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Mathematics 17 Online
OpenStudy (anonymous):

How do I do this?? http://static.k12.com/eli/bb/1212/3_69089/1_124143_5_69094/5b9ec945bcf24c6db230df7db66338de7234d2a9/media/1dbcc45fb1e71f8da92d03df7f82a8a28ce94906/AAHS_M8S2L3_1_teacher_graded_assignment_PROJECTGUIDE.pdf

OpenStudy (larseighner):

\[\Large {a \over b} \pm {c \over d} = {{ad \pm bc} \over bd}\]

OpenStudy (anonymous):

Can you show me how to do the first one?

OpenStudy (larseighner):

\[\Large {a \over b} \pm {c \over d} = {{ad \pm bc} \over bd}\] \[\Large {6 \over x} + {6 \over 3x} = {{6\cdot 3x + 6\cdot x} \over x\cdot 3x}={24x \over 3x^2} = {8 \over x}\]

OpenStudy (anonymous):

So I just multiply by 3 then add then divide?

OpenStudy (larseighner):

You have to express both numbers in the numerator with a common denominator. Then add them, then divide.

OpenStudy (anonymous):

oh okay. I think I got it. If I need help I'm going to message you okay?

OpenStudy (larseighner):

k

OpenStudy (anonymous):

Wait. So is the answer supposed to be 8/x?

OpenStudy (larseighner):

That is the answer to the fist part. Part b says x = 24 mph.

OpenStudy (anonymous):

oh okay.

OpenStudy (anonymous):

okay I just need help on the second one now. It looks different than the first one.

OpenStudy (anonymous):

@LarsEighner

OpenStudy (larseighner):

\[\Large {a \over b} \pm {c \over d} = {{ad \pm bc} \over bd} \] In the above, what should you replace a with?

OpenStudy (anonymous):

I just dont know what to do with the x+15 part

OpenStudy (larseighner):

That is just d. Follow the rule above. What is ad and bd?

OpenStudy (anonymous):

a=325 right??

OpenStudy (anonymous):

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