Answer :
Answer:
the appointment was 1 mile away from Alex's house for every minute of driving time at 50 mph.
Step-by-step explanation:
Let's call the distance from Alex's house to the appointment "d".
If Alex drives 50 mph, he will cover this distance in d/50 hours.
If he leaves 15 minutes (or 0.25 hours) early and drives 40 mph, he will cover the same distance in (d/40) + 0.25 hours.
Since both of these times are the same (since Alex arrives on time), we can set them equal to each other and solve for d:
d/50 = (d/40) + 0.25
Multiplying both sides by 200 (the least common multiple of 50 and 40) to get rid of the denominators, we get:
4d = 5d + 50
Subtracting 4d from both sides, we get:
d = 50
Therefore, the appointment is 50 miles away from Alex's house.
As you mentioned, the equation X/50 = the appointment was __ miles away from Alex house can be used to solve the problem. Plugging in X = 50 gives:
50/50 = 1
So the appointment was 1 mile away from Alex's house for every minute of driving time at 50 mph.
The appointment was _[tex]\[50 \text{ miles} \][/tex]_miles away from Alex house
Let's denote the distance from Alex's house to the appointment as [tex]\(X\)[/tex] miles.
First, let's calculate the time it would take for Alex to drive to his appointment at 50 mph:
[tex]\[ \text{Time at 50 mph} = \frac{X}{50} \, \text{hours} \][/tex]
Next, we need to determine the time it takes for Alex to drive to his appointment at 40 mph, considering he leaves 15 minutes (or [tex]\(\frac{1}{4}\)[/tex] hour) early.
Since Alex leaves 15 minutes early and still arrives on time, the total travel time when driving at 40 mph is the same as the travel time at 50 mph plus 15 minutes.
So, the time it takes Alex to drive to his appointment at 40 mph is:
[tex]\[ \text{Time at 40 mph} = \frac{X}{40} \, \text{hours} \][/tex]
According to the problem, the time taken at 40 mph plus the 15 minutes early start is equal to the time it would take at 50 mph:
[tex]\[ \frac{X}{40} + \frac{1}{4} = \frac{X}{50} \][/tex]
Now, we solve for [tex]\(X\)[/tex]:
First, let's get a common denominator for the fractions:
[tex]\[ \frac{X}{40} + \frac{1}{4} = \frac{X}{50} \][/tex]
[tex]\[ \frac{X}{40} + \frac{10}{40} = \frac{X}{50} \][/tex]
[tex]\[ \frac{X + 10}{40} = \frac{X}{50} \][/tex]
Cross-multiply to solve for [tex]\(X\)[/tex]:
[tex]\[ 50(X + 10) = 40X \][/tex]
[tex]\[ 50X + 500 = 40X \][/tex]
[tex]\[ 50X - 40X = -500 \][/tex]
[tex]\[ 10X = 500 \][/tex]
[tex]\[ X = 50 \][/tex]
Therefore, the distance from Alex’s house to his appointment is 50 miles.
So, if the appointment is [tex]\(X\)[/tex] miles away from Alex’s house, the equation is [tex]\(X = 50\)[/tex].
The final answer in the context of the original question would be:
[tex]\[ X/50 = 50 \text{ miles} \][/tex]
