solve

- mathmath333

solve

- schrodinger

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- mathmath333

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

- alekos

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

- alekos

Is this a test?

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## More answers

- mathmath333

how get that

- mathmath333

this is no test

- alekos

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

- anonymous

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

- alekos

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

- mathmath333

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

- alekos

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

- mathmath333

that can be if 4 values are not valid

- alekos

that's right. the 4 values are not valid

- alekos

just plot the graph and you'll see

- mathmath333

i look for a pen paper method

- alekos

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

- ParthKohli

Hey, yes, just do it casewise.

- mathmath333

u mean take 6 cases ?

- ParthKohli

You'd need five, so yes.

- mathmath333

how is it 5 cases

- anonymous

It's 4 cases

- ParthKohli

Oh, four.

- ParthKohli

|dw:1434037874987:dw|

- ParthKohli

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

- anonymous

|dw:1434037949803:dw|

- ParthKohli

Yes.

- alekos

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

- ParthKohli

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

- alekos

I've done it both ways. algebraically and graphically

- mathmath333

how u did algebriacally

- alekos

I showed you earlier on

- ParthKohli

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

- alekos

the other cases are not valid

- ParthKohli

|dw:1434038294034:dw|

- ParthKohli

|dw:1434038341589:dw|

- ParthKohli

|dw:1434038401841:dw|

- ParthKohli

|dw:1434038477315:dw|

- ParthKohli

|dw:1434038527838:dw|

- ParthKohli

|dw:1434038567104:dw|

- ParthKohli

Discard an answer if it doesn't fall in the given interval.

- mathmath333

here how would u identify useful and nonuseful regions

- 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

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

- ParthKohli

Did you see those drawings?

- mathmath333

yes

- ParthKohli

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

- ParthKohli

^ do you see why those are true?

- mathmath333

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

- ParthKohli

Exactly. So you can solve it now, yes?

- mathmath333

where did u learnt this method from

- ParthKohli

This is a standard method.

- ParthKohli

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

- alekos

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

- ParthKohli

What have you done?

- mathmath333

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

- mathmath333

*chose

- 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

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

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

How would you sketch the graph?

- alekos

You can use pen and paper or Desmos

- ParthKohli

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

- 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

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

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

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

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

- 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

Nah, takes only a couple of minutes.

- ParthKohli

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

- alekos

OK, no problem. Either method is valid.

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