Question 1149918
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First, a solution by logical reasoning....<br>
Let's change the statement of the problem a bit so that the two purchases are by different students.  That will make it easier to describe the method for solving the problem.<br>
So we have<br>
Student A: 3 boxes of pencils and 2 boxes of pens for $6.
Student B: 2 boxes of pencils and 4 boxes of pens for $8.<br>
Student B bought twice as many boxes of pens as student A.  So consider a third student C buying twice as much as student A, so that students B and C buy the same numbers of boxes of pens.<br>
Student C: 6 boxes of pencils and 4 boxes of pens for $12.<br>
Now compare the purchases by students B and C.  They bought the same number of boxes of pens; student C bought 4 more boxes of pencils than student B, and his cost was $4 more.  So each box of pencils costs $1.<br>
Then use the purchases of any one of the three students to find that the cost of each box of pens is $1.50.<br>
Now a typical algebraic solution, which follows the exact same path as the solution above.<br>
[1] {{{3x+2y = 6}}}  the cost of 3 boxes of pencils and 2 boxes of pens is $6
[2] {{{2x+4y = 8}}}  the cost of 2 boxes of pencils and 4 boxes of pens is $8<br>
[3] {{{6x+4y = 12}}}  double the first purchase<br>
[4] {{{4x = 4}}}  compare purchases [2] and [3]<br>
[5] {{{x = 1}}}  the cost of each box of pencils is $1<br>
[6] {{{3(1)+2y = 6}}}  substitute [5] into [1]
{{{3+2y = 6}}}
{{{2y = 3}}}
{{{y = 3/2 = 1.5}}}  the cost of each box of pens is $1.50<br>