<PRE><B>Question 1179258</B>
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The first tutor always solves systems of equations using substitution.  That might be her preferred method; but when both equations are given in Ax+By=C form, I think some sort of elimination is much easier.&lt;br&gt;
Below is an unusual method for solving the problem.  I first present the solution informally using logical reasoning; then I show the corresponding formal algebraic solution.  The method shown can make the solution of some similar problems easier than standard elimination; but for most problems a more straightforward solution using elimination will be fastest.&lt;br&gt;
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&quot;On the first day, Cindy sells 6 adult tickets and 5 children&#39;s tickets for a total of 112.50.&quot;
&quot;On the second day, she sells 8 adult tickets and 4 children&#39;s tickets for a total of 130.00&quot;&lt;br&gt;
The difference between the two sales is 2 more adult tickets and 1 fewer children&#39;s tickets for $17.50 more.&lt;br&gt;
Apply that difference four more times to make the number of children&#39;s tickets zero.  Doing that gives us 8+4(2) = 16 adult tickets and 4-4(1) =  0 children&#39;s tickets for a total of $130+4($17.50)=$200.&lt;br&gt;
So the cost of each adult ticket is $200/16 = $12.50.&lt;br&gt;
Then use the purchase on the second day to determine the cost of each children&#39;s ticket:
8($12.50)+4x = $130
4x = $30
x = $30/4 = $7.50&lt;br&gt;
ANSWER: Adult ticket price $12.50; Children&#39;s ticket price $7.50&lt;br&gt;
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Algebraically, this unusual solution method looks like this:&lt;br&gt;
{{{system(6a+5c=112.5,8a+4c=130)}}}&lt;br&gt;
Find the difference between the two equations:&lt;br&gt;
{{{2a-c = 17.50}}&lt;br&gt;
Now use this system:&lt;br&gt;
{{{system(8a+4c=130,2a-c=17.50)}}}&lt;br&gt;
Multiply the second equation by 4:&lt;br&gt;
{{{system(8a+4c=130,8a-4c=70)}}}&lt;br&gt;
Add the two equations to eliminate c:&lt;br&gt;
{{{16a=200}}}
{{{a = 200/16 = 12.50}}}&lt;br&gt;
Use that value for a to find c:&lt;br&gt;
{{{8(12.50)+4c = 130}}}
{{{4c = 30}}}
{{{c = 7.50}}}&lt;br&gt;
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So why use this unusual method for solving the problem?&lt;br&gt;
In a formal algebraic solution, finding the difference between the two original equations gives us an equation in which one of the coefficients is -1.  That will make solving the system by elimination easier, because it will keep the coefficients smaller.&lt;br&gt;
And if a formal algebraic solution is not required, this method of solving the problem makes an informal solution easier, for the same reason.&lt;br&gt;
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