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# What is the inverse of y=2x+3

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What is the inverse of y=2x + 3?  Please  show all steps.

gibsonj338  May 15, 2015

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+26399
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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
Sort:

#1
+26399
+10

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

.

Alan  May 15, 2015
#2
+1794
+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
+26399
+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
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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
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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
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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
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Consider the function y = x2

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 = y2 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!

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Alan  May 15, 2015
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Mmm Thanks Alan

I am just thinking here.

$$\\y=x^2\\ \pm\sqrt{y}=x\\ x=\pm\sqrt{y}\\ No inverse even with domain restricitions because all x in the domain map to 2 y values$$

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

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