Solving for distance using a system of linear equations

Question

Two trains leave towns 610km apart at the same time and travel toward each other. One train travels 15 /kmh slower than the other. If they meet in 2 hours, what is the rate of each train?

Answer 1

Let the speed of train 1 = r₁ 

 

Speed of train 2 = r₂

 

Therefore distances covered by each train:

 

Train 1 =2*r₁ and train 2 = 2*r₂

 

Total distance covered by both trains = 2r1+2r2 

 

Irrespective of their speeds, total distance covered = 610 km.

 

Let us assume that train 1 is 15 km/h slower than train 2. Therefore r₂ = r₁+15

 

Therefore, expressing the equation in terms of r₁, we have:

 

2r1+2(r1+15) = 610

 

2r1+2r1+30 = 610

 

4r1+30 = 610

 

Solving, we have:

 

4r1 = 580

 

r1 = 580/4

 

r₁ = 145

r₂ = 145+15=160

 

Solution:

Speed of train 1 = 145 km/h; Speed of train 2 = 160km/h

 

Check

145(2) + 160(2) = 290+320 =610

 

 

Answer 2

Suppose the rate of the slower train is X kmh, so the rate of the faster train would be (x + 15) kmh

 

the distance of the slower train traveled is 2 X km, the distance of the faster train would be 2(X+15) (D = R x H)

total distance is 610km, so 2X + 2(X+15) = 610.

 

2X + 2(X+15) = 610

2X + 2X + 30 = 610 (distributive property)

4X = 580 (combination and subtraction)

X = 145 (kmh)---The rate of the slower train

X + 15 = 145 + 15 = 160 kmh --- The rate of the faster train