Question

Find the inverse of the function
\(\huge\frac{4x-3}{2x+1}\)

Answer 1

Once again, I need someone to check out what I've done.

So first, I've multiplied both sides by 2x+1
As a result, I got this:

x(2y+1)=4y-3

Answer 2

Then, I distributed the x into the parentheses

2yx+x=4y-3

Answer 3

Then, I subtracted x and 4y from both sides, which left me with this:

2yx-4y=-x-3

Answer 4

from there, I condensed 2yx-4y, so the equation was left as follows:

2y(x-2)

And the expression as a whole looked like this:

2y(x-2)=-x-3

Answer 5

from here, I divided both sides by x-2. The point here is to isolate the y. Problem is, there is a two right in front of the y. Does this mean that I have to divide again by 2 in order to fully isolate the y?

Answer 6

Looks good to me! you should get (x+3)/2(2-x)

Answer 7

Wait. So it would look like

\(\huge\frac{x+3}{2(2-x)}\)

Answer 8

Yeah!

Answer 9

I don't quite understand how you would get from this:

\(\huge 2y(x-2)=-x-3\)


to this: \(\huge\frac{x+3}{2(2-x)}\)

Answer 10

The last clear step to me is dividing both sides by x-2

that would give you

\(\huge 2y=\frac{-x-3}{x-2}\)

Answer 11

\(\huge \color{red}2y=\frac{-x-3}{x-2}\)

This 2 is eluding me so much and I have no clue why.

Answer 12

multiply - sign on both sides !

Answer 13

negative sign

Answer 14

Why do you doubt?

Answer 15

I'm not sure xD

it's just when I see this:

\(\huge 2y=\frac{-x-3}{x-2}\)

I think that the whole thing is going to end up looking like this

\(\huge y=\frac{\frac{-x-3}{x-2}}{2}\)

Answer 16

And that kind of seems super messy to me. I have no idea how to make it appear clean and neat like ankit made it seem to be. I'm sure he's right, i just don't know how to get there

Answer 17

@Jamierox4ev3r dividing both sides by 2 is the same as multiplying both sides by 1/2

\[\Large 2x = 4\]

\[\Large {\color{red}{\frac{1}{2}}}*2x = {\color{red}{\frac{1}{2}}}*4\]

Answer 18

\(\dfrac{\dfrac{a}{b}}{2} = \dfrac{a}{b}\cdot\dfrac{1}{2} = \dfrac{a}{2b}\)

\(-x+2 = -(x-2) = 2-x\)

No mysteries in there.

Answer 19

Woah I did not see it that way. That makes sense, so basically it's like doing this:

\(\huge\frac{-x-3}{x-2}\times\frac{1}{2}\)

Answer 20

still don't get how the numerator becomes positive

Answer 21

Yeah! now multiply both numerator and denominator by -1

Answer 22

\(-\dfrac{a}{b} = -\dfrac{-a}{-b} = \dfrac{-a}{b} = \dfrac{a}{-b}\)

Look around. It's there.

Answer 23

I apologize, but I still don't see why the numerator and denominator should be multiplied by -1. If i multiply -x-3 / x-2 by 1/2, then wouldn't the final answer be

\(\huge\frac{-x-3}{2(x-2)}\)