solving for X in a rational equation...

[Melissaherman]

I'm doing some calc homework and am attempting to find inverse equations. I know how to do that, but first I need to solve for X.... this should be simple math but for the life of me I can't remember how to do it... it's been awhile since I've done math.... here is an example of what I mean:

y= 4x-1/2x+3
How do I solve for x? Any help would be appreciated!

 

[stapel]

y= 4x-1/2x+3

Your formatting is unclear. Does the above mean any of the following?

. . . . .\(\displaystyle \L y\, =\, 4x\, -\, \frac{1}{2x}\, +\, 3\)

. . . . .\(\displaystyle \L y\, =\, 4x\, -\, \frac{1}{2}x\, +\, 3\)

. . . . .\(\displaystyle \L y\, =\, 4x\, -\, \frac{1}{2x\, +\, 3}\)

. . . . .\(\displaystyle \L y\, =\, \frac{4x\, -\, 1}{2x\, +\, 3}\)

Or did you mean something else?

Thank you.

Eliz.

 

[Melissaherman]

The last one.... sorry about that, I wasnt exactly sure how to write it.
Thanks!

 

[Denis]

Next time, use brackets: y = (4x - 1) / (2x + 3)
Crisscross multiplication:
4x - 1 = y(2x + 3) ; complete the multiplication on right:
4x - 1 = 2xy + 3y ; get the x's together:
4x - 2xy = 3y + 1 ; factor out the x:
x(4 - 2y) = 3y + 1 ; divide by (4 - 2y) to isolate the x:
x = (3y + 1) / (4 - 2y)

Follow all that :?:

 

[Melissaherman]

ahhhhh yeah. Ok that was a lot easier than I thought.... it's all comin' back to me now, haha. Thanks a lot, I really appreciate the quick response!