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

Explanation:

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

Given $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$. Subtract $\frac{1}{5}$ from both sides: $\frac{2}{3}|2 - x|=\frac{4}{15}-\frac{1}{5}$. Since $\frac{1}{5}=\frac{3}{15}$, then $\frac{4}{15}-\frac{3}{15}=\frac{4 - 3}{15}=\frac{1}{15}$. So $\frac{2}{3}|2 - x|=\frac{1}{15}$, the first - option is correct.

Step2: Combine left - hand side terms

Starting from the original equation $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$. First, get a common denominator for the left - hand side. $\frac{1}{5}=\frac{3}{15}$ and $\frac{2}{3}|2 - x|=\frac{10}{15}|2 - x|$. Then $\frac{3}{15}+\frac{10}{15}|2 - x|=\frac{4}{15}$, which simplifies to $\frac{3 + 10|2 - x|}{15}=\frac{4}{15}$, or $3+10|2 - x| = 4$, and $10|2 - x|=1$, $|2 - x|=\frac{1}{10}$, the third option is correct.

Step3: Solve for $|2 - x|$ from the first correct step

From $\frac{2}{3}|2 - x|=\frac{1}{15}$, multiply both sides by $\frac{3}{2}$: $|2 - x|=\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$. Cross - multiply the original equation $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$ to get $3 + 10|2 - x|=4$, then $10|2 - x| = 1$, $|2 - x|=\frac{1}{10}$. If we start from $\frac{2}{3}|2 - x|=\frac{1}{15}$ and rewrite it in another way. Cross - multiply to get $30|2 - x| = 3$, or $|2 - x|=\frac{1}{10}$. If we go back to the original equation $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$ and solve for $|2 - x|$ directly: $\frac{2}{3}|2 - x|=\frac{4}{15}-\frac{1}{5}=\frac{4 - 3}{15}=\frac{1}{15}$, then $|2 - x|=\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$. If we rewrite the original equation as $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, and first simplify the left - hand side terms: $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{3 + 10|2 - x|}{15}=\frac{4}{15}$, so $3+10|2 - x| = 4$, $10|2 - x|=1$, $|2 - x|=\frac{1}{10}$. If we start from $\frac{2}{3}|2 - x|=\frac{1}{15}$ and multiply both sides by $\frac{3}{2}$ we get $|2 - x|=\frac{1}{10}$. If we rewrite the original equation $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$ and solve for $|2 - x|$: $\frac{2}{3}|2 - x|=\frac{4}{15}-\frac{1}{5}=\frac{1}{15}$, then $|2 - x|=\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$. If we consider the equation $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, and we want to check the second option. Starting from $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, we get $\frac{2}{3}|2 - x|=\frac{4}{15}-\frac{1}{5}=\frac{1}{15}$, not $\frac{13}{15}|2 - x|=\frac{4}{15}$. For the fourth option, $\frac{1}{5}+|\frac{4}{3}-\frac{3}{3}x|=\frac{4}{15}$. The original is $\frac{1}{5}+\frac{2}{3}|2 - x|=\frac{4}{15}$, and $|\frac{4}{3}-\frac{3}{3}x|$ is not equivalent to $\frac{2}{3}|2 - x|$. For the fifth option, from $\frac{2}{3}|2 - x|=\frac{1}{15}$, $|2 - x|=\frac{1}{10}\neq\frac{4}{13}$.

Answer:

A. $\frac{2}{3}|2 - x|=\frac{1}{15}$ C. $|2 - x|=\frac{1}{10}$