# Michael drove to a friend’s house at a rate of 40 mi/h. He returned by the same route at a rate of 45 mi/h. The driving time for the round trip was 4 h. What is the distance Michael traveled?

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### 1 Answer

Let `t_1` be the time travelled by Michael in going to his friend's house and let `t_2` be the time on his return trip.

Since the total time in his round trip is 4 hours, then the first equation is:

`t_1 + t_2= 4` (Let this be EQ1.)

Then, apply the formula of speed `(s=d/t)` to express the t's in terms of distance travelled.

In going to his friend's house, plug-in s=40 and t=t_1. And, solve for t_1.

`40=d/t_1`

`t_1=d/40` (Let this be EQ2.)

In his return trip, plug-in s=45 and t=t_2. Since Michael took the same route, then his distance travelled is still the same (d). So,

`45=d/t_2`

`t_2=d/45` (Let this be EQ3.)

Now that the t's are expressed in terms of d, substitute EQ2 and EQ3 to EQ1.

`t_1 + t_2= 4`

`d/40+d/45=4`

To simplify this, multiply both sides by the LCD of the two fractions which is 360.

`360(d/40+d/45)=4*360`

`9d+8d=1440`

Then, combine like terms.

`17d=1440`

And, divide both sides by 17 to get d only on the left side.

`(17d)/17=1440/17`

`d=84.71`

**Hence, the distance travelled by Michael is 84.7 miles.**