Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles. She can ride 9 mph faster than she can walk. How fast can she walk? Using r r as your variable to represent the rate at which she walks, write an equation using the information as it is given above that can be used to solve this problem.

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Answer:

[tex]\frac{48}{r+9}=\frac{12}{r}[/tex]

Step-by-step explanation:

Let r represent Linda's walking rate.                      

We have been given that Linda can ride 9 mph faster than she can walk, so Linda's bike riding rate would be [tex]t+9[/tex] miles per hour.

[tex]\text{Time}=\frac{\text{Distance}}{\text{Rate}}[/tex]

We have been given that Linda can bicycle 48 miles in the same time as it takes her to walk 12 miles.

[tex]\text{Time while riding}=\frac{48}{r+9}[/tex]

[tex]\text{Time taken while walking}=\frac{12}{r}[/tex]

Since both times are equal, so we will get:

[tex]\frac{48}{r+9}=\frac{12}{r}[/tex]

Therefore, the equation [tex]\frac{48}{r+9}=\frac{12}{r}[/tex] can be used to solve the rates for given problem.

Cross multiply:

[tex]48r=12r+108[/tex]

[tex]48r-12r=12r-12r+108[/tex]

[tex]36r=108[/tex]

[tex]\frac{36r}{36}=\frac{108}{36}[/tex]

[tex]r=3[/tex]

Therefore, Linda's walking at a rate of 3 miles per hour.

Linda's bike riding rate would be [tex]t+9\Rightarrow 3+9=12[/tex] miles per hour.

Therefore, Linda's riding the bike at a rate of 12 miles per hour.