Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 4.0 mi/h, the time to go one mile decreases by 11 s. What was your original speed?

I used the equation V=d/t; I rearranged it to V=(1/t-11)-4. But then I end up with 2 variables. Any help?

1mile=v*t
1mile= (v+4)(t-11)
solve for t in the first equation
t= 1/v
Put that into the second equation
1= (v+4)(1/v -11)
multiply both sides by v
v= (v+4)(1-11v)
multiply out the RHS
rearragne the terms to be a quadratic, and solve.

41 m/h

To solve this problem, let's start by using the equations you provided:

1 mile = v * t (Equation 1)
1 mile = (v + 4) * (t - 11) (Equation 2)

Now, let's solve Equation 1 for t by rearranging the equation:
t = 1 / v

Now, substitute this value of t into Equation 2:
1 = (v + 4) * (1 / v - 11)

Next, multiply both sides of the equation by v to get rid of the fraction:
v = (v + 4) * (1 - 11v)

Now, distribute the terms on the right-hand side:
v = v + 4 - 11v - 44v

Combine like terms:
v + 44v = v - 11v + 4
45v = -10v + 4

Now, rearrange the terms to form a quadratic equation:
45v + 10v = 4
55v = 4

Divide both sides by 55:
v = 4 / 55

Therefore, your original speed was 4/55 miles per hour.