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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 i