Question 806787
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Your equation is fundamentally correct, and given correct arithmetic you should have arrived at the correct numerical answer of 13 and 1/3 pounds of raisins and 6 and 2/3 pounds of nuts.  


Where you may be running afoul of the teacher's system is that you did not precisely follow the instructions in the problem.  You wrote a single equation and the problem asks for a <i><b>system</b></i> of equations.  In this sort of scenario, you can write two equations in two variables.  Here, we need a <i>quantity</i> equation and a <i>value</i> equation:


First, declare your variables.  Let *[tex \Large x] represent the number of pounds of raisins, and let *[tex \Large y] represent the number of pounds of nuts.


Quantity equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 20]


Value equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4.2x\ +\ 3.6y\ =\ 4(20)]


This particular problem is going to degenerate into the same thing you started with when you solve the quantity equation for *[tex \Large y] and then solve the system using the substitution method. But I suspect the teacher is in the process of introducing the idea of systems of equations, and the lesson here is for the student to understand the structure of linear systems and the process of solving them -- whether or not you can figure out how many pounds of raisins you need for the 20 lb mixture is immaterial.


Hope that helps.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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