College Algebra rate word problem

Question

two planes leave los angeles at the same time. One heads south to San Diego, while the other heads north to san francisco. San diego planes flies 50 mph slower than the san francisco plane. In 1/2 hour, the planes are 275 miles apart. What are their speeds?

Answer 1

let x mph be the speed of the plane to San Diego.

Then the speed of the San Francisco plane is x+50 mph.

 

So in 1/2 hour they fly (1/2)x+(1/2)(x+50) and we know that it is 275 miles.

 

Our equation then looks like this:

 

x/2 + (x+50)/2 = 275.

 

 

Could you solve it? 

 

If you need more help, let me know.

Answer 2

Rsd = Speed of the plane flying to San Diego. Rsf = Speed of the plane flying to San Fransisco. 

The second sentence tells us that Rsd = Rsf - 50, the San Fransisco plane is fifty miles per hour faster than the San Diego plane.

Now we just use the simple equation: rate * time = distance.

The time is 1/2 hour so we have (1/2) * Rsf + (1/2) * Rsd = 275 ==> (1/2) * (Rsf + Rsd) = 275. Now we substitute in our equation Rsd = Rsf - 50. ==> (1/2) * (Rsf + Rsf - 50) = 275 ==> (1/2) * (2Rsf - 50) = 275 ==> Rsf - 25 = 275 ==> Rsf = 300. Now just plug this back into our original equation. Rsd = 300 - 50 = 250.

Rsd = 250. Rsf = 300.