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Mathematics 15 Online
OpenStudy (anonymous):

PLEASE HELP! What is the inverse to the given relation? y = 3x + 12 I have NO idea how to do this. Can someone please set this up for me? I can solve with help! please!

OpenStudy (dtan5457):

y=3x+12 Switch x and y.

OpenStudy (anonymous):

What? x=3y+12?

OpenStudy (dtan5457):

Yeah. Switch the variables, and basically resolve for y. Make sense?

OpenStudy (dtan5457):

@PrincessTianaB

OpenStudy (anonymous):

I don't even know how to do that... I am 18, and was in the hospital for an illness I had for almost 3 years while I should've been in school.. I am just learning how to do all of this. I have no idea.

OpenStudy (dtan5457):

I'm so sorry to hear that, it's ok. We'll start from scratch.

OpenStudy (anonymous):

Oh no! Nothing to be sorry about, I was just explaining that I literally know none of this! But okay, great! That's what I need!

OpenStudy (dtan5457):

So, you did switch the variables correctly. x=3y+12 To solve for y again, like in any normal solve for variable equation. You must isolate the term of y. 3y=x-12 Here, you subtracted the 12.

OpenStudy (dtan5457):

Making sense so far?

OpenStudy (anonymous):

Yes! So that made the 12, negative?

OpenStudy (dtan5457):

x=3y+12 -12 -12 x-12=3y or 3y=x-12

OpenStudy (dtan5457):

good so far?

OpenStudy (anonymous):

Yes! Made much sense!

OpenStudy (dtan5457):

Good! Now, there is one more step. 3y=x-12 Y has be alone, and you can do that by dividing by 3.

OpenStudy (dtan5457):

Wanna give me your final inversed equation?

OpenStudy (anonymous):

So y=x-4 ?

OpenStudy (dtan5457):

Hmm, no. If the equation was 3y=-12 y=-4 However there is a x here.

OpenStudy (dtan5457):

\[y=\frac{ x-12 }{ 3 }\]

OpenStudy (dtan5457):

You literally just put it as a fraction ^^

OpenStudy (dtan5457):

That is the final inverse for your function :)

OpenStudy (anonymous):

Wait, that's literally it? I was trying to think way to deep into that! Lol. Thank you so much!

OpenStudy (dtan5457):

No problem. Feel free to fan me so you can see when I'm online. Your welcome to tag me into any more questions :)

OpenStudy (anonymous):

I fanned you and actually am about to post another question, my last one for the night if you don't mind helping! Ill tag you in it! (:

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