A grocer wants to mix two kinds of coffee. One kind sells for 
1.45 per pound, and other sells for 2.20 per pound. He wants to mix a total of 21 pounds and sell it for 1.70Per pound. How many pounds of each kind should he use in the new mix? (ROund off the answers to the nearest hundredth)
Detailed Solution:
Given that there are two kinds of coffee.
Let us say that “X pounds” of coffee is sold at 1.45 per pound.
Therefore the cost of the “X pounds” of coffee is 1.45*X
The other kind of coffee “Y” pounds” is sold at 2.20 per pound.
Therefore the cost of “Y pounds” of coffee would be 2.20*Y.
Since there are two variables X and Y, we need to frame two equations in X and Y to solve them.
Given that the grocer wants to mix two kinds of coffee and the total weight of the mixture is 21 pounds.
Hence the first equation can be
X+Y = 21——–>Equation1
We are also given that the total cost of the mixture is 1.70 per pound.
The cost of 21 pounds of the mixture would be 21*1.7=35.7
Hence the second equation would be
1.45X+2.20Y=35.7 ——–> Equation 2
Solving the equations 1 and 2:
X+Y = 21–>Equation1
1.45X+2.20Y=35.7 –> Equation 2
From the equation1, we can say that Y = 21-X.
Let us substitute this new value of Y = 21-x in the second equation 1.45X+2.20Y =1.7
1.45X+2.20(21-x) =35.7
1.45X+46.2-2.2X=35.7
-0.75X =35.7-46.2
-0.75X=-10.5
Therefore X= 
X=14 pounds
Therefore the value of Y would be Y= 21-X=21-14
Y= 7 Pounds
Therefore, the grocer needs to mix 14 pounds of coffee which sells for 1.45 and 7pounds of coffee which sells for 2.20 pounds.
Verification:
We can cross-check the answers we got for the X =14 and Y =7 values.
When the values of X and Y are plugged back into the original equations, they must hold good.
Let us take up the equation1
x+y=21
14+7=21
21=21.Therefore, the solutions we arrived work for the Equation1
Let us cross-check with the equation2.
1.45X+2.2Y =35.7
1.45(14)+2.2(7) =35.7
20.3+15.4=35.7
Therefore, the solutions we arrived work for the Equation2 as well.
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