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An airplane makes a round-trip supply run that takes a total of 6 hours and is 350 miles each direction. The air current going to the destination is in the direction of the plane at 20 miles per hour. The air current traveling back to the starting point is against the plane at 20 miles per hour. Let [tex][tex]$x$[/tex][/tex] represent the speed of the airplane, in miles per hour, when there is no wind.

Replace the values of [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] to create the equation that describes this situation.

[tex]a \cdot (x + b) + a \cdot (x - b) = c[/tex]



Answer :

GinnyAnswer
To solve the given problem, we need to set up the equation correctly based on the airplane's speed and the influence of the wind on its travel time.

Let [tex]\( x \)[/tex] be the speed of the airplane in still air, in miles per hour (mph).

### Step-by-Step Solution:

1. Determine the speed of the airplane with and against the wind:
- With the wind: [tex]\( x + 20 \)[/tex] mph
- Against the wind: [tex]\( x - 20 \)[/tex] mph

2. Determine the time taken for each leg of the trip:
- The distance for each leg is 350 miles.
- Time taken to travel with the wind: [tex]\( \frac{350}{x + 20} \)[/tex] hours
- Time taken to travel against the wind: [tex]\( \frac{350}{x - 20} \)[/tex] hours

3. Set up the equation for the total round-trip time:
- The total time for the round trip is given as 6 hours.
- Therefore, we write the equation:
[tex]\[ \frac{350}{x + 20} + \frac{350}{x - 20} = 6 \][/tex]

### Final Equation:
To rewrite it by replacing [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ a = \frac{350}{x + 20}, \quad b = \frac{350}{x - 20}, \quad c = 6 \][/tex]

Thus, the equation in terms of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] becomes:
[tex]\[ a + b = c \][/tex]
where:
[tex]\[ \frac{350}{x + 20} + \frac{350}{x - 20} = 6 \][/tex]

Now, let's explicitly state [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

[tex]\[ a = \frac{350}{x + 20}, \quad b = \frac{350}{x - 20}, \quad c = 6 \][/tex]

Inserting the values into the final form:
[tex]\( \boxed{ \frac{350}{x + 20} + \frac{350}{x - 20} = 6 } \)[/tex]

This is the correct equation that describes the situation.

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