Question

which expression is equivalent to \\(\frac{1}{3}\\)? \\(\frac{3d}{3d + 1} + \frac{1}{3d + 1}\\) \\(\frac{d + 5}{3d + 3} - \frac{4}{3d + 3}\\) \\(\frac{2}{3d} - \frac{1}{3d}\\) \\(\frac{d}{d + 3} + \frac{1}{d + 3}\\)

Explanation:

Step1: Simplify Option 1

$\frac{3d}{3d+1} + \frac{1}{3d+1} = \frac{3d+1}{3d+1} = 1$

Step2: Simplify Option 2

$\frac{d+5}{3d+3} - \frac{4}{3d+3} = \frac{d+5-4}{3d+3} = \frac{d+1}{3(d+1)} = \frac{1}{3}$

Step3: Simplify Option 3

$\frac{2}{3d} - \frac{1}{3d} = \frac{2-1}{3d} = \frac{1}{3d}$

Step4: Simplify Option 4

$\frac{d}{d+3} + \frac{1}{d+3} = \frac{d+1}{d+3}$

Answer:

B. $\frac{d+5}{3d+3} - \frac{4}{3d+3}$