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Mathematics 23 Online
OpenStudy (anonymous):

a) Write the value of the variable that make the denominator zero b) Keeping the restrictions in mind, solve the equation (x-2/2x)+1 = x+1/x

OpenStudy (anonymous):

a) x = 0 b) x= 2

OpenStudy (anonymous):

the answer in the book is x=0 x=4

OpenStudy (anonymous):

a) is obvious but im having trouble finding b because my answer was zero for b

OpenStudy (lgbasallote):

\[\frac{x-2}{2x} + 1 = \frac{x+1}{x}\]

OpenStudy (anonymous):

then you didnt respected the math rules when you copied the equation. is it -2/(2x) or (-2/2)*x

OpenStudy (anonymous):

exactly how igb put it

OpenStudy (anonymous):

@lgbasallote , ahh... Now we're getting somewhere

OpenStudy (anonymous):

x= 4 indeed

OpenStudy (anonymous):

idk how you could have taken it any other way...

OpenStudy (lgbasallote):

you just multiply everything by 2x \[x - 2 + 2x = 2(x+1)\] \[3x - 2 = 2x + 2\] \[3x - 2x = 2+2\] \[x = 4\]

OpenStudy (lgbasallote):

do you get that?

OpenStudy (lgbasallote):

tell me if a step is unclear @dylanlamoreaux :)

OpenStudy (anonymous):

if you multiply everything by 2x arent you going to get 2x^2 for the right side?

OpenStudy (lgbasallote):

nope \[\frac{x+1}{\cancel x} \times 2\cancel x\]

OpenStudy (lgbasallote):

x just cancels out

OpenStudy (lgbasallote):

\[\frac{x+1}{x} \times 2x = 2(x+1)\]

OpenStudy (anonymous):

oh got it

OpenStudy (lgbasallote):

nice!!

OpenStudy (anonymous):

cool thanks!

OpenStudy (lgbasallote):

<tips hat>

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