Let's solve each of the given equations step by step.

a) $3x + 10 = 6x + 31$

1. Subtract $3x$ from both sides to move the $x$-terms to one side: $10 = 3x + 31$
2. Subtract 31 from both sides to move the constants to one side: $10 - 31 = 3x$ $-21 = 3x$
3. Divide both sides by 3: $x = -7$

Solution: $x = -7$

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b) $\frac{3}{5}x = 10$

1. Multiply both sides by 5 to eliminate the fraction: $3x = 50$
2. Divide both sides by 3: $x = \frac{50}{3}$

Solution: $x = \frac{50}{3}$

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c) $\frac{3}{5}x + 1 = 2$

1. Subtract 1 from both sides: $\frac{3}{5}x = 1$
2. Multiply both sides by 5: $3x = 5$
3. Divide both sides by 3: $x = \frac{5}{3}$

Solution: $x = \frac{5}{3}$

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d) $5(x + 2) = 30$

1. Expand the left side: $5x + 10 = 30$
2. Subtract 10 from both sides: $5x = 20$
3. Divide both sides by 5: $x = 4$

Solution: $x = 4$

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e) $-3x + 5 > 5x + 25$

1. Add $3x$ to both sides: $5 > 8x + 25$
2. Subtract 25 from both sides: $-20 > 8x$
3. Divide both sides by 8: $x < \frac{-20}{8}$ $x < -2.5$

Solution: $x < -2.5$

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Let me know if you would like any further clarification on these solutions!

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5 related questions:

1. How do you solve equations involving more complex fractions?
2. What is the process of solving inequalities with absolute values?
3. Can you show how to graph the inequality $-3x + 5 > 5x + 25$?
4. What happens if there is more than one solution to a quadratic equation?
5. How do we handle systems of equations with two unknowns?

Tip: Always check your solution by substituting the value of $x$ back into the original equation to verify it satisfies the equation.