Rahul solved the equation [tex]$2\left(x-\frac{1}{8}\right)-\frac{3}{5} x=\frac{55}{4}$[/tex]. In which step did he use the addition property of equality?

Rahul's Solution

\begin{tabular}{|c|c|}
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Steps & Resulting Equations \\
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1 & [tex]$2 x-\frac{1}{4}-\frac{3}{5} x=\frac{55}{4}$[/tex] \\
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2 & [tex][tex]$\frac{7}{5} x-\frac{1}{4}=\frac{55}{4}$[/tex][/tex] \\
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3 & [tex]$\frac{7}{5} x=\frac{56}{4}$[/tex] \\
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4 & [tex]$x=10$[/tex] \\
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\end{tabular}



Answer :

Sure, let's analyze Rahul's solution step by step to find out where he used the addition property of equality.

1. Original Equation:
[tex]\[ 2\left(x-\frac{1}{8}\right)-\frac{3}{5} x=\frac{55}{4} \][/tex]
Expanding the first term:
[tex]\[ 2x - 2\left(\frac{1}{8}\right) - \frac{3}{5} x=\frac{55}{4} \][/tex]
Simplifying:
[tex]\[ 2x - \frac{1}{4} - \frac{3}{5} x=\frac{55}{4} \][/tex]

2. Combining Like Terms:
[tex]\[ 2x - \frac{3}{5} x - \frac{1}{4}=\frac{55}{4} \][/tex]
Finding a common denominator for the terms involving [tex]\( x \)[/tex]:
[tex]\[ 2x = \frac{10}{5}x \implies \frac{10}{5}x - \frac{3}{5}x = \left(\frac{10-3}{5}\right)x = \frac{7}{5}x \][/tex]
Thus, we have:
[tex]\[ \frac{7}{5} x - \frac{1}{4}=\frac{55}{4} \][/tex]

3. Using the Addition Property of Equality:
To isolate the term with [tex]\( x \)[/tex], we add [tex]\( \frac{1}{4} \)[/tex] to both sides of the equation:
[tex]\[ \frac{7}{5} x - \frac{1}{4} + \frac{1}{4} = \frac{55}{4} + \frac{1}{4} \][/tex]
Simplifying:
[tex]\[ \frac{7}{5} x = \frac{55 + 1}{4} = \frac{56}{4} \][/tex]

4. Solving for [tex]\( x \)[/tex]:
[tex]\[ \frac{7}{5} x = \frac{56}{4} \implies x = \left(\frac{5}{7}\right) \left(\frac{56}{4}\right) \][/tex]
Simplifying further:
[tex]\[ x = \frac{5}{7} \times 14 = 10 \][/tex]

Therefore, Rahul used the addition property of equality in Step 3.