document.write( "Question 1135543: For a trip, one high school rented and filled 6 vans and 8 buses with 202 students. Another high school instead fit its 68 students into 4 vans and 2 buses. With each bus and van seating the same number of students, how many students can a bus carry? How many students can a van carry? \n" ); document.write( "

Algebra.Com's Answer #753174 by ikleyn(52788) $\"\"$ $\"About$

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document.write( "Let x = capacity of a van and y = capacity of a bus.\r\n" );
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document.write( "From the condition, you have these two equations\r\n" );
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document.write( "    6x + 8y = 202   (1)\r\n" );
document.write( "    4x + 2y =  68   (2)\r\n" );
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document.write( "In this case, I will apply the Elimination method. I multiply eq(2) by 4 (both sides), keeping equation (1) as is.\r\n" );
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document.write( "    6x + 8y = 202    (1')\r\n" );
document.write( "   16x + 8y = 4*68   (2')\r\n" );
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document.write( "Now subtract eq(1') from eq(2').  The terms \" 8y \" will cancel each other, and you will get\r\n" );
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document.write( "    16x - 6x = 4*68 - 202,    or\r\n" );
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document.write( "    10x = 70   ================>  x = 70/10 = 7.\r\n" );
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document.write( "Now substitute this found value of x into eq(2) to find y. You will get\r\n" );
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document.write( "    4*7 + 2y = 68,\r\n" );
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document.write( "    2y = 68 - 28 = 40  ============>  y = 40/2 = 20.\r\n" );
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document.write( "ANSWER.  Van's capacity is 7 students;  bus' capacity is 20 students.\r\n" );
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