\(\bf \begin{array}{lccclll}
&amount&price&cost\\
&-----&-----&-----\\
\textit{\$7/lb coffee}&x&7&7x\\
\textit{\$4/lb coffee}&y&4&4y\\
-----&-----&-----&-----\\
blend&14&5&14\cdot 5
\end{array}\)
so.. whatever "x" and "y" amounts are, we know that, added together, they must yield 14lbs
thus x + y = 14
and whatever the cost of each is, 7x + 4y must be 14*5
7x+4y = 70
thus \(\bf \begin{cases}
x+y=14\to \boxed{y}=14-x\\\\
7x+4y=70\\
----------\\
7x+4\left( \boxed{14-x}\right)=70
\end{cases}\)
solve for "x", to see how much of the $7/lb type will be needed
what about the $4/lb one? well, y = 14 - x