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This equation shows how the accuracy of Dakota's watch is related to how long it's been since she last set it. $\newline$ $s = 5d$ $\newline$ The variable $d$ represents the number of days passed since Dakota last set her watch, and the variable $s$ represents how many seconds behind the watch is. How far behind is Dakota's watch if she last set it $3$ days ago? $\newline$ _____ seconds

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Find a value using two-variable equations: word problems

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Q. This equation shows how the accuracy of Dakota's watch is related to how long it's been since she last set it.

\newline

s = 5d

\newline

The variable

d

represents the number of days passed since Dakota last set her watch, and the variable

s

represents how many seconds behind the watch is. How far behind is Dakota's watch if she last set it

3

days ago?

\newline

_____ seconds

Identify Variable: Identify the variable that $3$ days corresponds to. $\newline$ $d$ represents the number of days since the watch was last set. So, $3$ days gives the value for $d$ .
Substitute Value: We know that $d = 3$ . $\newline$ Plug in the value for $d$ in $s = 5d$ . $\newline$ $s = 5d$ $\newline$ Substitute $d$ with $3$ . $\newline$ $s = 5 \times 3$
Solve for $s$ : $s = 5 \times 3$ . $\newline$ Solve for $s$ . $\newline$ $s = 5 \times 3 = 15$

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