1. **Problem:** Check the work for the given problems from the test. 2. **Additive inverse of $-\frac{2}{3}$:** - The additive inverse of a number $a$ is $-a$. - Given $-\frac{2}{3}$, its additive inverse is $\frac{2}{3}$. - Check: $-\frac{2}{3} + \frac{2}{3} = 0$. 3. **Multiplicative inverse of $-\frac{2}{3}$:** - The multiplicative inverse of a number $a$ (nonzero) is $\frac{1}{a}$. - Given $-\frac{2}{3}$, its multiplicative inverse is $-\frac{3}{2}$. - Check: $-\frac{2}{3} \times -\frac{3}{2} = 1$. 4. **Values of $x$ in $\{-6, -4, -3, 2, 3\}$ where $x > -3$:** - Check each: $2 > -3$ true, $3 > -3$ true. - So, $\{2, 3\}$. 5. **Set builder notation for integers less than 5:** - Integers less than 5 are $\{..., -3, -2, -1, 0, 1, 2, 3, 4\}$. - Set builder: $\{x \mid x \in \mathbb{Z}, x < 5\}$. 6. **Simplify $4[x - 4(y - 2[5y + 2])]$:** - Calculate inside brackets: $5y + 2 = 5y + 2$. - Multiply by 2: $2(5y + 2) = 10y + 4$. - Then $y - (10y + 4) = y - 10y - 4 = -9y - 4$. - Substitute back: $x - 4(-9y - 4) = x + 36y + 16$. - Multiply by 4: $4(x + 36y + 16) = 4x + 144y + 64$. 7. **Solve $|3x - 2| < 8$ and express in interval notation:** - Inequality splits: $-8 < 3x - 2 < 8$. - Add 2: $-6 < 3x < 10$. - Divide by 3: $-2 < x < \frac{10}{3}$. - Interval notation: $(-2, \frac{10}{3})$. 8. **Solve equation $\frac{x - 2}{4} - \frac{x + 5}{6} = \frac{5x - 2}{9}$:** - LCD is 36. - Multiply both sides by 36: $$36 \times \left(\frac{x - 2}{4} - \frac{x + 5}{6}\right) = 36 \times \frac{5x - 2}{9}$$ - Distribute: $$9(x - 2) - 6(x + 5) = 4(5x - 2)$$ - Expand: $$9x - 18 - 6x - 30 = 20x - 8$$ - Simplify left: $$3x - 48 = 20x - 8$$ - Rearrange: $$3x - 20x = -8 + 48$$ $$-17x = 40$$ - Solve for $x$: $$x = \frac{40}{-17} = -\frac{40}{17}$$ 9. **Simplify $\frac{3(7 - 2)^2}{2^2} \div \frac{6 - 4^2}{10 - 4 \cdot 3}$:** - Calculate powers and operations: $$(7 - 2)^2 = 5^2 = 25$$ $$2^2 = 4$$ $$6 - 16 = -10$$ $$10 - 12 = -2$$ - Division becomes multiplication by reciprocal: $$\frac{3 \times 25}{4} \times \frac{-2}{-10} = \frac{75}{4} \times \frac{1}{5} = \frac{75}{20} = \frac{15}{4}$$ 10. **Mixture problem:** - Table: - Cashews: amount $x$, cost per unit 8, value $8x$. - Pistachios: amount 20, cost per unit 16, value 320. - Mixed nuts: amount $x + 20$, cost per unit 12, value $12(x + 20)$. - Equation: $$8x + 320 = 12(x + 20)$$ - Expand right: $$8x + 320 = 12x + 240$$ - Rearrange: $$320 - 240 = 12x - 8x$$ $$80 = 4x$$ - Solve for $x$: $$x = 20$$ **All work is correct and consistent with the problem statements.**