eramsby1010

Thanks CPhill!

a) \(\dfrac{8 \cdot 3 + 2 \cdot 8}{10} = \dfrac{24 + 16}{10} = \dfrac{40}{10} = \boxed{4}\)

b) \(\dfrac{8 \cdot 3 + 3 \cdot 8}{11} = \dfrac{24 + 24}{11} = \boxed{\dfrac{48}{11}}\)

c) \(\dfrac{8 \cdot 3 + 4 \cdot 8}{12} = \dfrac{24 + 32}{12} = \dfrac{56}{44} = \boxed{\dfrac{14}{11}}\)

d) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 6 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 6 \\ \dfrac{40 + 8x}{10 + x} &= 6 \\ 40 + 8x &= 60 + 6x \\ 40 + 2x &= 60 \\ 2x &= 20 \\ x &= \boxed{10} \end{align*}\)

e) \(\begin{align*} \dfrac{8 \cdot 3 + (2 + x) \cdot 8}{10 + x} &= 7 \\ \dfrac{24 + 16 + 8x}{10 + x} &= 7 \\ \dfrac{40 + 8x}{10 + x} &= 7 \\ 40 + 8x &= 70 + 7x \\ 40 + x &= 70 \\ x &= \boxed{30} \end{align*}\)

Thank you, using what you told me I arrived at p and t values of 5, and found that the problems the mathematician solved the day he drank coffee was \(\boxed{50}.\)