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I have $x-y=3$ and $y\le1$ and $x\ge\frac12$.

I proved that $\sqrt{(2x-1)^2}+\sqrt{(2y-2)^2}=7$ and that $-\frac52\le y\le 1$ and $\frac12\le x\le4$.

How can I prove that $|x+y-5|+|x+y+2|=7$?

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  • $\begingroup$ please see my answer @user114328 $\endgroup$ Dec 8, 2013 at 14:53

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ok let us consider following cases :

first side is just $-x-y+5$,second case is $x+y+2$,so sum is $5+2=7$

to be more deeply,let us take such situations

  1. $y=1$

2.$x=4$

we have

$|x+y-5]+|x+y+2|=x+y-5+x+y+2=2*x+2*y-3=2*(x+y)-3 =2*(5)-3=7$

can you continue from this?

just consider this situation when $x<0$ then $|x|=-x$,else if $x>=0$ then $|x|=x$.ok let us consider following case,when

$y=-5/2$ and $x=1$ ,we have that $-5/2+1-5<0$,that why we will have

$-x-y+5$,

for second we have $-5/2+1+2>0$ ,

that why we have second as $x+y+2$,together is is still $7$

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  • $\begingroup$ oups it is x-y=3 not x+y=3 :((( $\endgroup$
    – user114328
    Dec 8, 2013 at 14:42
  • $\begingroup$ Just a suggestion: to denote multiplication, you just need to juxtapose constants with variables, or when multiplying two constants, use $\cdot$ which is coded \cdot, and refrain from using the programmer's $*$. $\endgroup$
    – amWhy
    Dec 8, 2013 at 14:52
  • $\begingroup$ i am new in latex,thanks for suggestion $\endgroup$ Dec 8, 2013 at 14:52
  • $\begingroup$ I understand! I was in the same position a while back, @dato. And I learned a lot just by these sorts of comments from other helpful users. $\endgroup$
    – amWhy
    Dec 8, 2013 at 14:54
  • $\begingroup$ yes yes thanks,i will learn it,i know some commands,thanks for suggestion $\endgroup$ Dec 8, 2013 at 14:55

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