## vera_ewing one year ago How do you solve this?

1. vera_ewing

2. jim_thompson5910

They provided the definition of $$\Large \boxed{x}$$ $\Large \boxed{x} = \frac{x+3}{x-1}$ subtract 1 from both sides to get $\Large \boxed{x}-1 = \frac{x+3}{x-1}-1$ now simplify the right hand side

3. vera_ewing

Wait why do I subtract 1 from both sides?!

4. jim_thompson5910

because they want to know what $$\large \boxed{x}-1$$ is equivalent to

5. vera_ewing

So we just put a -1 on both sides? That makes no sense...

6. jim_thompson5910

you start off with $$\large \boxed{x}$$ and do that operation to get $$\large \boxed{x}-1$$

7. vera_ewing

But there is no -1 originally. Where did it come from?

8. jim_thompson5910

maybe it might make more sense if y = 2x was given to you if that's the case, then you would subtract 1 from both sides to get y-1 = 2x-1

9. jim_thompson5910

I'm doing that operation of subtract 1 from both sides to turn $$\large \boxed{x}$$ into $$\large \boxed{x}-1$$ basically it's like turning y into y-1

10. vera_ewing

I'm so sorry, but why are we allowed to subtract 1 from both sides? There is no 1 on either side of the equation....

11. jim_thompson5910

This is what the full step by step picture looks like $\Large \boxed{x} = \frac{x+3}{x-1}$ $\Large \boxed{x}-1 = \frac{x+3}{x-1}-1$ $\Large \boxed{x}-1 = \frac{x+3}{x-1}-1*\frac{x-1}{x-1}$ $\Large \boxed{x}-1 = \frac{x+3}{x-1}-\frac{x-1}{x-1}$ $\Large \boxed{x}-1 = \frac{x+3-(x-1)}{x-1}$ $\Large \boxed{x}-1 = \frac{x+3-x+1}{x-1}$ $\Large \boxed{x}-1 = \frac{4}{x-1}$

12. jim_thompson5910

I'm using the subtraction property of equality if a = b, then a-c = b-c

13. vera_ewing

Ohh okay. And why is there a box around the x on the left?

14. jim_thompson5910

because that's how they set up the notation in the problem

15. vera_ewing

Does it mean anything?

16. jim_thompson5910

it's just their fancy way of defining a function

17. jim_thompson5910

they could have easily said "let y = ..." or "let f(x) = ..."

18. jim_thompson5910

why they chose a box, who knows

19. vera_ewing

Okay. Can you please give me a similar problem to solve?

20. vera_ewing

I need to be able to do these on my own.

21. jim_thompson5910

ok what is $\Large \boxed{x} +2x$ equivalent to? (same function from the previous problem)

22. vera_ewing

I'm not sure :(

23. jim_thompson5910

replace the box x with what it is equal to then simplify

24. vera_ewing

4/x-1 ?

25. jim_thompson5910

no

26. jim_thompson5910

box x was equal to (x+3)/(x-1)

27. jim_thompson5910

so saying $\Large \boxed{x}+2x$is the same as saying (x+3)/(x-1) + 2x. Simplify from here

28. vera_ewing

What would the first step be?

29. jim_thompson5910

you need to get the denominators the same

30. jim_thompson5910

2x = 2x/1 has a denominator of 1

31. jim_thompson5910

how can you transform that 1 into (x-1)

32. vera_ewing

So (x+3)/(x-1) + 2x/1 ?

33. jim_thompson5910

how can you get 2x/1 to have a denominator of x-1

34. vera_ewing

Multiply 1 on both sides?

35. jim_thompson5910

why not multiply top and bottom by (x-1) ?

36. jim_thompson5910

why not multiply top and bottom of 2x/1 by (x-1) ?

37. jim_thompson5910

you might be mixing up + and =

38. vera_ewing

Show me what that step would look like please.

39. jim_thompson5910

|dw:1434499487201:dw|

40. jim_thompson5910

|dw:1434499531547:dw|

41. jim_thompson5910

|dw:1434499550135:dw|

42. jim_thompson5910

|dw:1434499580374:dw|

43. jim_thompson5910

I'll let you finish

44. vera_ewing

Hold on, this all makes so much sense, but I just don't understand what property lets us multiply (x-1) on the top and bottom.

45. jim_thompson5910

I'm essentially multiplying by 1 (x-1)/(x-1) = 1

46. jim_thompson5910

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