document.write( "Question 1160648: Jen Butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 111. Two adults and three children must pay $ 79. Find the price of the adult's ticket and the price of a child's ticket. \n" ); document.write( "

Algebra.Com's Answer #784043 by ikleyn(52794) $\"\"$ $\"About$

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document.write( "From the condition, you have these two equations\r\n" );
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document.write( "    3A + 4C = 111    (1)\r\n" );
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document.write( "    2A + 3C =  79    (2)\r\n" );
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document.write( "To solve this system, I will apply the Elimination method.\r\n" );
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document.write( "For it, I multiply equation (1) by 2  (both sides), and multiply equation (2) by 3 (both sides).\r\n" );
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document.write( "My goal is to make coefficients at \"A\" equal.\r\n" );
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document.write( "You will get then\r\n" );
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document.write( "    6A + 8C = 222    (3)\r\n" );
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document.write( "    6A + 9C = 237    (4)\r\n" );
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document.write( "Now subtract equation (3) from equation (4).  The terms \"6A\" will cancel each other,\r\n" );
document.write( "and you will get single equation for only one unknown \"C\"\r\n" );
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document.write( "         9C - 8C = 237 - 222\r\n" );
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document.write( "          C       = 15.\r\n" );
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document.write( "So, we just made half of the job: we found the price of the child ticket, C = 15.\r\n" );
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document.write( "Now substitute this value of \"C\" into equation (1) to get \"A\"\r\n" );
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document.write( "    3A + 4*15 = 111,\r\n" );
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document.write( "which gives you\r\n" );
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document.write( "    3A = 111 - 5*15 = 111 - 60 = 51.\r\n" );
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document.write( "Hence,  A = 51/3 = 17.\r\n" );
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document.write( "ANSWER.  The child ticket price is $15;  the adult ticket price is $17.\r\n" );
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\n" ); document.write( "\n" ); document.write( "You may check the solution by substituting the found values into the original equations.\r
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