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14) Rohan solves a linear equation. His work is shown below:

[tex]\[
\begin{aligned}
2x - 3 & = \frac{x}{2} - 5 \\
\text{Or, } 2(2x - 3) & = x - 5 \ldots \ldots \text{ step 1 } \\
\text{Or, } 4x - 6 & = x - 5 \ldots \ldots \text{ step 2 } \\
\text{Or, } 4x - x & = 6 - 5 \ldots \ldots \text{ step 3 } \\
\text{Or, } 3x & = 1 \ldots \ldots \text{ step 4 } \\
\text{Or, } x & = \frac{1}{3}
\end{aligned}
\][/tex]

(i) In which step did he make a mistake?



Answer :

GinnyAnswer
Let's carefully examine Rohan's solution step-by-step to identify where he made a mistake.

Given equation:

[tex]\[ 2x - 3 = \frac{x}{2} - 5 \][/tex]

Step 1: Eliminating the fraction

To eliminate the fraction, multiply both sides of the equation by 2:
[tex]\[ 2 \cdot (2x - 3) = 2 \cdot \left(\frac{x}{2} - 5 \right) \][/tex]

Step 2: Simplifying both sides

Carrying out the multiplication:
[tex]\[ 4x - 6 = x - 10 \][/tex]

At this stage, Step 2 is already different from Rohan’s:

Rohan’s Step 2:
[tex]\[ 4x - 6 = x - 5 \][/tex]

Therefore, Rohan made an error here. Correctly simplifying both sides should give:
[tex]\[ 4x - 6 = x - 10 \][/tex]

Step 3: Rearranging terms to isolate [tex]\(x\)[/tex]

Moving [tex]\(x\)[/tex] to one side and constants to the other:
[tex]\[ 4x - x = -10 + 6 \][/tex]
[tex]\[ 3x = -4 \][/tex]

Notice in Rohan’s work:

Rohan’s Step 3:
[tex]\[ 4x - x = 6 - 5 \][/tex]
[tex]\[ 3x = 1 \][/tex]

Rohan’s Step 3 is incorrect. He mistakenly added 6 to the right side instead of subtracting it correctly.

Correct Step 4: Solving for [tex]\(x\)[/tex]

Now, divide both sides by 3 to isolate [tex]\(x\)[/tex]:
[tex]\[ x = \frac{-4}{3} \][/tex]

Therefore, Rohan made a mistake in Step 3 when he incorrectly handled the constants. His final solution [tex]\(x = \frac{1}{3}\)[/tex] is incorrect due to this mistake. The correct solution is [tex]\( x = \frac{-4}{3} \)[/tex].

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