1. **Solve the equation** $\frac{x - 6}{4x} = \frac{1}{2}$. 2. Multiply both sides by $4x$ to eliminate the denominator: $$x - 6 = \frac{1}{2} \times 4x = 2x$$ 3. Rearrange to isolate $x$: $$x - 6 = 2x \implies x - 2x = 6 \implies -x = 6 \implies x = -6$$ 4. **Check for restrictions:** The denominator $4x \neq 0 \Rightarrow x \neq 0$. Since $x = -6$ is allowed, this is the solution. --- **Problem:** Solve $\frac{2x}{3} = \frac{8}{3}$. 1. Multiply both sides by 3 to clear denominators: $$2x = 8$$ 2. Divide both sides by 2: $$x = 4$$ --- **Problem:** Solve $\frac{x + 5}{2x} = \frac{3}{4}$. 1. Cross multiply: $$(x + 5) \times 4 = 3 \times 2x$$ 2. Simplify: $$4x + 20 = 6x$$ 3. Rearrange: $$4x - 6x = -20 \implies -2x = -20 \implies x = 10$$ 4. Check denominator $2x \neq 0 \Rightarrow x \neq 0$, so $x=10$ is valid. --- **Problem:** Solve $\frac{x + 2}{x + 4} = \frac{x - 2}{x + 2}$. 1. Cross multiply: $$(x + 2)(x + 2) = (x - 2)(x + 4)$$ 2. Expand both sides: $$x^2 + 4x + 4 = x^2 + 4x - 8$$ 3. Subtract $x^2 + 4x$ from both sides: $$4 = -8$$ 4. This is a contradiction, so no solution exists. 5. Check restrictions: denominators $x+4 \neq 0 \Rightarrow x \neq -4$, and $x+2 \neq 0 \Rightarrow x \neq -2$. --- **Problem:** Find vertical and horizontal asymptotes of $f(x) = \frac{4}{x - 1}$. 1. Vertical asymptote occurs where denominator is zero: $$x - 1 = 0 \implies x = 1$$ 2. Horizontal asymptote is $y = 0$ because degree numerator (0) < degree denominator (1). --- **Problem:** Find vertical and horizontal asymptotes of $f(x) = \frac{5x + 2}{2x + 1}$. 1. Vertical asymptote where denominator zero: $$2x + 1 = 0 \implies x = -\frac{1}{2}$$ 2. Horizontal asymptote: degrees numerator and denominator both 1, so ratio of leading coefficients: $$y = \frac{5}{2}$$