In today's blog we will discuss a more complicated scenario of what I mentioned in Section 1.0. For the same problem (4 apples need to be distributed between 2 people), if we change the numbers and the question a little, we will need additional applications of Algebra to solve it. Let's say now we have 11 apples, and there are 3 people, but 2 of them are rotten and need to be thrown away before the distribution, how many will each one get now. We begin the same way, first by choosing a variable, let's say 'y' this time and creating an imaginary situation. If 'y' is the unknown value each person will receive, then we can safely assume that the total number of apples distributed will equal to 3 x 'y'. So we can begin the equation by completing the left hand side of it and writing 3y=
Now we need a number for the right hand side. We have 11 apples in total, so for now let's right 11, but oh wait! Two of them have gone bad, so we need to throw them away. So we must subtract those 2 apples from the total number of apples that is 11. Hence, we can safely arrive at the number 9. This will therefore be the total number of apples used in the equation and we can put '9' in the right hand side of the equation. Now, we get 3y=9.
To arrive at the value of y, we can now carry out the same procedure of dividing both sides of the equation by 3 so that we can deduce, y=3. Therefore, the number of apples each person will receive now is 3.
This Algebraic equation this time used 'subtraction' along with just multiplication and division.