50 POINTS! :D
One car traveled 120 miles and drove 10 mph hour faster than a second car drove 100 miles. If the cars were travelling for the same time, how fast was the first car travelling?

90 mph
50 mph
60 mph
80 mph

The speed of the first car is 60 mph

Step-by-step explanation:

speed = distance/time

Solve the above equation for distance to get

distance = speed * time

or simply

d = st

Now we use this formula for distance to write an equation for each car.

Let s = speed of second car

Then since the speed of the first car is 10 mph faster, the first car's speed is s + 10.

The time the two cars traveled is equal but unknown, so let the time = t.

First car: speed = s + 10; time = t; distance = 120 miles

d = st

120 = (s + 10)t    

(s + 10)t = 120      Equation 1

Second car: speed = s; time = t; 100 miles

d = st

100 = st

st = 100       Equation 2

Equations 1 and 2 form a system of 2 equations in 2 unknowns.

(s + 10)t = 120

st = 100

Distribute t in the first equation.

st + 10t = 120

From the second equation we know st = 100, so substitute 100 for st.

100 + 10t = 120

10t = 120

t = 2

The time traveled was 2 hours.

Equation 2:

st = 100

Substitute t with 2.

s * 2 = 100

s = 50

The speed of the second car was 50 mph.

The speed of the first car is s + 10.

s + 10 = 50 + 10 = 60

The speed of the first car is 60 mph

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: