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You can use the fact that number of breads purchased cannot be negative since a customer either buys them or not and usually do not sell to the shopkeeper.(if somehow they end up selling to shop owner, then yes that will go in negative, but we'll assume it is wrong in most cases as generally shop owners are there to sell stuffs).
The third table of values matches the equation and includes only viable solutions.
<h3>What is a viable solution here?</h3>It is talking about those solutions which are seen in real world. As stated above, a customer either buys the bread or not, thus number of breads sold will be either positive or 0(in case of no selling). Thus, we cannot have number of breads as negative.
Such solutions which are correct in the real world context here are called here as viable solutions.
<h3>Checking one by one all the tables for them being matched with table and viability</h3>For first table, the number of breads are in negative, thus it is not going to have viable solution.
For second table, we have:
b = 0 thus c = 3.5b = 3.5 times 0 = 0 which is correctly given in second column.
b = 0.5, thus c = 3.5b = 3.5 times 0.5 =1.75 which is correctly given.
b = 1, thus c= 3.5 times 1 = 3.5 which is correctly given
b = 2001.5 thus c = 3.5 times 2001.5 = 7005.25 which is not correctly given, thus wrong.
For third table, we have:
b = 0, thus , correctly given in second column.
b = 3, thus , correctly given.
b = 6, thus , correctly given.
b = 9, thus , correctly given.
Thus, the third table of values matches the equation and includes only viable solutions.
Learn more about purchasing to cost relation here:
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