How does one go about solving 5/6 < n/7, 1/2 > n/10 ?  TOPIC_SOLVED

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How does one go about solving 5/6 < n/7, 1/2 > n/10 ?

Postby FrustratedDad on Tue Mar 10, 2009 12:10 am

How does one go about solving these fractional problems?

5/6 < n/7 n=

1/2 > n/10 n=

Am I too old for this? Any help would be appreciated! :confused:
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Re: How does one go about solving 5/6 < n/7, 1/2 > n/10 ?  TOPIC_SOLVED

Postby stapel_eliz on Tue Mar 10, 2009 2:10 am

FrustratedDad wrote:How does one go about solving these fractional problems?

5/6 < n/7
1/2 > n/10

Use the exact same methods you used for solving linear equations. For instance:

linear analog:
5/6 = n/7
(5/6)(7/1) = (n/7)(7/1)
(5*7)/(6*1) = n
35/6 = n

Now do the same thing for the inequalities, remembering to keep the inequality signs pointing the right way. (In your case, they'll stay pointing the way they are, since you aren't multiplying or dividing through by negatives during your solution process.)

If you get stuck, or if you are unsure of your steps or solutions, please reply showing your work and reasoning so far. Thank you! :D
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