mathmath333
  • mathmath333
solve
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} |x-3|+|x+4|+|x-1|=14\hspace{.33em}\\~\\ \end{align}}\)
alekos
  • alekos
I get x = 14/3 and -14/3
alekos
  • alekos
Is this a test?

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mathmath333
  • mathmath333
how get that
mathmath333
  • mathmath333
this is no test
alekos
  • alekos
well first we take all the positive values within the modulus (x-3) + (x+4) + (x-1) =14 and solve for x
anonymous
  • anonymous
best way to do this problem is partitioning the x's
alekos
  • alekos
then we take all the negative values -(x-3) - (x+4) - (x-1) =14 and solve for x
mathmath333
  • mathmath333
there will be total of 6 cases but that is a long way
alekos
  • alekos
if you plot the graph you'll see that there are only two values
mathmath333
  • mathmath333
that can be if 4 values are not valid
alekos
  • alekos
that's right. the 4 values are not valid
alekos
  • alekos
just plot the graph and you'll see
mathmath333
  • mathmath333
i look for a pen paper method
alekos
  • alekos
that's fine, you can do it that way. But Desmos is easier
ParthKohli
  • ParthKohli
Hey, yes, just do it casewise.
mathmath333
  • mathmath333
u mean take 6 cases ?
ParthKohli
  • ParthKohli
You'd need five, so yes.
mathmath333
  • mathmath333
how is it 5 cases
anonymous
  • anonymous
It's 4 cases
ParthKohli
  • ParthKohli
Oh, four.
ParthKohli
  • ParthKohli
|dw:1434037874987:dw|
ParthKohli
  • ParthKohli
For some reason, I thought there were four absolute values. Oops.
anonymous
  • anonymous
|dw:1434037949803:dw|
ParthKohli
  • ParthKohli
Yes.
alekos
  • alekos
I'll say it again. Plot the graph and you will all see that there are only two values
ParthKohli
  • ParthKohli
If we're gonna take the graphing route, might as well just solve the equation directly.
alekos
  • alekos
I've done it both ways. algebraically and graphically
mathmath333
  • mathmath333
how u did algebriacally
alekos
  • alekos
I showed you earlier on
ParthKohli
  • ParthKohli
Alekos, you covered only two of the four cases...
alekos
  • alekos
the other cases are not valid
ParthKohli
  • ParthKohli
|dw:1434038294034:dw|
ParthKohli
  • ParthKohli
|dw:1434038341589:dw|
ParthKohli
  • ParthKohli
|dw:1434038401841:dw|
ParthKohli
  • ParthKohli
|dw:1434038477315:dw|
ParthKohli
  • ParthKohli
|dw:1434038527838:dw|
ParthKohli
  • ParthKohli
|dw:1434038567104:dw|
ParthKohli
  • ParthKohli
Discard an answer if it doesn't fall in the given interval.
mathmath333
  • mathmath333
here how would u identify useful and nonuseful regions
ParthKohli
  • 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.
mathmath333
  • mathmath333
ok so i have to check x by putting x in original equation,ok
ParthKohli
  • ParthKohli
Did you see those drawings?
mathmath333
  • mathmath333
yes
ParthKohli
  • ParthKohli
In the interval \(\left[-4, 1\right]\):\[|x-3| = -(x-3)\]\[|x+4| = x+4\]\[|x-1 | = - (x-1) \]
ParthKohli
  • ParthKohli
^ do you see why those are true?
mathmath333
  • mathmath333
because \(|x|=x,\ \ if\ \ x>0\\~\\and\\~\\|x|=-x\ \ if\ \ x<0\)
ParthKohli
  • ParthKohli
Exactly. So you can solve it now, yes?
mathmath333
  • mathmath333
where did u learnt this method from
ParthKohli
  • ParthKohli
This is a standard method.
ParthKohli
  • ParthKohli
If you go to coaching, they do teach it there.
alekos
  • alekos
Tedious and unnecessary. If you follow what I've done it will give you the most efficient answer
ParthKohli
  • ParthKohli
What have you done?
mathmath333
  • mathmath333
@alekos u didnt defined why u choosed the two particular cases
mathmath333
  • mathmath333
*chose
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
So for example if you want to solve\[|x+3| + |1 - x| = 4\]It reduces to the inequality\[(x+3)(1-x) \ge 0\]
alekos
  • 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
ParthKohli
  • ParthKohli
How would you sketch the graph?
alekos
  • alekos
You can use pen and paper or Desmos
ParthKohli
  • ParthKohli
I'll take the pen-and-paper approach. How did you sketch the graph here?
ParthKohli
  • 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.
ParthKohli
  • 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?
alekos
  • 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.
ParthKohli
  • 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.
ParthKohli
  • ParthKohli
I'm pretty sure that he knows computers exist too.
alekos
  • 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.
ParthKohli
  • ParthKohli
Nah, takes only a couple of minutes.
ParthKohli
  • ParthKohli
There are four cases and each case should take you around 20-25 seconds.
alekos
  • alekos
OK, no problem. Either method is valid.

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