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Mathematics 12 Online
OpenStudy (anonymous):

If x ≠ 0, then u/x +5u/x - u/5x =

OpenStudy (kenljw):

u/x +5u/x - u/5x = (1/x) (u + 5u -u/5)

OpenStudy (anonymous):

^ haha um, wrong thread.?

OpenStudy (anonymous):

but, (1/x) (u + 5u -u/5) should the u+5u be over the u/5?

OpenStudy (kenljw):

u/x +5u/x - u/5x = (1/x) (u + 5u -u/5) = (u/x) (1 + 5 - 1/5) = 31/5

OpenStudy (kenljw):

29/5

OpenStudy (anonymous):

Hold on, it's easier if I write this on paper to see the fractions clearer,

OpenStudy (anonymous):

I am so lost on how you got 31

OpenStudy (kenljw):

change it to 29/5 (5X6 -1)/5

OpenStudy (anonymous):

Okay, now I'm lost on how you got (5x6-1)/5...

OpenStudy (kenljw):

u/x +5u/x - u/5x = (1/x) (u + 5u -u/5) = (u/x) (1 + 5 - 1/5) 6 - 1/5 = (30 -1)/5 I found common denominator 1/a + 1/b = (1/a) (b/b) + (1/b) (a/a) = (a + b)/ab

OpenStudy (kenljw):

final answer 29u/(5x)

OpenStudy (anonymous):

I'm still so lost.. you got the 6 fro adding 1+5?

OpenStudy (kenljw):

true but you must subtract 1/5

OpenStudy (anonymous):

so 6/1 - 1/5.?

OpenStudy (kenljw):

yes

OpenStudy (anonymous):

Ok, and why would it be 31 instead of 29.? do I have to do the opposite and add instead of subtract

OpenStudy (kenljw):

it should be 29, I made error first time. original equation subtracted 1/5 not added 1/5

OpenStudy (anonymous):

ah okay, thanks for the help.! I'm re-taking my mathematics placement testing tomorrow, so I really appreciate it

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