document.write( "Question 271852: A machine produces 30 articles in t hours. If the machine were to produce 5 more articles each hour, it would take half an hour less to produce 30 articles. Find t. \n" ); document.write( "
Algebra.Com's Answer #198993 by jsmallt9(3758)\"\" \"About 
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The basic fact we need to understand to solve this problem is that if
\n" ); document.write( "n = number of articles produced
\n" ); document.write( "r = number of articles produced per hour (rate)
\n" ); document.write( "t = number of hours of production
\n" ); document.write( "then
\n" ); document.write( "n = rt

\n" ); document.write( "We have two sets of data. First it says 30 articles are produced in t hours. So
\n" ); document.write( "30 = rt

\n" ); document.write( "Then it says that it would take 30 minutes less time to produce 30 if the rate was 5 more articles per hour. So
\n" ); document.write( "\"30+=+%28r%2B5%29%28t-1%2F2%29\"
\n" ); document.write( "(Note the use of 1/2. The time is supposed to be in hours so 30 minutes was converted to 1/2 hour.) We now have a system of two equations in two variables. Since these are not linear equations we'll use the Substitution Method to solve the system. Solving the first equation for r we get:
\n" ); document.write( "\"30%2Ft+=+r\"
\n" ); document.write( "Substituting this into the second equation we get:
\n" ); document.write( "\"30+=+%2830%2Ft+%2B+5%29%28t+-+1%2F2%29\"
\n" ); document.write( "Now we solve this one variable equation. Multiplying out the right side we get:
\n" ); document.write( "\"30+=+30+-+15%2Ft+%2B+5t+-+5%2F2\"
\n" ); document.write( "Subtracting 30 from each side we get:
\n" ); document.write( "\"0+=+-15%2Ft+%2B+5t+-+5%2F2\"
\n" ); document.write( "Next we'll get rid of the fractions by multiplying both sides by the Lowest Common Denominator (LCD). The LCD here is 2t:
\n" ); document.write( "\"2t%280%29+=+2t%28-15%2Ft+%2B+5t+-+5%2F2%29\"
\n" ); document.write( "On the right side we need to use the Distributive Property:
\n" ); document.write( "\"2t%280%29+=+2t%2A%28-15%2Ft%29+%2B+2t%2A5t+-+2t%2A%285%2F2%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"0+=+-30+%2B+10t%5E2+-+5t\"
\n" ); document.write( "This is a quadratic equation. To solve it we'll first get it in the correct order:
\n" ); document.write( "\"0+=+10t%5E2+-+5t+-+30\"
\n" ); document.write( "Now we'll factor it (or use the Quadratic Formula):
\n" ); document.write( "\"0+=+5%28t%5E2+-+t+-+6%29\"
\n" ); document.write( "\"0+=+5%28t%2B2%29%28t-3%29\"
\n" ); document.write( "From the Zero Product Property we know that this product is zero only if one of the factors is zero. 5 can never be zero. But (t+2) and (t-3) could be zero with the \"right\" values of y. So we solve the following:
\n" ); document.write( "\"t%2B2+=+0\" or \"t+-+3+=+0\"
\n" ); document.write( "or
\n" ); document.write( "\"t+=+-2\" or \"t+=+3\"
\n" ); document.write( "Since t is a number of hours we will reject the negative solution. So the only practical, real-life solution for t is 3. \n" ); document.write( "
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