The ratio of the numerator to the denominator of a fraction is 2 to 3 . If both the numerator and the denominator are increased by 2 , the fraction becomes \( 3 / 4 \). What is the original fraction? Which of the following systems of equations can be used to solve the problem? \( 3 n-2 d=0 \) and \( 4 n+2=3 d+2 \) \( 3 n=2 d \) and \( 4 n+8=3 d+6 \)

Let’s break this down! The ratio of the numerator to the denominator can be represented as \( n = 2k \) and \( d = 3k \) for some constant \( k \). When both are increased by 2, the new fraction becomes \( \frac{2k + 2}{3k + 2} = \frac{3}{4} \). Cross-multiplying gives \( 4(2k + 2) = 3(3k + 2) \). Simplifying leads us to a single equation that can easily be manipulated to find \( k \). For the system of equations, look at that first option. The equation \( 3n - 2d = 0 \) correctly represents the initial ratio \( \frac{n}{d} = \frac{2}{3} \), and the second equation \( 4n + 8 = 3d + 6 \) corresponds to the condition after increasing both parts. You’re on the right track to solve it!