Question 670863
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A comment about the two responses you have received to this point, one using three variables and one using a single variable.<br>
The easiest way to set the problem up is directly from the given information, which means using three variables and three equations; but then solving the problem means solving a system of three equations in three variables.<br>
It takes a bit of effort (not much, really) to set the problem up using a single variable and a single equation; but then the effort needed to solve the problem is much less.<br>
So the most efficient way to solve the problem using formal algebra is using a single variable and a single equation.<br>
The given information compares the number of dimes to the number of nickels, and it compares the number of nickels to the number of quarters.  The common item there is the number of nickels, so the clear logical choice for the single variable is the number of nickels.<br>
So proceed as shown in the response from the tutor who uses a single variable:<br>
let x = # of nickels
then 2x = # of dimes (there are twice as many dimes as nickels)
and x+8 = # of quarters (... and 8 fewer nickels than quarters -- i.e., 8 more quarters than nickels)<br>
The equation then says the total number of coins is 44:<br>
(x) + (2x) + (x+8) = 44<br>
The solution from there is easy....<br>