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Mathematics 7 Online
OpenStudy (anonymous):

The Ortegas have an apple orchard that contains 100 trees. The number of trees in each row is 10 more than twice the number of rows. Find the number of rows and the number of trees per row.

OpenStudy (mikeymaximum):

The answer is 10

OpenStudy (kinggeorge):

I think the best way to do this, is to let \(x=\) number of rows and \(y=\) number of trees in each row. This then gives you some equations which you can use to find \(x\) and \(y\). Do you have any guesses as to what these equations might be?

OpenStudy (anonymous):

Would the equation be \[2x^2+10x-150\]

OpenStudy (kinggeorge):

That's actually really close. I think the equation you need to solve should be\[2x^2+10x-100=0.\]So the only difference is the 100 instead of 150

OpenStudy (kinggeorge):

So how would you go about solving that equation?

OpenStudy (anonymous):

Thats where Im stuck at

OpenStudy (kinggeorge):

Well the first thing I would do, is divide by 2. Then we have\[\frac{2x^2+10x-100}{2}=x^2+5x-50=\frac02=0.\]Now you have a couple ways to approach it. First you could try and use the quadratic formula, or you could try and factor.

OpenStudy (kinggeorge):

Let's try factoring it since it'll turn out nicely. To factor, we need to find two numbers \(a\) and \(b\) so that \(ab=-50\) and \(a+b=5\). Since \(ab\) has to be a negative number, one of them has to be negative. That means our factorization will look like\[(x-?)(x+?).\]

OpenStudy (kinggeorge):

The next step is to start figuring out the factors of 50. We have three possibilities:\[50=1*50\]\[50=2*25\]\[50=5*10\]Does one of these factorizations meet our requirements?

OpenStudy (anonymous):

Would it be (x+10)(x-5)

OpenStudy (kinggeorge):

That's exactly it! Nice job. So that means that \(x=5\) or \(x=-10\)

OpenStudy (kinggeorge):

But, we can't really have a negative number of rows now can we? So it must be that \(x=5\). So we have 5 rows, and \(100/5=20\) trees per row.

OpenStudy (anonymous):

Ok, I got it, Thank you!

OpenStudy (kinggeorge):

You're welcome :)

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