Question

Solve the following rational equations.

Answer 1

\[\frac{2}{2b+1}+\frac{2b^2-b+4}{2b^2-7b-4}=\frac{b}{b-4}\]

Answer 2

Can I give you the first step...? I'm still solving it.

Answer 3

Sure.

Answer 4

First Factorise \[2b^2-7b-4\]
It gives you (2b+1)(b-4)

After that, you can add the two fractions on the left hand side, since they have 2b+1 in common.

\[{2(b-4) + (2b^2-b+4)\over (2b+1)(b-4)}= {b \over b-4}\]

Answer 5

Now, you can knock off (b-4) altogether, as if comes over to the left hand side.

Answer 6

So you're left with\[{2(b-4)+(2b^2-b+4)\over2b+1}=b\]

Answer 7

Hold on...

Answer 8

ok.

Answer 9

See if you can somehow carry on... I'm still trying to figure it out, but I know that's the way to go.

Answer 10

(2b^2-b+4 )is not factorisable... So I tried adding 2b -8 to it, but then it becomes (2b^2+b-4) and that's not factorisable too.. Just to let you know... Did you use quadratic formula yet?

Answer 11

No...hmmm i am not sure what to do.

Answer 12

Wait... Let me see...

Answer 13

are you stuck anywhere so far? The ones I've mentioned?

Answer 14

No i just do not know how to finish factoring it.

Answer 15

Hmmm. I have a problem... see\[{2(b-4)+(2b^2-b+4)\over2b+1}=b\]
\[2(b-4)+(2b^2-b+4)=b(2b+1)\]
\[2b^2+b-4=2b^2+b\]
\[-4=0\]
And this doens't make sense at all!

Either I've made a mistake in my calculation... or there's something wrong with the question?
Can you double check my calculations and your question?

Answer 16

I double checked my method, and it is either a wrong solution, or the question was incorrectly written in either book/paper or here..?

Answer 17

My solution comes out as 0=-4
which is obviously wrong...

Answer 18

Seems like : cannot be solved for b

Answer 19

I got 0 = 4

Answer 20

-4*

Answer 21

The problem has no solution.

Answer 22

That's what I thought :) I didn't think that was an option... But it must have been!

Answer 23

Correct! ^