Atrain travels train travels 288 kilometre at a uniform speed. if the speed had been 4 kilometre per hour more it would have taken 1 hour less for the same journey. find the speed of the train

Let the speed of the train be x km/h.

Case 1:

Distance = 288 km

Speed = x km/h

Time = Distance/Speed

= 288/x h

Case 2:

Distance = 288 km

Speed = (x+4) km/h

Time = 288/x + 4 h

Since 288/x > 288/x + 4

288/x - 288/x+4 = 1

288[1/x - 1/x+4 ] = 1

[ x + 4 - x / x(x + 4) ] = 1/288

[4 / x^2 + 4x ] = 1/288

x^2 + 4x = 1152

x^2 + 4x - 1152 = 0

x^2 + 36x - 32x - 1152 = 0

x(x + 36) - 32(x + 36) = 0

(x + 36)(x - 32) = 0

x + 36 = 0 , x - 32 = 0

x = -36 , x = 32

x = -36 , rejected since speed cannot be negative.

Therefore , speed of the train = 32 km/h

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)