document.write( "Question 1199791: A train ticket in a certain city is $1.50. People who use the train also have the option of purchasing a frequent rider pass for $16.50 each month. With the pass, each ticket costs only $0.75. Determine the number of times in a month the train must be used so that the total monthly cost without the pass is the same as the total monthly cost with the pass. \n" ); document.write( "

Algebra.Com's Answer #833757 by math_tutor2020(3817) $\"\"$ $\"About$

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\n" ); document.write( "Option A: No monthly pass
\n" ); document.write( "Option B: Monthly pass\r
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\n" ); document.write( "\n" ); document.write( "x = number of times the person rides the train
\n" ); document.write( "y = total cost after riding x times\r
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\n" ); document.write( "\n" ); document.write( "The cost equation for option A is
\n" ); document.write( "y = 1.50x\r
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\n" ); document.write( "\n" ); document.write( "The cost equation for option B is
\n" ); document.write( "y = 0.75x+16.50\r
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\n" ); document.write( "\n" ); document.write( "Use substitution to solve for x.
\n" ); document.write( "1.50x = 0.75x+16.50
\n" ); document.write( "1.50x-0.75x = 16.50
\n" ); document.write( "0.75x = 16.50
\n" ); document.write( "x = 16.50/0.75
\n" ); document.write( "x = 22
\n" ); document.write( "The costs are the same when the person rides the train 22 times.\r
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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "y = 1.50x = 1.50*22 = 33 dollars is the cost of option A
\n" ); document.write( "y = 0.75x+16.50 = 0.75*22+16.50 = 33 dollars is the cost of option B
\n" ); document.write( "Each option yields the same cost, which confirms the answer.\r
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\n" ); document.write( "\n" ); document.write( "Answer: 22
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