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3. Simplify the algebraic expressions:

a. $(a - b) + (2a + b)$: $a - b + 2a + b = 3a$

b. $0.5x - 7.9y - (1.2x - 3.8y)$: $0.5x - 7.9y - 1.2x + 3.8y = -0.7x - 4.1y$

c. $x - (x - 3x) - (x + 2x)$: $x - x + 3x - x - 2x = x$

d. $3m - 4n + [- (4m - 5n) - (7m + n)]$: $3m - 4n - 4m + 5n - 7m - n = -8m$

4. Perform the following multiplications:

a. $a(a - b)$: $a^2 - ab$

b. $6a^2b(a^2 - 9)$: $6a^2b(a^2 - 9) = 6a^4b - 54a^2b$

c. $\frac{1}{3}t\left(\frac{2}{3}t - \frac{1}{4}\right)$: $\frac{1}{3}t\left(\frac{2}{3}t - \frac{1}{4}\right) = \frac{2}{9}t^2 - \frac{1}{12}t$

d. $m^3(m - 1)$: $m^4 - m^3$

e. $(3a^3b - 5a)(-2a^3b^3)$: $(3a^3b - 5a)(-2a^3b^3) = -6a^6b^4 + 10a^4b^3$

f. $\frac{5}{4}x\left(\frac{9}{8}y - \frac{1}{4}m\right)$: $\frac{5}{4}x\left(\frac{9}{8}y - \frac{1}{4}m\right) = \frac{45}{32}xy - \frac{5}{16}xm$

5. Solve the following equations:

a. $-300 + X = -658$: $X = -658 + 300 = -358$

b. $-560 + X = -87$: $X = -87 + 560 = 473$

c. $- (210 - X) = -2$: $-210 + X = -2 \implies X = 208$

d. $- (X - 78) = -7$: $-X + 78 = -7 \implies X = 85$

6. Solve each equation:

a. $2x = 6$: $x = 3$

b. $2x - 3 = 6 + x$: $2x - x = 9 \implies x = 9$

c. $\frac{3}{4}(2x - 4) = x + 19$: $\frac{3}{4}(2x - 4) = x + 19 \implies \frac{3}{2}x - 3 = x + 19 \implies \frac{1}{2}x = 22 \implies x = 44$

d. $5x + 3x - 4 = 2x + 5$: $8x - 4 = 2x + 5 \implies 6x = 9 \implies x = \frac{3}{2}$

e. $b - (1 - 2b) + (b - 3) = -4$: $b - 1 + 2b + b - 3 = -4 \implies 4b - 4 = -4 \implies b = 0$

f. $-4(2a + 1) - (a - 2) = -11$: $-8a - 4 - a + 2 = -11 \implies -9a - 2 = -11 \implies -9a = -9 \implies a = 1$

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Expanding Questions:

1. How are terms combined when simplifying algebraic expressions?
2. What strategies are useful for solving equations involving fractions?
3. What is the importance of factoring in simplifying multiplications like in Question 4?
4. How can you verify your solution to an equation is correct?
5. How are distributive properties applied to equations in algebra?

Tip:

When solving equations, always double-check your work by substituting the solution back into the original equation to ensure it satisfies all conditions!