which equations are equivalent to $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$? check all that apply…

Explanation:

Step1: Subtract $\frac{1}{5}$ from both sides

Given $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, subtracting $\frac{1}{5}=\frac{3}{15}$ from both sides gives $\frac{2}{3}|2 - x|=\frac{4}{15}-\frac{3}{15}=\frac{1}{15}$.

Step2: Multiply both sides by $\frac{3}{2}$

Starting from $\frac{2}{3}|2 - x|=\frac{1}{15}$, multiplying both sides by $\frac{3}{2}$: $|2 - x|=\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$.

Step3: Analyze other options

For $\frac{13}{15}|2 - x|=\frac{4}{15}$, starting from the original $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, $\frac{1}{5}=\frac{3}{15}$ and $\frac{2}{3}|2 - x|=\frac{2\times5}{3\times5}|2 - x|=\frac{10}{15}|2 - x|$, so $\frac{3}{15}+\frac{10}{15}|2 - x|\neq\frac{13}{15}|2 - x|$. For $\frac{1}{5}+|\frac{4}{3}-\frac{2}{3}x|=\frac{4}{15}$, since $\frac{2}{3}|2 - x|=\frac{2}{3}|-(x - 2)|=\frac{2}{3}|x - 2|$, and $|\frac{4}{3}-\frac{2}{3}x|=\frac{2}{3}|2 - x|$, but we cannot just remove the $\frac{2}{3}$ - coefficient inside the absolute - value like this from the original equation. For $|2 - x|=\frac{4}{13}$, it does not match the result of our simplification.

Answer:

$\frac{2}{3}|2 - x|=\frac{1}{15}$, $|2 - x|=\frac{1}{10}$