Inverse of a function

D

dmurdoch

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Given a function [MATH]y = (x + 10) / 3x[/MATH] can someone please show me the steps to invert this to the formula for [MATH]x =[/MATH]?

David
 
You must try something. Give us a clue where you are.

Multiple both sides by something that is not zero. Surely, that will help.
 
I really don't like fractions so I would multiply both sides by that denominator. If you multiply both sides by 3x you get 3xy= x+ 10. Can you solve that linear equation for x?
 
dmurdoch said:
Given a function [MATH]y = (x + 10) / 3x[/MATH] can someone please show me the steps to invert this to the formula for [MATH]x =[/MATH]?

David
Click to expand...
Is it:

\(\displaystyle y = \frac{x+10}{3}*x\) ...............................This is what you wrote - or -

\(\displaystyle y = \frac{x+10}{3*x}\)
 
Thanks everyone for your help. Yes, I meant solve the following for [MATH]x[/MATH]
[MATH]y=(x+10)/(3x)[/MATH]
Yes, I had missed the second set of parentheses, and fyi the correct answer is:

[MATH]x=10/(3y+1)[/MATH]
 
dmurdoch said:
Thanks everyone for your help. Yes, I meant solve the following for [MATH]x[/MATH]
[MATH]y=(x+10)/(3x)[/MATH]
Yes, I had missed the second set of parentheses, and fyi the correct answer is:

[MATH]x=10/(3y+1)[/MATH]
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.
 
Okay, you have verified that the original equation is \(\displaystyle y= \frac{x+ 10}{3x}\). As I suggested before, multiplying both sides by 3x gives \(\displaystyle 3xy= x+ 10\). Can you solve that equation for x? I suggest subtracting x from both sides.
 
dmurdoch said:
Thanks everyone for your help. Yes, I meant solve the following for [MATH]x[/MATH]
[MATH]y=(x+10)/(3x)[/MATH]
Yes, I had missed the second set of parentheses, and fyi the correct answer is:

[MATH]x=10/(3y+1)[/MATH]
Click to expand...
Are you sure that is the answer given? Please confirm.
 
dmurdoch said:
Thanks everyone for your help. Yes, I meant solve the following for [MATH]x[/MATH]
[MATH]y=(x+10)/(3x)[/MATH]
Yes, I had missed the second set of parentheses, and fyi the correct answer is:

[MATH]x=10/(3y+1)[/MATH]
Click to expand...
fyi, that is not "the correct answer"! In the original equation, if [math]x= 1[/math] then [math]y= \frac{1+ 10}{3(1)}= \frac{10}{3}[/math]. In your equation, if [math]y= \frac{10}{3}[/math] then [math]x= \frac{10}{3\left(\frac{10}{3}\right)+ 1}= \frac{10}{11}[/MATH] not [math]1[/math].

Multiplying both sides of y= (x+ 10)/(3x) gives 3xy= x+ 10. Subtract x from both sides: 3xy- x= (3y- 1)x= 10. You, apparently, added x to the left side while subtracting it from the right.
 
HallsofIvy said:
fyi, that is not "the correct answer"! In the original equation, if [math]x= 1[/math] then [math]y= \frac{1+ 10}{3(1)}= \frac{10}{3}[/math]. In your equation, if [math]y= \frac{10}{3}[/math] then [math]x= \frac{10}{3\left(\frac{10}{3}\right)+ 1}= \frac{10}{11}[/MATH] not [math]1[/math].

Multiplying both sides of y= (x+ 10)/(3x) gives 3xy= x+ 10. Subtract x from both sides: 3xy- x= (3y- 1)x= 10. You, apparently, added x to the left side while subtracting it from the right.
Click to expand...

Yes, you're correct, I broke algebra's golden rule of doing the same to both sides. Correct answer is:

[math]x= \frac{10}{3y - 1}[/math]
Thanks for your help.
 
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