Find the inverse of F(x)?

**WashingtonStudent** · September 8th 2009, 07:27 PM

f(x) = (2x+3)/(4x-5)

So, I let f(x) = y and then I get y = (2x+3) / (4x-5) and then I figure I solve for x... But I am having difficulty in this, any suggestions? Thank you

**Rapha** · September 8th 2009, 07:36 PM

Hi WashingtonStudent!

Originally Posted by WashingtonStudent

f(x) = (2x+3)/(4x-5)

So, I let f(x) = y and then I get y = (2x+3) / (4x-5) and then I figure I solve for x... But I am having difficulty in this, any suggestions? Thank you

Do you know this method: "Polynomial long division"

You get $\frac{2x+3}{4x-5} = 0.5 + \frac{11}{2*(4x-5)}$

thus you got to solve $y = 0.5 + \frac{11}{2*(4x-5)}$

I
Yours
Rapha

**zodiacbrave** · September 10th 2009, 03:11 PM

Could you go over the steps in a bit more detail? I am familiar with polynomial long divison.. but when I make 2x + 3 the divided and 4x - 5 the divisor then I end up with..

0.5*(4x-5) + 5.5

**pickslides** · September 10th 2009, 03:33 PM

I found

$\frac{2x+3}{4x-5}= \frac{1}{2(4x-5)}+\frac{1}{2}$

Now $y = \frac{1}{2(4x-5)}+\frac{1}{2}$

To find the inverse, swap y and x and solve for y.

$x = \frac{1}{2(4y-5)}+\frac{1}{2}$

$x-\frac{1}{2} = \frac{1}{2(4y-5)}$

$4y-5 = \frac{1}{2(x-\frac{1}{2})}$

$4y = \frac{1}{2(x-\frac{1}{2})}+5$

$y = \frac{1}{8(x-\frac{1}{2})}+\frac{5}{4}$

$y = \frac{1}{8x-4}+\frac{5}{4}$

**DarthBlood** · September 10th 2009, 03:39 PM

to find inverse, i'm pretty sure you have to exchange the Y with the X, and then solve for Y.

original equation:
$y=\frac{2x+3}{4x-5}$

exchange variables:
$x=\frac{2y+3}{4y-5}$

solve for y:
$x(4y-5)=2y+3$ <----move the denominator on the right to the left
$4xy-5x=2y+3$ <----factor
$4xy-2y=5x+3$ <----move all the y's to one side and the single x to the other
$2y(2x-1)=5x+3$ <----factor out the 2y from the left side
$2y=\frac{5x+3}{2x-1}$ <----get the 2y by itself
$y=\frac{5x+3}{4x-2}$ <----get the y by itself

EDIT: I think my way is easier, pickslides...

**pickslides** · September 10th 2009, 03:47 PM

yes, i like it.

I was only making note of the previous poster's reference to long division.

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