SOLUTION: 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.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 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.      Log On


   

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.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The basic fact we need to understand to solve this problem is that if
n = number of articles produced
r = number of articles produced per hour (rate)
t = number of hours of production
then
n = rt

We have two sets of data. First it says 30 articles are produced in t hours. So
30 = rt

Then it says that it would take 30 minutes less time to produce 30 if the rate was 5 more articles per hour. So
30+=+%28r%2B5%29%28t-1%2F2%29
(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:
30%2Ft+=+r
Substituting this into the second equation we get:
30+=+%2830%2Ft+%2B+5%29%28t+-+1%2F2%29
Now we solve this one variable equation. Multiplying out the right side we get:
30+=+30+-+15%2Ft+%2B+5t+-+5%2F2
Subtracting 30 from each side we get:
0+=+-15%2Ft+%2B+5t+-+5%2F2
Next we'll get rid of the fractions by multiplying both sides by the Lowest Common Denominator (LCD). The LCD here is 2t:
2t%280%29+=+2t%28-15%2Ft+%2B+5t+-+5%2F2%29
On the right side we need to use the Distributive Property:
2t%280%29+=+2t%2A%28-15%2Ft%29+%2B+2t%2A5t+-+2t%2A%285%2F2%29%29
which simplifies to:
0+=+-30+%2B+10t%5E2+-+5t
This is a quadratic equation. To solve it we'll first get it in the correct order:
0+=+10t%5E2+-+5t+-+30
Now we'll factor it (or use the Quadratic Formula):
0+=+5%28t%5E2+-+t+-+6%29
0+=+5%28t%2B2%29%28t-3%29
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:
t%2B2+=+0 or t+-+3+=+0
or
t+=+-2 or t+=+3
Since t is a number of hours we will reject the negative solution. So the only practical, real-life solution for t is 3.