Question

Solving rational equations ???

Answer 1

1.(1/6x2 ) = ( 1/3x2) - (1/x)

2. (1/n2) + (1/n) = (1/2n2)

3. (1/x) = (6/3x) + 1

4. (1/6x2) = (1/2x) + (7/ 6x2)

5. (1/6b2) + (1/6b) = (1/b2)

Answer 2

If you explain your process

Answer 3

nope \[\frac{1}{6x^2}=\frac{1}{3x^2}-\frac{1}{x}\]

Answer 4

6x on the first fraction

Answer 5

Correct

Answer 6

What..?

Answer 7

someone helps him, please!

Answer 8

to avoid confusion, can you use the draw tool to write the 1st question.

Answer 9

ok... an easy solution is take the reciprocal of each fraction so you have

but 1st you have the condition that x cannot be zero

\[6x^2 = 3x^2 - x\]

which becomes

\[3x^2 + x = 0\]

which can be factorised

\[x(3x + 1) = 0\]

solve for x

Answer 10

got it?

Answer 11

see mine, quite easy.
\[\frac{1}{3x^2}=\frac{2}{6x^2}\], got it?

Answer 12

so, \[\frac{1}{6x^2}=\frac{2}{6x^2}-\frac{1}{x}\]

Answer 13

\[minus~ both~ sides~ by ~\frac{1}{6x^2}\\and~plus ~both~sides~by~\frac{1}{x}\]
you will have \[\frac{1}{6x^2}=\frac{1}{x}\] ooooh, switch now, it turns 6x^2 =x
then \[6x^2-x=0\\x(6x-1)=0\\then~ x=0 ~or~ x=\frac{1}{6}\]

Answer 14

Thanks to you guys !

Answer 15

@loser66 can you explain again im not getting the right answer