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mathmath333
 one year ago
solve
mathmath333
 one year ago
solve

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} x3+x+4+x1=14\hspace{.33em}\\~\\ \end{align}}\)

alekos
 one year ago
Best ResponseYou've already chosen the best response.0I get x = 14/3 and 14/3

alekos
 one year ago
Best ResponseYou've already chosen the best response.0well first we take all the positive values within the modulus (x3) + (x+4) + (x1) =14 and solve for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0best way to do this problem is partitioning the x's

alekos
 one year ago
Best ResponseYou've already chosen the best response.0then we take all the negative values (x3)  (x+4)  (x1) =14 and solve for x

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1there will be total of 6 cases but that is a long way

alekos
 one year ago
Best ResponseYou've already chosen the best response.0if you plot the graph you'll see that there are only two values

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1that can be if 4 values are not valid

alekos
 one year ago
Best ResponseYou've already chosen the best response.0that's right. the 4 values are not valid

alekos
 one year ago
Best ResponseYou've already chosen the best response.0just plot the graph and you'll see

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i look for a pen paper method

alekos
 one year ago
Best ResponseYou've already chosen the best response.0that's fine, you can do it that way. But Desmos is easier

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Hey, yes, just do it casewise.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u mean take 6 cases ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4You'd need five, so yes.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434037874987:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4For some reason, I thought there were four absolute values. Oops.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434037949803:dw

alekos
 one year ago
Best ResponseYou've already chosen the best response.0I'll say it again. Plot the graph and you will all see that there are only two values

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4If we're gonna take the graphing route, might as well just solve the equation directly.

alekos
 one year ago
Best ResponseYou've already chosen the best response.0I've done it both ways. algebraically and graphically

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how u did algebriacally

alekos
 one year ago
Best ResponseYou've already chosen the best response.0I showed you earlier on

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Alekos, you covered only two of the four cases...

alekos
 one year ago
Best ResponseYou've already chosen the best response.0the other cases are not valid

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038294034:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038341589:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038401841:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038477315:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038527838:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434038567104:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Discard an answer if it doesn't fall in the given interval.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1here how would u identify useful and nonuseful regions

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4First 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
 one year ago
Best ResponseYou've already chosen the best response.1ok so i have to check x by putting x in original equation,ok

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Did you see those drawings?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4In the interval \(\left[4, 1\right]\):\[x3 = (x3)\]\[x+4 = x+4\]\[x1  =  (x1) \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4^ do you see why those are true?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1because \(x=x,\ \ if\ \ x>0\\~\\and\\~\\x=x\ \ if\ \ x<0\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Exactly. So you can solve it now, yes?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1where did u learnt this method from

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4This is a standard method.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4If you go to coaching, they do teach it there.

alekos
 one year ago
Best ResponseYou've already chosen the best response.0Tedious and unnecessary. If you follow what I've done it will give you the most efficient answer

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4What have you done?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1@alekos u didnt defined why u choosed the two particular cases

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Some problems have neat tricks involved in them.\[a + b = a + b \]implies that \(a\) and \(b\) have the same sign. Similarly,\[a + b = ab\]implies that \(a\) and \(b\) have the opposite sign.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4So for example if you want to solve\[x+3 + 1  x = 4\]It reduces to the inequality\[(x+3)(1x) \ge 0\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.0That'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
 one year ago
Best ResponseYou've already chosen the best response.4How would you sketch the graph?

alekos
 one year ago
Best ResponseYou've already chosen the best response.0You can use pen and paper or Desmos

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4I'll take the penandpaper approach. How did you sketch the graph here?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4The point is... the only way you can sketch a graph is casewise. So you first consider cases, find an expression for each straightline, and then draw straightlines 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
 one year ago
Best ResponseYou've already chosen the best response.4But 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.4The only reason he's been given that problem is to be able to solve it himself in spite of the existence of computers.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4I'm pretty sure that he knows computers exist too.

alekos
 one year ago
Best ResponseYou've already chosen the best response.0That'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
 one year ago
Best ResponseYou've already chosen the best response.4Nah, takes only a couple of minutes.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4There are four cases and each case should take you around 2025 seconds.

alekos
 one year ago
Best ResponseYou've already chosen the best response.0OK, no problem. Either method is valid.
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