solve this word problem
Algebra

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type a coffe cost 4.45 a lb and type b cost 5.80 lb. to make a mixture you use twice as many pounds of type b than type a for a total cost of 625.95. how many pounds of type a coffee were used?
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This is a classic system of equations problem, just set up as a word problem, so let's first these words into math.
We have two unknowns, the pounds of coffee a and the pounds of coffee b, which we will denote simply as "a" and "b" respectively. So:
a: pounds of coffee a
b: pounds of coffee b
Coffee a costs 4.45 a lb, so multiplying "a" by 4.45 will give us the total cost of coffee a. Coffee b costs 5.80 a lb, so multiplying "b" by 5.80 will give us the total cost of coffee b. Adding these two terms together will give us the overall total costs, which we know to be 625.95. So our first equation looks like this:
4.45a + 5.80b = 625.95
We need another equation to solve this. The problem said that you used twice as many pounds of type b than type a, so b is 2 times a. Now we have our second equation:
b = 2a
We know what be is in terms of a, so lets plug that new definition into the first equation:
4.45a + 5.80(2a) = 625.95
Solve for a:
a(4.45 + 11.6) = 625.95
a = 39
Therefore, we used 39 pounds of coffee type a.
To find how many pounds of coffee type b was used, just plug a back into the second equation:
b = 2a = 2*39 = 78
So, we used 39 pounds of coffee a, and 78 pounds of coffee b.
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