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Let's go through each of the problems step-by-step. I'll start with the first one:

### 1. \( \frac{1}{x} = \frac{2}{3x - 1} \)

**Step 1:** Cross-multiply to eliminate the fractions:
\[
1 \cdot (3x - 1) = 2 \cdot x
\]
This simplifies to:
\[
3x - 1 = 2x
\]

**Step 2:** Solve for \( x \):
\[
3x - 2x = 1
\]
\[
x = 1
\]

**Solution for Problem 1:** \( x = 1 \)

---

### 2. \( \frac{7 - 2x}{x + 3} \leq 5 \)

**Step 1:** Multiply both sides by \( x + 3 \) (assuming \( x + 3 \neq 0 \)):

\[
7 - 2x \leq 5(x + 3)
\]

**Step 2:** Distribute the 5 on the right-hand side:
\[
7 - 2x \leq 5x + 15
\]

**Step 3:** Bring all terms involving \( x \) to one side and constants to the other side:
\[
7 - 15 \leq 5x + 2x
\]
\[
-8 \leq 7x
\]

**Step 4:** Solve for \( x \):
\[
x \geq -\frac{8}{7}
\]

**Solution for Problem 2:** \( x \geq -\frac{8}{7} \)

---

### 3. \( \frac{1}{x} + \frac{1}{2x} = \frac{2}{3} \)

**Step 1:** Combine the fractions on the left-hand side:
\[
\frac{1}{x} + \frac{1}{2x} = \frac{2 + 1}{2x} = \frac{3}{2x}
\]

**Step 2:** Set this equal to the right-hand side:
\[
\frac{3}{2x} = \frac{2}{3}
\]

**Step 3:** Cross-multiply:
\[
3 \cdot 3 = 2 \cdot 2x
\]
\[
9 = 4x
\]

**Step 4:** Solve for \( x \):
\[
x = \frac{9}{4}
\]

**Solution for Problem 3:** \( x = \frac{9}{4} \)

---

### 4. \( \frac{x - 3}{x + 4} \leq 2 \)

**Step 1:** Move all terms to one side:
\[
\frac{x - 3}{x + 4} - 2 \leq 0
\]

**Step 2:** Combine into a single fraction:
\[
\frac{x - 3 - 2(x + 4)}{x + 4} \leq 0
\]
\[
\frac{x - 3 - 2x - 8}{x + 4} \leq 0
\]
\[
\frac{-x - 11}{x + 4} \leq 0
\]

**Step 3:** Solve the inequality \( \frac{-x - 11}{x + 4} \leq 0 \). The critical points are at \( x = -4 \) and \( x = -11 \).

Analyzing the sign of the expression around these points:
- For \( x < -11 \), the expression is positive.
- For \( -11 < x < -4 \), the expression is negative.
- For \( x > -4 \), the expression is positive.

So, the solution is \( -11 \leq x \leq -4 \) (but \( x \neq -4 \) since that makes the denominator zero).

**Solution for Problem 4:** \( -11 \leq x < -4 \)

---

### 5. \( \frac{1}{x^2 + 4} + \frac{1}{x^2 - 4} = \frac{18}{x^4 - 16} \)

**Step 1:** Recognize that \( x^4 - 16 = (x^2 + 4)(x^2 - 4) \), so the right-hand side simplifies to:
\[
\frac{18}{(x^2 + 4)(x^2 - 4)}
\]

**Step 2:** Combine the left-hand side into a single fraction:
\[
\frac{(x^2 - 4) + (x^2 + 4)}{(x^2 + 4)(x^2 - 4)} = \frac{2x^2}{(x^2 + 4)(x^2 - 4)}
\]

**Step 3:** Set this equal to the right-hand side:
\[
\frac{2x^2}{(x^2 + 4)(x^2 - 4)} = \frac{18}{(x^2 + 4)(x^2 - 4)}
\]

**Step 4:** Cancel the denominators and solve:
\[
2x^2 = 18
\]
\[
x^2 = 9
\]
\[
x = \pm 3
\]

**Solution for Problem 5:** \( x = 3 \) or \( x = -3 \)

---

Would you like more details on any step or have any other questions? Here are five questions to deepen your understanding:

1. How do you determine the critical points in inequalities with rational expressions?
2. Why is it important to check for excluded values in rational equations?
3. How does cross-multiplying help in solving fractional equations?
4. How can you simplify complex rational expressions before solving them?
5. What is the significance of analyzing signs in rational inequalities?

**Tip:** When solving inequalities, always analyze the critical points carefully, and check which intervals satisfy the inequality.

Solve the following equations/inequalities: 
1. (1/x) = 2 / (3x - 1), 
2. (7 - 2x) / (x + 3) ≤ 5, 
3. (1/x) + (1/2x) = 2/3, 
4. (x - 3)/(x + 4) ≤ 2, 
5. (1/(x^2 + 4)) + (1/(x^2 - 4)) = 18/(x^4 - 16)

Step-by-step solutions to solving rational equations and inequalities involving cross-multiplication. Includes methods for solving complex rational equations, inequalities, and critical point analysis for Grade 9-12 students.

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