Please HELP!

[rachealjohnson]

I'm stumped with this problem and keep comming up with the wrong answers. I solve the problem and then when I check it, it turn out to be incorrect.


x/a-1 = ax+3

solve for X


I was thinking we would probably have to solve for X and A to check it at least that is what I was doing.

 

[stapel]

Your formatting is ambiguous. Do you mean this...?


. . . . .\(\displaystyle \large{\frac{x}{a}\,-\,1\,=\,ax\,+\,3}\)


...or this...?


. . . . .\(\displaystyle \large{\frac{x}{a\,-\,1}\,=\,ax\,+\,3}\)


...or something else? (And I apologize in advance for asking.)

When you reply, if you wouldn't mind, please include the steps you have done in trying to solve the equation.

Thank you.

Eliz.

 

[rachealjohnson]

Here is the correct way...I think

x / (a-1) = ax + 3

How do you get your equations to look like that?

thanks for the help!


-rj

 

[Denis]

Re: Here is the correct way...I think

x / (a-1) = ax + 3

Rule: if a/b = c, then a = bc

x = (a - 1)(ax + 2)
x = xa^2 + 3a - xa - 3
x - xa^2 + xa = 3a - 3

Can you finish it?

 

[rachealjohnson]

x - xa^2 + xa = 3a - 3

x-xa^2 + xa -3 = 3a

(x-xa^2 + xa - 3) / 3 = a

I'm just not sure if I can reduce the (.... -3) / 3 or not. I am thinking not b/c it is not being multiplied or divided on top. correct? or does it reduce/cancel?

Thanks for all the help.

--rj
[/quote]

 

[Unco]

The objective is to solve for \(\displaystyle \mbox{x}\).

From
\(\displaystyle \mbox{ x - xa^2 + xa = 3a - 3}\)

Factor out \(\displaystyle \mbox{x}\) from the left-hand side:

\(\displaystyle \mbox{ x(1 - a^2 + a) = 3a - 3}\)

\(\displaystyle \mbox{ \Rightarrow x = \frac{3a - 3}{1 - a^2 + a}}\)