a) y = 5 + 7n
b) y = 15 + 5n
c) 5 books
d) $40
a) Here, we want to write an equation representing the first club's total cost
Let the number of books bought be n
Then, mathematically;
[tex]y\text{ = 5 + 7n}[/tex]b) For the second club's total cost
Let the number of books bought be n
Then mathematically;
[tex]y\text{ = 15 + 5n}[/tex]c) For the cost to be the same, we equate y in both cases so asto find n
Thus, we have;
[tex]\begin{gathered} 15\text{ + 5n = 5 + 7n} \\ \\ 15-5\text{ = 7n-5n} \\ \\ 10\text{ = 2n} \\ \\ n\text{ = }\frac{10}{2} \\ \\ n\text{ = 5 books} \end{gathered}[/tex]d) Here, we simply substitute n for 5 in any of the equations
We have this as;
[tex]\begin{gathered} 5\text{ + 7n} \\ \\ 5\text{ + 7(5)} \\ 5\text{ + 35 = 40} \end{gathered}[/tex]