if alexandra increases her usual driving speed by 15 km/h, it will take her 2 hours less to make a trip to her gradnparents house. If she decreaseshe rusualy speed by 15 km/h, it will take her 3 h more than usualy to make the trip. How long is the trip?

Respuesta :

Answer:

Therefore,

Distance of the trip is

[tex]Distance =900\ km[/tex]

Step-by-step explanation:

Let "x" be the Speed of the car and,

"y" be the time to cover the trip to her grandparents,

To Find:

How long is the trip = ?

Solution:

We Know that,

[tex]Distance = Speed\times Time[/tex]

According to the First given condition,

[tex]Speed = x + 15[/tex]

[tex]Time = y - 2[/tex]

According to the Second given condition,

[tex]Speed = x - 15[/tex]

[tex]Time = y + 3[/tex]

As we  know Distance will remain same so we have

[tex]Distance = Speed\times Time= x\times y[/tex]

Substituting the values we get

[tex](x+15)\times (y-2)=(x-15)\times (y+3)=x\times y[/tex]

So on  Substituting we have

[tex](x+15)\times (y-2)=(x-15)\times (y+3)[/tex]

[tex]5x-30y=15\\Divding\ by\ 3\\x-6y=3[/tex]..............Equation ( 1)

Again on  Substituting we have

[tex](x-15)\times (y+3)=x\times y[/tex]

[tex]3x-15y=45\\Dividing\ by\ 5\\x-5y=15[/tex]...........Equation ( 2)

Subtracting Equation 1 from equation 2 we get

[tex]x- x -5y--6y=15-3\\\\y=12[/tex]

Substitute y = 12 in equation 1 we get

[tex]x-6\times 12=3\\x = 72+3=75[/tex]

So we have now,

[tex]Speed = 75\ km/h[/tex]

[tex]Time = 12\ h[/tex]

Therefore the Distance of the trip is

[tex]Distance = x\times y=75\times 12=900\ km[/tex]

Therefore,

Distance of the trip is

[tex]Distance =900\ km[/tex]