Question

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.

Answer 1

Answer:

$\frac{48}{r+9}=\frac{12}{r}$

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 $t+9$ miles per hour.

$\text{Time}=\frac{\text{Distance}}{\text{Rate}}$

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

$\text{Time while riding}=\frac{48}{r+9}$

$\text{Time taken while walking}=\frac{12}{r}$

Since both times are equal, so we will get:

$\frac{48}{r+9}=\frac{12}{r}$

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

Cross multiply:

$48r=12r+108$

$48r-12r=12r-12r+108$

$36r=108$

$\frac{36r}{36}=\frac{108}{36}$

$r=3$

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

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

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