What is the value of y−5x−3for x = 2 and y = -4
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Answer:
x = 20.
Step-by-step explanation:
First, you should remember the relation:
Distance = Speed*Time.
First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.
We can write this as:
20km = (x km/h)*T
We know that the time is shortened by 12 minutes if the speed is increased by 5km/h
Rewriting these 12 minutes in hours (remember that 60min = 1 hour)
12 min = (12/60) hours = 0.2 hours
Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h
We can write this as:
20km = (x + 5 km/h)*(T - 0.2 h)
Then we have a system of two equations, and we want to find the value of x:
20km = (x km/h)*T
20km = (x + 5 km/h)*(T - 0.2 h)
First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:
20km/(x km/h) = T
Replacing that in the other equation we get:
20km = (x + 5 km/h)*(T - 0.2 h)
20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)
Now we can solve this for x.
Removing the units (that we know that are correct) so the math is easier to read, we get:
20 = (x + 5)*(20/x - 0.2)
We only want to solve this for x.
20 = x*20/x - x*0.2 + 5*20/x - 5*0.2
20 = 20 - 0.2*x + 100/x - 1
subtracting 20 in both sides we get:
20 - 20 = 20 - 0.2*x + 100/x - 1 - 20
0 = -0.2*x + 100/x - 1
If we multiply both sides by x we get:
0 = -0.2*x^2 + 100 - x
-0.2*x^2 - x + 100 = 0
This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:
Then the two solutions are:
x = (1 + 9)/-0.4 = -25
x = (1 - 9)/-0.4 = 20
As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.
x = 20
then the average speed initially is 20 km/h