h) \( \frac{2 x}{3}=\frac{x}{7} \) i) \( \frac{2 x-5}{12}=\frac{-x}{4}-\frac{5}{3} \) i) \( x+\frac{x}{x}-1=\frac{x}{l} \)

To solve the equation \( \frac{2 x}{3}=\frac{x}{7} \), you can cross-multiply to eliminate the fractions: \( 2x \cdot 7 = x \cdot 3 \), which simplifies to \( 14x = 3x \). To isolate \( x \), subtract \( 3x \) from both sides: \( 14x - 3x = 0 \) gives \( 11x = 0 \). Thus, \( x = 0 \). For the second equation \( \frac{2 x-5}{12}=\frac{-x}{4}-\frac{5}{3} \), start by finding a common denominator. The common denominator for the fractions on the right is 12. Rewriting gives \( \frac{2x-5}{12} = \frac{-3x - 20}{12} \). Since the denominators are equal, set the numerators equal: \( 2x - 5 = -3x - 20 \). Rearranging yields \( 5x = -15 \), so \( x = -3 \). As for the equation \( x+\frac{x}{x}-1=\frac{x}{l} \), note that \( \frac{x}{x} = 1 \) for \( x \neq 0 \). This simplifies to \( x + 1 - 1 = \frac{x}{l} \) or just \( x = \frac{x}{l} \). To solve this, multiply both sides by \( l \): \( lx = x \). If \( x \neq 0 \), divide by \( x \) to find \( l = 1 \). If \( x = 0 \), any \( l \) is valid as both sides equal 0.