## Jamierox4ev3r one year ago Find the inverse of the function $$\huge\frac{4x-3}{2x+1}$$

1. Jamierox4ev3r

Once again, I need someone to check out what I've done. So first, I've multiplied both sides by 2x+1 As a result, I got this: x(2y+1)=4y-3

2. Jamierox4ev3r

Then, I distributed the x into the parentheses 2yx+x=4y-3

3. Jamierox4ev3r

Then, I subtracted x and 4y from both sides, which left me with this: 2yx-4y=-x-3

4. Jamierox4ev3r

from there, I condensed 2yx-4y, so the equation was left as follows: 2y(x-2) And the expression as a whole looked like this: 2y(x-2)=-x-3

5. Jamierox4ev3r

from here, I divided both sides by x-2. The point here is to isolate the y. Problem is, there is a two right in front of the y. Does this mean that I have to divide again by 2 in order to fully isolate the y?

6. ankit042

Looks good to me! you should get (x+3)/2(2-x)

7. Jamierox4ev3r

Wait. So it would look like $$\huge\frac{x+3}{2(2-x)}$$

8. ankit042

Yeah!

9. Jamierox4ev3r

I don't quite understand how you would get from this: $$\huge 2y(x-2)=-x-3$$ to this: $$\huge\frac{x+3}{2(2-x)}$$

10. Jamierox4ev3r

The last clear step to me is dividing both sides by x-2 that would give you $$\huge 2y=\frac{-x-3}{x-2}$$

11. Jamierox4ev3r

$$\huge \color{red}2y=\frac{-x-3}{x-2}$$ This 2 is eluding me so much and I have no clue why.

12. CGGURUMANJUNATH

multiply - sign on both sides !

13. CGGURUMANJUNATH

negative sign

14. tkhunny

Why do you doubt?

15. Jamierox4ev3r

I'm not sure xD it's just when I see this: $$\huge 2y=\frac{-x-3}{x-2}$$ I think that the whole thing is going to end up looking like this $$\huge y=\frac{\frac{-x-3}{x-2}}{2}$$

16. Jamierox4ev3r

And that kind of seems super messy to me. I have no idea how to make it appear clean and neat like ankit made it seem to be. I'm sure he's right, i just don't know how to get there

17. jim_thompson5910

@Jamierox4ev3r dividing both sides by 2 is the same as multiplying both sides by 1/2 $\Large 2x = 4$ $\Large {\color{red}{\frac{1}{2}}}*2x = {\color{red}{\frac{1}{2}}}*4$

18. tkhunny

$$\dfrac{\dfrac{a}{b}}{2} = \dfrac{a}{b}\cdot\dfrac{1}{2} = \dfrac{a}{2b}$$ $$-x+2 = -(x-2) = 2-x$$ No mysteries in there.

19. Jamierox4ev3r

Woah I did not see it that way. That makes sense, so basically it's like doing this: $$\huge\frac{-x-3}{x-2}\times\frac{1}{2}$$

20. Jamierox4ev3r

still don't get how the numerator becomes positive

21. ankit042

Yeah! now multiply both numerator and denominator by -1

22. tkhunny

$$-\dfrac{a}{b} = -\dfrac{-a}{-b} = \dfrac{-a}{b} = \dfrac{a}{-b}$$ Look around. It's there.

23. Jamierox4ev3r

I apologize, but I still don't see why the numerator and denominator should be multiplied by -1. If i multiply -x-3 / x-2 by 1/2, then wouldn't the final answer be $$\huge\frac{-x-3}{2(x-2)}$$

24. mathstudent55

|dw:1440997453887:dw|

25. mathstudent55

Multiply your fraction by 1, but write 1 as (-1)/(-1)

26. Jamierox4ev3r

Why do I need to multiply by -1? I'm still not seeing where that came from

27. mathstudent55

You are not multiplying by -1. You are multiplying by 1, but 1 is written as -1/-1 just to change the form of the fraction.

28. Jamierox4ev3r

So in effect, are you saying that $$\huge\frac{x+3}{2(2-x)}$$ and $$\huge\frac{-x-3}{2(x-2)}$$ are the same? I can see that the signs have all changed, but what's the point of doing that?

29. mathstudent55

Factor -1 out of the numerator: |dw:1440997706075:dw|

30. mathstudent55

Now factor -1 out of the denominator. |dw:1440997764411:dw|

31. mathstudent55

-1/-1 cancel out. -x + 2 = x - 2

32. mathstudent55

|dw:1440997831190:dw|

33. Jamierox4ev3r

okay, I see how you get there now. But my question to you now is would it technically be correct to leave the problem as -x-3/2(x-2) Basically, I'm still not sure what the purpose of factoring out the -1 from both numerator/denominator is.

34. mathstudent55

Technically and actually you are correct. Mathematicians are lazy and want to write the least amount possible.

35. Jamierox4ev3r

Ohh okay. I was just very confused by why I had to factor out the -1. In reality, it's just to get this in the simplest form possible. So not only did I find out what I needed to do with the 2y (all I had to do was multiply by 1/2), but I also learned the mathematicians are lazy and how I should deal with that. My goodness, thank you so much! I feel like I have a really clear understanding now

36. mathstudent55

|dw:1440998072399:dw|

37. mathstudent55

It also looks nicer to see $$\dfrac{x + 3}{2(2 - x)}$$ then $$\dfrac{-x - 3}{2(x - 2)}$$

38. Jamierox4ev3r

Yes I see. Now I need to be on the lookout for an abundance of signs. It's kind of silly, but yet it appears to be necessary. In any case, it is more aesthetically pleasing to see less things going on in a function. I reiterate, thank you so much for clearing that up.

39. mathstudent55

You are very welcome.

40. mathstudent55

Why write 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 when you can write 9 * 5? Why write 8 * 8 * 8 * 8 * 8 * 8 when you can write 8^6? Why write $$\dfrac{-x - 3}{2(x - 2)}$$ when you can write $$\dfrac{x + 3}{2(2 - x) }$$ ?

41. Jamierox4ev3r

Very true. The reason why things like exponents and radicals were created for the purpose of simplifying things as much as possible. Also to allow for as much accuracy as possible. Either way, you make an excellent point and I completely understand now. Gracias!

42. tkhunny

Unfortunately, students spend far to much time trying to match some arbitrary form of the answer. More unfortunately, many teachers accept only the arbitrary form. Just learn your algebra. Don't waste time on silliness.

43. mathstudent55

@tkhunny I agree that matching an arbitrary form of the solution accomplishes nothing, but if a student solves a problem, and his solution does not match exactly the solution in the book, the student may think it is incorrect. By manipulating a perfectly acceptable solution to make it look like the book's solution the student confirms that he did solve the problem correctly.

44. tkhunny

...and wastes a ton of time, and fails to enhance personal confidence, and drops substantially in learning and increases substantially in frustration if the book is wrong. I just don't like it. Personal Opinion: Use the clarity of your notation, the directness of your process, the confidence of your manipulation to "verify" your answer. Don't use the back of the book. Disclaimer: Online checking is very unforgiving. This may be more frustrating than the back of the book. Personal Opinion: Absolutely learn to manipulate algebraic expressions, but don't get too hung up on it. One day, you will have to come to terms with the fact that these are the same answer: $$\int\;something = \cos^{2}(x) + C$$ $$\int\;something = \sin^{2}(x) + C$$ :-)