Let's go through the given problem step-by-step and identify the mistake made.
The original equation is:
[tex]\[ |d - 600| = 400 \][/tex]
To solve an absolute value equation, we need to set up two separate cases because the absolute value of a number is always non-negative. Therefore, this absolute value equation can be split into two linear equations without the absolute value:
1. [tex]\( d - 600 = 400 \)[/tex]
2. [tex]\( d - 600 = -400 \)[/tex]
Now, let's solve each equation separately:
Case 1:
[tex]\[ d - 600 = 400 \][/tex]
Add 600 to both sides to isolate [tex]\( d \)[/tex]:
[tex]\[ d - 600 + 600 = 400 + 600 \][/tex]
[tex]\[ d = 1000 \][/tex]
Case 2:
[tex]\[ d - 600 = -400 \][/tex]
Add 600 to both sides to isolate [tex]\( d \)[/tex]:
[tex]\[ d - 600 + 600 = -400 + 600 \][/tex]
[tex]\[ d = 200 \][/tex]
Therefore, the correct solutions to the equation [tex]\( |d - 600| = 400 \)[/tex] are:
[tex]\[ d = 1000 \text{ and } d = 200 \][/tex]
Now, given the student's work:
[tex]\[
\begin{tabular}{|c|c|c|}
\hline Step One & \multicolumn{2}{|c|}{$|d-600|=400$} \\
\hline Step Two & $d-600=400$ & $d+600=-400$ \\
\hline Step Three & $+600+600$ & $-600-600$ \\
\hline Step Four & $d=1,000$ & $d=-1,000$ \\
\hline
\end{tabular}
\][/tex]
In Step Two, the student made a mistake in setting up the equations. The student incorrectly wrote the second equation as [tex]\( d + 600 = -400 \)[/tex] instead of [tex]\( d - 600 = -400 \)[/tex].
Therefore, the correction should be:
- The first equation should indeed be [tex]\( d - 600 = 400 \)[/tex].
- The second equation should be [tex]\( d - 600 = -400 \)[/tex].
This leads us to the conclusion that the student should have written the equations as:
[tex]\[ d - 600 = 400 \][/tex]
[tex]\[ d - 600 = -400 \][/tex]
The correct answer provided is:
The mistake made was in step two. The first equation should be [tex]\( d - 600 = 400 \)[/tex], and the second equation should be [tex]\( d - 600 = -400 \)[/tex].
