Which of the following expressions is the inverse of the function y = x - 2 / 3? y = 3x + 2 y = 2x + 3 y = x - 3 / by 2 y = x + 2 / by 3

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Which of the following expressions is the inverse of the function y = x - 2 / 3? y = 3x + 2 y = 2x + 3 y = x - 3 / by 2 y = x + 2 / by 3

Mathematics
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Is the original \[\Large y = x - \frac{2}{3}\] or is the original \[\Large y = \frac{x-2}{3}\]
The second one

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Other answers:

lack of sleep .. T_T
\[\Large y = \frac{x-2}{3}\] swap x and y \[\Large x = \frac{y-2}{3}\] now solve for y
hmmm I would get rid of the denominator by multiplying 3 all the way first and then swap x and y.
either way works
I'm not sure what to start with when solving for y.
\[\Large 3y = (3)\frac{x-2}{3} \rightarrow 3y=x-2\]
then swap x and y . In other words switch places ...
\[\large 3x=y-2\]
and then solve for y...this would be much easier because I got rid of the fraction first by multiplying the entire equation by 3
so what do we need to do to have y by itself
3x=y-2 +2 +2 5x=y
3x and 2 are NOT like terms
ummmm 3x+2 is not combinable
you cannot say 3x+2 turns into 5x or 5
Oh. Sorry.
because 3x has a number and variable 2 is just a number... it can't be combined
Ok... how do I isolate y.
y has to be by itself... \[\large 3x=y-2 \] so what do we need to do to have y by itself... y is destined to be alone xD
unheard, when you added 2 to both sides, you were on the right path
3x = y - 2 +2 +2 3x + 2 = y y = 3x + 2
:( ! why jim why
you just leave 3x+2 as it is since you cannot combine those terms
Oh, thank you Jim.
she already had it usukidoll
Thank you both.
jim's version would have required fractions to be split up \[\Large x = \frac{y-2}{3} \rightarrow x = \frac{y}{3}-\frac{2}{3}\] \[\Large x + \frac{2}{3} = \frac{y}{3}\] and then multiply 3 throughout the equation... too long...but another way to check \[ \Large 3x+2=y\]

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