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For each step in the solution to the given equation, provide the appropriate property of equality:

[tex]\[
\begin{tabular}{|c|c|}
\hline
\textbf{Step} & \textbf{Justification} \\
\hline
\frac{17}{3}-\frac{3}{4} x = \frac{1}{2} x + 5 & Given \\
\hline
\frac{17}{3} - \frac{3}{4} x - \frac{17}{3} = \frac{1}{2} x + 5 - \frac{17}{3} & Subtraction property of equality \\
\hline
-\frac{3}{4} x = \frac{1}{2} x - \frac{2}{3} & Simplification \\
\hline
-\frac{3}{4} x - \frac{1}{2} x = \frac{1}{2} x - \frac{2}{3} - \frac{1}{2} x & Subtraction property of equality \\
\hline
-\frac{5}{4} x = -\frac{2}{3} & Simplification \\
\hline
-\frac{5}{4} x \cdot -\frac{4}{5} = -\frac{2}{3} \cdot -\frac{4}{5} & Division property of equality \\
\hline
x = \frac{8}{15} & Simplification \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
To solve the given equation \(\frac{17}{3} - \frac{3}{4} x = \frac{1}{2} x + 5\) step-by-step while incorporating the subtraction and division properties of equality, here’s a detailed breakdown:

[tex]\[ \begin{array}{|c|c|} \hline \text{Step} & \text{Justification} \\ \hline \frac{17}{3} - \frac{3}{4} x = \frac{1}{2} x + 5 & \text{Given equation} \\ \hline \frac{17}{3} - \frac{3}{4} x - \frac{17}{3} = \frac{1}{2} x + 5 - \frac{17}{3} & \text{Subtraction property of equality: Subtract } \frac{17}{3} \text{ from both sides} \\ \hline -\frac{3}{4} x = \frac{1}{2} x + 5 - \frac{17}{3} & \text{Simplification: cancel } \frac{17}{3} - \frac{17}{3} \text{ on the left} \\ \hline -\frac{3}{4} x = \frac{1}{2} x - \frac{2}{3} & \text{Simplification: convert 5 into a common denominator and combine terms on the right} \\ \hline -\frac{3}{4} x - \frac{1}{2} x = \frac{1}{2} x - \frac{2}{3} - \frac{1}{2} x & \text{Subtraction property of equality: Subtract } \frac{1}{2} x \text{ from both sides} \\ \hline -\frac{5}{4} x = -\frac{2}{3} & \text{Combine like terms on both sides} \\ \hline x = \frac{-\frac{2}{3} \cdot -\frac{4}{5}}{-\frac{5}{4}\cdot\frac{4}{5} } & \text{Division property of equality: Multiply both sides by } -\frac{4}{5}\\ \hline x =\frac {-2 \cdot -4}{15} \cdot \frac{-5 }{-20} & \\ \hline x = 0.5333333333333333 = \frac{8}{15} & \text{Simplified final result} \\ \hline \end{array} \][/tex]

This solution verifies each step methodically from the initial equation to the simplified solution, ensuring that all properties of equality and algebraic manipulations are properly applied.
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