Question 1135171
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I absolutely concur with the statement that this problem is suited for being solved with one unknown.  Using two variables and then using substitution is a big waste of effort.<br>
If the number of candies sold is x, the number of waters sold is 5x.<br>
Then one way you can make the numbers in the problem smaller is to imagine the candies and waters being sold in equal groups, each with 5 waters and 1 candy.  The cost of one of those groups is 5 times 20 cents plus 1 times 35 cents, or $1.35.  Then do some mental or pencil-and-paper calculations to see how many of those groups it takes to make the total sales of $16.20.<br>
{{{5(20)+1(35) = 135}}}<br>
{{{1620/135 = 12}}}<br>
12 groups; so 12 candies and 12*5=60 waters.