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

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- Sep 8th 2009, 07:27 PMWashingtonStudentFind the inverse of F(x)?
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 - Sep 8th 2009, 07:36 PMRapha
- Sep 10th 2009, 03:11 PMzodiacbrave
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 - Sep 10th 2009, 03:33 PMpickslides
I found

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

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

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

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

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

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

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

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

$\displaystyle y = \frac{1}{8x-4}+\frac{5}{4}$ - Sep 10th 2009, 03:39 PMDarthBlood
to find inverse, i'm pretty sure you have to exchange the Y with the X, and then solve for Y.

original equation:

$\displaystyle y=\frac{2x+3}{4x-5}$

exchange variables:

$\displaystyle x=\frac{2y+3}{4y-5}$

solve for y:

$\displaystyle x(4y-5)=2y+3$ <----move the denominator on the right to the left

$\displaystyle 4xy-5x=2y+3$ <----factor

$\displaystyle 4xy-2y=5x+3$ <----move all the y's to one side and the single x to the other

$\displaystyle 2y(2x-1)=5x+3$ <----factor out the 2y from the left side

$\displaystyle 2y=\frac{5x+3}{2x-1}$ <----get the 2y by itself

$\displaystyle y=\frac{5x+3}{4x-2}$ <----get the y by itself

EDIT: I think my way is easier, pickslides... - Sep 10th 2009, 03:47 PMpickslides
yes, i like it.

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