Algebra 1 Word Problem #3

The main mathematical relationship we need for this is that speed is equal to distance over time (because it is measured in miles per hour). We will notate this with $S = \frac{D}{T} $, where S stands for speed, D for distance, and T for time. 

We're given that D = 20 miles. So $S = \frac{20}{T} $. We're also given that if S was 1 more, T would be 1 less. In equation form we can express this as $S + 1 = \frac{20}{T-1} $. 

Now we have a system of 2 equations with 2 unknowns, which we can solve with for example substitution. I'll start by writing the first equation in terms of T, which we can then substitute into the second equation.
Multiply both sides by T to get $ST = 20$, divide both sides by S to get $T = \frac{S}{20} $. Now substitute this into the second equation so that

$S+1 = \frac{20}{\frac{20}{s}-1} $

Multiply both sides by $\frac{20}{s}-1 $ to get $(S+1)(\frac{20}{S}-1) = 20 $. We can multiply the left side using the distributive property:  $20 - S + \frac{20}{S} - 1 = 20$, subtract 20 from both sides and move the S/20 over to get $-S - 1= -\frac{20}{S}$, now all we have to do is multiply by S and we have reduced the problem to a quadratic equation! (although you have to be careful with steps like this, multiplying by a variable can introduce false solutions, which you have to weed out later.)

Multiplying by S, we get $-S^{2} - S +20 = 0 $. To solve this let's first calculate the discriminant X = $b^{2} - 4ac $, it's X = $(-1)^{2} -4(-1)(20) = 81 $. A discriminant larger than zero means that the quadratic equation has two real roots. The two roots are then given by $\frac{-b \pm \sqrt{X} }{2a} $. Plugging in our numbers we get $\frac{1 \pm 9}{-2} = $ 4 or -5.

-5 is obviously a false solution in this case, since speed cannot be negative. The answer to the question is 4 miles per hour. To check, we can plug S=4 into the original equation. $T = \frac{20}{S} $. At 4mph the time taken is 20/4 = 5 hours. If the cyclist had biked 1 mph faster (so 5mph) we should get 4 hours. Plugging in 5 we get 20/5 = 4 hours, which is indeed 1 hour shorter.