#1**+10 **

Best Answer

First get x on its own:

Subtract 3 from each side

y - 3 = 2x

Divide through by 2

y/2 - 3/2 = x

or x = y/2 - 3/2

Now simply switch labels: x↔y

y = x/2 - 3/2

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Alan
May 15, 2015

#2**+5 **

y=2x+3

First switch x and y around to get x=2y+3

Now solve for y

x-3=2y

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

gibsonj338
May 15, 2015

#3**+5 **

If you look carefully gibsonj338 you will see that the two results are the same - they are just expressed slightly differently.

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Alan
May 15, 2015

#4**+5 **

**Please correct me if I am wrong Alan but**

I think that you have always made a big point of telling me that I am not allowed to swap the x and y over and rearrange afterwards.

I have never understood why but i think it is because you have to be very careful that the inverse is still a function.

That is, in the original function every x value must map to a unique y value (vertical line test)

and if you are going to get an inverse function then every original y value must map to a unique original x value.

So if you take an inverse of a function you may have to limit the domain of the new inverse function or else it may not be a function at all.

Most times I do not think that there is any problem with doing it this way but you have to be careful.

**Did I get this it right Alan ?**

Melody
May 15, 2015

#5**+5 **

Yes, but in this case of course it is a simple linear function defined over all the real numbers for both x and y, so it works out ok either way.

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Alan
May 15, 2015

#6**0 **

Thanks Alan,

I thought it would work anyway so long as I am careful about the restrictions.

It still confuses me. I don't understand why you say I can't do it.

Melody
May 15, 2015

#7**+5 **

Consider the function y = x^{2}

This doesn't have an inverse (in the real number domain) because there isn't a unique value of x for a given y, but by writing x = y^{2} before doing anything else you imply there is (to qualify as a function an expression must map an input to a *unique* output).

However, as you say, the important thing is being careful about the restrictions. As long as you keep this in mind I will not say you can't do it Melody!

.

Alan
May 15, 2015

#8**0 **

Mmm **Thanks Alan**

I am just thinking here.

**Your way**

$$\\y=x^2\\

\pm\sqrt{y}=x\\

x=\pm\sqrt{y}\\

$No inverse even with domain restricitions because all x in the domain map to 2 y values$$

**My usual way**

**$$\\y=x^2\\ x=y^2\\ y=\pm\sqrt{x}\\ $But this is not a function so there is no inverse.$$$**

**Ok I can see that your way is a little tidier :)**

Melody
May 15, 2015