Question 7106
The first thing we need to nail down here is how to express how many pounds of each type of coffee there will be. We first need to agree, then, that the total number of pounds of coffee will be 90 pounds. Let's just call them coffee A and coffee B, and we'll let A and B represent how many pounds of their kind there are to make up those 90 total pounds. So far, we've got {{{ A + B = 90 }}} BUT we can only express it in terms of one variable.


We'll let A be the number of pounds of coffee A. That makes the leftover from the 90 pounds be the amount (in pounds) of coffee B. So, the entire 90 pounds, take away A pounds leaves you with B pounds. In equation form, {{{ 90 - A }}} = B


The cost of coffee A would be $7.00*A, which is $7.00 times how many pounds of A there are. The cost of coffee B would have to be $4.00(90 - A). When both coffees are mixed, we want the resulting mixture to weigh 90 pounds and we want each pound of the new mixture to cost $6.00. So, the price of our entire mixture should be $540.00 (The 90 pounds times the $6 per pound).


We have the equation{{{ 7A + 4(90-A) = 540 }}} <---- We'll now solve for A.


{{{ 7A + 360 - 4A = 540 }}}


{{{ 3A = 180 }}} <---- like terms and subtr 360 from both sides in one shot.


{{{ A = 60 }}} <---- There had to be 60 pounds of coffee A that cost $7 per pound (as pure coffee A)