mathmath333 one year ago solve

1. mathmath333

\large \color{black}{\begin{align} |x-3|+|x+4|+|x-1|=14\hspace{.33em}\\~\\ \end{align}}

2. alekos

I get x = 14/3 and -14/3

3. alekos

Is this a test?

4. mathmath333

how get that

5. mathmath333

this is no test

6. alekos

well first we take all the positive values within the modulus (x-3) + (x+4) + (x-1) =14 and solve for x

7. anonymous

best way to do this problem is partitioning the x's

8. alekos

then we take all the negative values -(x-3) - (x+4) - (x-1) =14 and solve for x

9. mathmath333

there will be total of 6 cases but that is a long way

10. alekos

if you plot the graph you'll see that there are only two values

11. mathmath333

that can be if 4 values are not valid

12. alekos

that's right. the 4 values are not valid

13. alekos

just plot the graph and you'll see

14. mathmath333

i look for a pen paper method

15. alekos

that's fine, you can do it that way. But Desmos is easier

16. ParthKohli

Hey, yes, just do it casewise.

17. mathmath333

u mean take 6 cases ?

18. ParthKohli

You'd need five, so yes.

19. mathmath333

how is it 5 cases

20. anonymous

It's 4 cases

21. ParthKohli

Oh, four.

22. ParthKohli

|dw:1434037874987:dw|

23. ParthKohli

For some reason, I thought there were four absolute values. Oops.

24. anonymous

|dw:1434037949803:dw|

25. ParthKohli

Yes.

26. alekos

I'll say it again. Plot the graph and you will all see that there are only two values

27. ParthKohli

If we're gonna take the graphing route, might as well just solve the equation directly.

28. alekos

I've done it both ways. algebraically and graphically

29. mathmath333

how u did algebriacally

30. alekos

I showed you earlier on

31. ParthKohli

Alekos, you covered only two of the four cases...

32. alekos

the other cases are not valid

33. ParthKohli

|dw:1434038294034:dw|

34. ParthKohli

|dw:1434038341589:dw|

35. ParthKohli

|dw:1434038401841:dw|

36. ParthKohli

|dw:1434038477315:dw|

37. ParthKohli

|dw:1434038527838:dw|

38. ParthKohli

|dw:1434038567104:dw|

39. ParthKohli

40. mathmath333

here how would u identify useful and nonuseful regions

41. ParthKohli

First thing: you have to check for all intervals. There is no shortcut. Second thing: nonuseful intervals are those which don't contain the answer obtained in that interval's case. If you're considering an interval, say, [1,10] and the answer comes out to be 20, then you don't accept 20 as an answer.

42. mathmath333

ok so i have to check x by putting x in original equation,ok

43. ParthKohli

Did you see those drawings?

44. mathmath333

yes

45. ParthKohli

In the interval $$\left[-4, 1\right]$$:$|x-3| = -(x-3)$$|x+4| = x+4$$|x-1 | = - (x-1)$

46. ParthKohli

^ do you see why those are true?

47. mathmath333

because $$|x|=x,\ \ if\ \ x>0\\~\\and\\~\\|x|=-x\ \ if\ \ x<0$$

48. ParthKohli

Exactly. So you can solve it now, yes?

49. mathmath333

where did u learnt this method from

50. ParthKohli

This is a standard method.

51. ParthKohli

If you go to coaching, they do teach it there.

52. alekos

Tedious and unnecessary. If you follow what I've done it will give you the most efficient answer

53. ParthKohli

What have you done?

54. mathmath333

@alekos u didnt defined why u choosed the two particular cases

55. mathmath333

*chose

56. ParthKohli

Some problems have neat tricks involved in them.$|a| + |b| = |a + b |$implies that $$a$$ and $$b$$ have the same sign. Similarly,$|a| + |b| = |a-b|$implies that $$a$$ and $$b$$ have the opposite sign.

57. ParthKohli

So for example if you want to solve$|x+3| + |1 - x| = 4$It reduces to the inequality$(x+3)(1-x) \ge 0$

58. alekos

That's right I didn't define it. But if you sketch the graph you'll see that there is no need. It's self apparent

59. ParthKohli

How would you sketch the graph?

60. alekos

You can use pen and paper or Desmos

61. ParthKohli

I'll take the pen-and-paper approach. How did you sketch the graph here?

62. ParthKohli

The point is... the only way you can sketch a graph is casewise. So you first consider cases, find an expression for each straight-line, and then draw straight-lines all around the graph (which is a hard task) keeping the scale in mind, then you draw y = 14 and see where it intersects the graph. Looks like it intersects in two places: the first interval and the last interval. So you go to the first and last intervals and solve their respective equations.

63. ParthKohli

But if you find the linear expression, why not just equate it to 14 and get done with the problem instead of sketching all of them?

64. alekos

I agree. Pen and paper does take a little bit of time. But, we are all using computers nowadays so we can graph using technology and come to a solution very quickly.

65. ParthKohli

The only reason he's been given that problem is to be able to solve it himself in spite of the existence of computers.

66. ParthKohli

I'm pretty sure that he knows computers exist too.

67. alekos

That's fine. As I said it can be solved using pen and paper . Just takes a little longer. I know what I would rather use.

68. ParthKohli

Nah, takes only a couple of minutes.

69. ParthKohli

There are four cases and each case should take you around 20-25 seconds.

70. alekos

OK, no problem. Either method is valid.