Inverse Function: Solving for y in 2(x-3)/y = 3

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The discussion focuses on solving the equation 2(x-3)/y = 3 for y. Participants clarify that the correct approach involves manipulating the equation to isolate y, emphasizing the importance of maintaining equality on both sides. A key point is that mistakes were made in canceling terms improperly and misunderstanding the distribution of variables. The final correct form of y is expressed as y = 3/(2x) + 3, highlighting the need to properly handle fractions and algebraic properties. The conversation underscores the significance of following algebraic rules to arrive at the correct solution.
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Homework Statement




y = 3
___
2(x-3)

Homework Equations





The Attempt at a Solution



Ive tried several times, I can't get the steps down. They say use reverse bedmass, and I don't understand.

Here's what i did.

y= 3/2(x-3)

x= 3/2(y-3)

x= 3/2y-6
x+6=3/2y

x+6
___ = 3
2

Which is already wrong, as you can see. Can someone please tell me the step so i can stop being stuck on this for 2 hours? =)
 
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"x= 3/2y-6"

You were correct up to here. If it helps, write parenthesis in the denominator.

x=\frac{3}{(2y-6)}

What you did was just canceled out the 6 in the denominator which you cannot do. Our objective is to get y by itself. Do that by multiplying 2y-6 to both sides. That should get rid of the fraction and leave you with.

x(2y-6)=3

Try getting y by itself from here.
 
Ok...

x = 3 / (2y-6)

x(2y-6) 3
_______ = _____
(2y-6) (2y-6)

2xy-6x 3
______=_____
2y-6 2y-6

Pretty sure I am already wrong. .. .
 
When I mean multiply both sides you don't do it like this:

\frac{2y-6}{2y-6} x=\frac{3}{2y-6}\frac{2y-6}{2y-6}

\uparrow\uparrow\uparrow\uparrow That is a no no.

You do it like this:

x(\frac{2y-6}{1})=\frac{3}{2y-6}(\frac{2y-6}{1})

Notice how the (2y-6) in the right side would cancel out. That would leave us with this.

x(2y-6)=3

Remember this: If you do anything to one side, you MUST do it to the other side of the equation.
 
So... the x still factors into 2y-6 on both sides..? your eq would give 2xy-6x/x on the left side..
 
What do you mean "factors on both sides)? My equation would not leave you with a fraction (though the answer would be a fraction). Try distributing the x term to (2y-6). Then just solve for y.

*Update: I see what you did. No that is wrong. Remember your algebraic properties:

a*\frac{b}{c}=\frac{ab}{c}
 
So, i did it like this..


2x(y-3) = 3

2x*y -6x =3

2x*y = 6x+3

y= 6x+3 /2x

y= 3x +3/2x

The answer is 2/3x +3 , however. Why is the x on the bottom?
 
Starting from here:
"2xy = 6x+3"

We would divide 2x to get y by itself.

\frac{2xy}{2x}=\frac{6x+3}{2x}

I see you split the fraction on the right side to divide 6x/2x, however you forgot to cancel out the x. However you cannot divide 3/(2x), so that is why 2x is in the denominator. The answer would look like this.

y=\frac{3}{2x}+3


I do not see how the right answer is: "2/3x +3" (assuming this is (2/(3x))+3) the 3 is supposed to be in the numerator and 2x is supposed to be in the denominator.
 
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