1. ## Rational Equation

#.. Erika and Andy are planning to meet at a location 5 miles from their house. Erika is running and Andy is riding his bike. Andy rides his bike at a speed that is 10 miles per hour faster than Erika runs. If it takes Erika 40 minutes (2/3 of an hour) longer to reach the location, set up and solve an equation to find Erika's speed.

no idea how to setup, or solve. please show work to help me understand. thanks much.

2. Hello overflow

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Originally Posted by overflow
#.. Erika and Andy are planning to meet at a location 5 miles from their house. Erika is running and Andy is riding his bike. Andy rides his bike at a speed that is 10 miles per hour faster than Erika runs. If it takes Erika 40 minutes (2/3 of an hour) longer to reach the location, set up and solve an equation to find Erika's speed.

no idea how to setup, or solve. please show work to help me understand. thanks much.
Suppose Erika runs at $x$ mph. Then Andy rides at $(x+10)$ mph. Using the formula
$\text{time} =\frac{\text{distance}}{\text{speed}}$
we find that, to cover the $5$ miles, Erika takes $\frac{5}{x}$ hours, and Andy takes $\frac{5}{x+10}$ hours. Since Erika's time is $\frac23$ hour more than Andy's we can say
$\frac{5}{x}=\frac{5}{x+10}+\frac23$
The easiest way to handle an equation like this is to get rid of fractions as soon as possible. So we multiply both sides by $3x(x+10)$, and get
$15(x+10)=15x+2x(x+10)$

$\Rightarrow 15x+150=15x+2x^2+20x$

$\Rightarrow 2x^2 +20x-150=0$

$\Rightarrow x^2+10x-75=0$

$\Rightarrow (x+15)(x-5)=0$

$\Rightarrow x=5$, since $x$ cannot be negative.
So Erika runs at $5$ mph.