solve $$\frac{2^{t - y} \cdot 2^{y - 1}}{2^{y + 2}} = 8^{2y}$$ Simplify the left side by adding exponents with the same base 2: $$\frac{2^{(t - y) + (y - 1)}}{2^{y + 2}} = 8^{2y}$$ $$\frac{2^{t - 1}}{2^{y + 2}} = 8^{2y}$$ $$2^{t - 1 - (y + 2)} = 8^{2y}$$ $$2^{t - 1 - y - 2} = 8^{2y}$$ $$2^{t - y - 3} = 8^{2y}$$ Rewrite $8$ as $2^3$: $$2^{t - y - 3} = (2^3)^{2y} = 2^{6y}$$ Since bases are equal, exponents are equal: $$t - y - 3 = 6y$$ $$t - 3 = 7y$$ Solve for $y$: $$y = \frac{t - 3}{7}$$ 3. Solve for $x$: $$25^{2x} + 3(5^{2x}) = 4$$ Rewrite $25$ as $5^2$: $$ (5^2)^{2x} + 3(5^{2x}) = 4$$ $$5^{4x} + 3 imes 5^{2x} = 4$$ Set substitution $a = 5^{2x}$, then $a^2 = 5^{4x}$: $$a^2 + 3a - 4 = 0$$ Factor quadratic: $$(a + 4)(a - 1) = 0$$ So $a = -4$ (not valid since $5^{2x} > 0$) or $a = 1$. Solve $5^{2x} = 1$: $$5^{2x} = 5^0$$ Therefore: $$2x = 0$$ $$x = 0$$ 4. Solve inequality: $$90^x - 4(3^{2x + 1}) + 27 \geq 0$$ Rewrite $90 = 9 \times 10$ and realize $9 = 3^2$: Though not easily factorable, express parts in base 3: $$90^x = (9 \times 10)^x = 9^x \cdot 10^x = (3^2)^x \cdot 10^x = 3^{2x} \cdot 10^x$$ This mixes bases, so instead test critical points or rewrite inequality carefully. Alternatively, denote $a = 3^x$. Then: $$90^x = (9 \times 10)^x = 9^x \cdot 10^x = (3^2)^x \cdot 10^x = a^2 \cdot 10^x$$ Equation get complicated; better to handle numerical or graphing approach here. Since a direct algebraic solution is complex, numerical or graphical methods are recommended. **Final answers:** - For $n$: $n = 3$ - For bonus $y$: $y = \frac{t - 3}{7}$ - For $x$ in the third problem: $x = 0$ is the solution - For the inequality, numerical or graphing solution is needed.