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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
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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
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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
\[x - 2 + 2x = 2(x+1)\]
\[3x - 2 = 2x + 2\]
\[3x - 2x = 2+2\]
\[x = 4\]
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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
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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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