Opgave 2:
a) Løs ligningen:
3𝑥𝑥 + 10 = 6𝑥𝑥 + 31
b) Løs ligningen:
3
5 𝑥𝑥 = 10
c) Løs ligningen:
3
5𝑥𝑥 + 1 = 2
d) Løs ligningen:
5(𝑥𝑥 + 2) = 30
e) Løs uligheden:
−3𝑥𝑥 + 5 > 5𝑥𝑥 + 25

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:** \( x = -7 \)

---

### 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:** \( x = \frac{50}{3} \)

---

### 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:** \( x = \frac{5}{3} \)

---

### 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:** \( x = 4 \)

---

### 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:** \( x < -2.5 \)

---

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

---

**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.

Opgave 2: a) Løs ligningen: 3x + 10 = 6x + 31 b) Løs ligningen: 3/5 x = 10 c) Løs ligningen: 3/5x + 1 = 2 d) Løs ligningen: 5(x + 2) = 30 e) Løs uligheden: -3x + 5 > 5x + 25

Learn how to solve linear equations and inequalities with fractions through step-by-step solutions. Problems cover key algebraic concepts including solving linear equations, manipulating inequalities, and handling fractional terms. Suitable for students in Grades 7-9.

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