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Answer :
The equation relating the accuracy of Colin's watch to the number of days since he last set it is given by
[tex]$$
1.3d + 18.6 = 97.9,
$$[/tex]
where [tex]$d$[/tex] is the number of days and [tex]$97.9$[/tex] is the number of seconds by which the watch is behind.
Step 1: Subtract the constant term from both sides to isolate the term with [tex]$d$[/tex]:
[tex]$$
1.3d = 97.9 - 18.6.
$$[/tex]
Calculating the right side, we have
[tex]$$
1.3d = 79.3.
$$[/tex]
Step 2: Divide both sides of the equation by [tex]$1.3$[/tex] to solve for [tex]$d$[/tex]:
[tex]$$
d = \frac{79.3}{1.3}.
$$[/tex]
Performing the division gives
[tex]$$
d = 61.
$$[/tex]
Thus, Colin set his watch [tex]$\boxed{61}$[/tex] days ago.
[tex]$$
1.3d + 18.6 = 97.9,
$$[/tex]
where [tex]$d$[/tex] is the number of days and [tex]$97.9$[/tex] is the number of seconds by which the watch is behind.
Step 1: Subtract the constant term from both sides to isolate the term with [tex]$d$[/tex]:
[tex]$$
1.3d = 97.9 - 18.6.
$$[/tex]
Calculating the right side, we have
[tex]$$
1.3d = 79.3.
$$[/tex]
Step 2: Divide both sides of the equation by [tex]$1.3$[/tex] to solve for [tex]$d$[/tex]:
[tex]$$
d = \frac{79.3}{1.3}.
$$[/tex]
Performing the division gives
[tex]$$
d = 61.
$$[/tex]
Thus, Colin set his watch [tex]$\boxed{61}$[/tex] days ago.
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