In class we went over the problems in chapters, 6:12; 7:1; and 7:18.

The first problem reads, A good runner can go 100 paces while a poor runner covers 60 paces. The poor runner has covered a distance of 100 paces before the good runner sets off in pursuit. How many paces does it take the good runner before he catches up the poor runner.

The second problem reads, Certain items are purchased jointly. If each person pays 8 coins, the surplus is 3 coins, and if each person gives 7 coins, the deficiency is 4 coins. Find the number of people and the total cost of the items.

The third reads, There are two piles, one containing 9 gold coins and the other 11 coins. The two piles of coins weigh the same. One coin is taken from each pile and put into the other. It is now found that the pile of mainly gold coins weighs 13 units less than the pile of mainly silver coins. Find the weight of a silver coin and of a gold coin.

Below I have inserted a picture of the scratch work I did in class.

In the first problem, I first created a chart to kind of find a rough estimate of where the good runner would catch the poor runner. After seeing the range I needed I could make sure that my x value made sense. Following, I came up with the formula 60x+100=100x to find how long it would take the second runner to catch the first. The 60x represents the amount of paces the poor runner runs per time value, the +100 represents the amount of the head start, and the 100x represents the amount of paces the good runner would need to catch the second runner. Therfore, to find the total amount of paces for the good runner needs to run to catch the poor runner, I needed to solve for x. I did so by subtracting 60x from both sides to get 100=40x. Then I divided both sides by 40 to get 2.5 hours. Lastly I determined how many paces were in 2.5 hours for the good runner to get 250 paces.

In the second problem, I took both amounts and made equations. If each person paid we knew that we had 3 leftover, thus we have 8 cents per person with 3 cents left over, same follows for when the amount is split that everyone pays 7 cents. They are 4 cents short. Therefore we know that these are equal to one another so we set the equations equal to one another. We therefore find that there are 7 people total and that collectively they spend 53 cents.

In the last case we are given two weights that are equal to one another. We know that we unbalance the equation therefore we add 13 to the left side representing the difference of 13. We are able to determine the ratio of the gold by getting x by itself, therefore we can plug this into our x value to find that silver weighs 29.25 and gold weighs 36.75.

Each individual problem was solved algebraically using variables that represent numbers that can be altered. We were able to discover solutions by setting equations equal to one another in order to determine what value the variable needs to be in each case. Through the mathematics shown, we see how creative this new kind of mathematics was to create a functional generic system that can be applied to other values in the same type of problem. For example in the first problem, because the poor runner had a 100 pace head start, we knew our equation had to look like 60x+100=100x. To generalize this, we can represent our 100 as c, for constant. It can be replaced with any amount of head start and you could calculate how many paces it would take the good runner to catch the poor runner. Same follows by the second problem. The 3 and 4 in the problems could be replaced with any number and you could still take the same process to determine the total value and the number of people paying. Lastly the third problem follows suit. If you change the number of pieces of gold and silver to any other constants, you would use the same operations to determine the weight of silver and gold