Question 1160648
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Although not typical, here is an alternative solution method using elimination.<br>
(1) {{{3a+4c = 111}}}
(2) {{{2a+3c = 79}}}<br>
Subtract (2) from (1):
(3) {{{a+c = 32}}}<br>
Now double (3) and subtract from (2):
{{{2a+3c = 79}}}
{{{2a+2c = 64}}}
{{{c = 15}}}<br>
Plug c=15 into (3) to find a=17.<br>
ANSWER: adult ticket costs $17; child ticket costs $15.<br>
This method can be used -- either formally, as above, of informally, as below, because the difference between the two given cases is 1 adult and 1 child.<br>
This variation of a formal algebraic solution using elimination exactly follows an informal solution obtained by logical reasoning:<br>
3 adults and 4 children cost $111
2 adults and 3 children cost $79<br>
Therefore, continuing the "pattern" of 1 less adult and 1 less children for $32 less...<br>
1 adult and 2 children cost $47
0 adult and 1 child cost $15<br>
ANSWER: child $15; adult $32-$15 = $17<br>