Liam went on a bike ride of 48 miles. He realized that if he had gone four miles per hour faster, then he would have arrived six hours sooner. How fast did he actually ride?
We know that distance equals speed multiplied by time which means we have data from Liam’s actual bike ride, his distance, time, and speed. And then we have some information about an increased speed ride, the distance, time, and speed. Let’s fill in what we know about the actual bike ride. The distance was 48 miles. We don’t know how long it took. So we can just call the time 𝑡. We also don’t know the speed of his actual bike ride. Let’s call that value 𝑠. On Liam’s increased speed ride, the distance would be the same, 48 miles. And the time would be Liam’s first time minus six hours. So we can use the variable 𝑡 and then subtract six.
We’ll follow a similar approach when it comes to the speed of the increased ride. It’s his original speed plus four miles per hour. We’re interested in how fast he actually was riding. That’s our 𝑡 variable. Because we know that distance equals speed multiplied by time, we can also say that time is equal to the distance divided by the speed. In the actual bike ride, 𝑡 is equal to 48 divided by 𝑠. And in the increased speed ride, we have the original time minus six is equal to the same distance 48 over 𝑠 plus four. We know that 𝑡 equals 48 over 𝑠.
So we’ll take information from our first equation, what we know that 𝑡 equals, and plug it into our second equation. And we’ll have something that looks like this. 48 over 𝑠 minus six equals 48 over 𝑠 plus four. To subtract six from 48 over 𝑠, we need a common denominator. We can multiply six by 𝑠 over 𝑠, which gives us six 𝑠 over 𝑠. And that means we can subtract 48 minus six 𝑠 and put it all over 𝑠. And that is equal to 48 over 𝑠 plus four.
Now, we really want to solve for 𝑠. And to do that, we’ll need to get 𝑠 by itself. There is an 𝑠 in three places in this equation. So we need to start by eliminating these fractions. And we’ll do that by cross-multiplying. 48 minus six 𝑠 times 𝑠 plus four will be equal to 𝑠 times 48, 48𝑠. We need to multiply 𝑠 plus four times 48 minus six 𝑠. And that means we need to foil 𝑠 times 48 equals 48𝑠, 𝑠 times negative six 𝑠 equals negative six 𝑠 squared, four times 48 equals 192, and four times negative six 𝑠 equals negative 24𝑠, all equal to 48𝑠.
We can combine our two like terms, 48𝑠 minus 24𝑠 equals 24𝑠. And we’ll have negative six 𝑠 squared plus 24𝑠 plus 192 is equal to 48𝑠. We wanna set this equation equal to zero. And we can do that by subtracting 48𝑠 from that side. And if we subtract from the right side, we need to subtract from the left side. 24𝑠 minus 48𝑠 equals negative 24𝑠.
Now, we have an equation equal to zero. But we also have a negative leading coefficient. When we’re factoring, we never want a negative leading coefficient. We can solve this problem by multiplying the whole equation by negative one. Negative one times negative six 𝑠 squared equals six 𝑠 squared. Negative one times negative 24𝑠 positive 24𝑠, and negative one times 192 is negative 192. And the zero doesn’t change.
The next thing we notice is that this entire equation is divisible by six. We can divide every term by six. And we now have this. 𝑠 squared plus four 𝑠 minus 32 equals zero. We want to factor this equation. And that means we need two factors of negative 32 that when added together equal positive four. One and negative 32 will be far too big, two and negative 16 when added together equal negative 14. We’re getting closer, but we’re still not there. Four and negative eight when added together equal negative four. We’re almost there. If we use negative four and positive eight, they add together to equal positive four.
The missing terms are then negative four and positive eight. 𝑠 minus four times 𝑠 plus eight equal zero. We have to set 𝑠 minus four equal to zero. And then set 𝑠 plus eight equal to zero to solve for 𝑠. Add four to both sides and 𝑠 will equal four. Subtract eight from both sides and 𝑠 will equal negative eight. But our 𝑠 value represents a speed. And speed cannot be negative. We know that 𝑠 equals negative eight is not a valid option. And that means that 𝑠 must be equal to four. And 𝑠 in our key is equal to the speed of the actual bike ride. And that would be four miles per hour.
Actual speed equals four miles per hour. If we plug that in to our first equation, we find that the time it took Liam would be equal to 48 divided by four, 12 hours. We could plug in what we know to find out how quickly he would have completed the race, if he increased his speed. 12 minus six equals six hours. If Liam increased his speed from four miles per hour to eight miles per hour, he would have completed the 48 miles in six hours. However, the only piece of information this question was actually looking for is this one. Liam’s actual speed was four miles per hour.