Ask your own question, for FREE!
Mathematics 18 Online
OpenStudy (anonymous):

Help me please solve this rational expression? (1/3+x) + (1/x) = (1/2)

OpenStudy (anonymous):

Here is the original word problem:

OpenStudy (anonymous):

i posted the same thing about 25 minutes ago and for some reason no one could solve.. so.. o.o

OpenStudy (anonymous):

\[\frac{ 1 }{ t+3 }+\frac{ 1 }{t }=\frac{ 1 }{2 }\] \[\frac{ t+t+3 }{t \left( 1+t \right) }=\frac{ 1 }{ 2 }\] \[2\left( 2t+3 \right)=t ^{2}+t,t ^{2}+t-4t-6=0,~t ^{2}-3t-6=0\] solve it.

OpenStudy (anonymous):

Woah. o.o What's that bottom part. o.e

OpenStudy (anonymous):

correction ,write 3+t in place of 1+t

OpenStudy (anonymous):

I'm confused. I'm not supposed to simplify here. I'm finding an actual numerical value.

OpenStudy (nincompoop):

WHO CAME UP WITH THE EXPRESSION?

OpenStudy (anonymous):

I HAVE NO IDEA. stupid common core.. @nincompoop

OpenStudy (nincompoop):

OKAY

OpenStudy (anonymous):

I've seen it solved before. Somehow, the x needs to get isolated but I don't know how to do that with rational expressions.

OpenStudy (anonymous):

\[4t+6=3t+t ^{2},t ^{2}-t-6=0\]

OpenStudy (anonymous):

\[t ^{2}-3t+2t-6=0,t \left( t-3 \right)+2\left( t-3 \right)=0\] \[\left( t-3 \right)\left( t+2 \right)=0,t=3,t=-2\] rejecting t=-2 t=3

OpenStudy (anonymous):

OH, thank you!

OpenStudy (anonymous):

So if I plugged 3 into the original equation, it should satisfy?

OpenStudy (anonymous):

you are correct.

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

\[\frac{ 1 }{6 }+\frac{ 1 }{3 }=\frac{ 1+2 }{ 6 }=\frac{ 3 }{6 }=\frac{ 1 }{ 2 }\]

OpenStudy (anonymous):

np

OpenStudy (anonymous):

So the expression above, it simplifies to 2(2t+3)? this one: x+x+3/x(3+t)

OpenStudy (anonymous):

i mean 2x. My teacher doens't like it when I use anything other than x, y or z lol

OpenStudy (anonymous):

I have used t only because it is time.Otherwise you can use any variable x,y,z,a,b,...etc.

OpenStudy (anonymous):

But that's what it simplifies to?

OpenStudy (anonymous):

Nvm, got it! Thanks!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!