Last month Maria hiked a total of 90 miles on two trails: a 5-mile mountain trail and a 10-mile canal trail. Let [tex]$x$[/tex] represent the number of times Maria hiked the mountain trail, and let [tex]$y$[/tex] represent the number of times Maria hiked the canal trail.

Which equation can be used to find the number of times Maria hiked each trail?

A. [tex]$x + y = 90$[/tex]

B. [tex][tex]$5x - 10y = 90$[/tex][/tex]

C. [tex]$90 - 10y = 5x$[/tex]

D. [tex]$90 + 10y = 5x$[/tex]



Answer :

Certainly! Let's break down the problem step-by-step to find the correct equation representing the number of times Maria hiked each trail.

1. Understanding the Variables and Total Miles:
- Maria hiked a total of 90 miles.
- There are two types of trails:
- A 5-mile mountain trail.
- A 10-mile canal trail.

2. Introducing Variables:
- Let [tex]\( x \)[/tex] represent the number of times Maria hiked the 5-mile mountain trail.
- Let [tex]\( y \)[/tex] represent the number of times Maria hiked the 10-mile canal trail.

3. Formulating the Problem:
- The total distance hiked on the mountain trail would be [tex]\( 5x \)[/tex] miles since each hike on this trail covers 5 miles.
- The total distance hiked on the canal trail would be [tex]\( 10y \)[/tex] miles since each hike on this trail covers 10 miles.

4. Setting Up the Equation:
- The sum of these distances should equal the total distance hiked, which is 90 miles. Hence, the equation representing this relationship would be:
[tex]\[ 5x + 10y = 90 \][/tex]

5. Rearranging the Equation:
- We can rearrange this equation to isolate one of the variables if necessary. For instance, solving for [tex]\( 5x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ 5x = 90 - 10y \][/tex]

This gives us the equation:
[tex]\[ 90 - 10y = 5x \][/tex]

Therefore, the correct equation to use in order to find the number of times Maria hiked each trail is:
[tex]\[ 90 - 10y = 5x \][/tex]

So, among the given choices, the correct equation is:
[tex]\[ 90 - 10y = 5x \][/tex]

Thus, the third option is the correct one.