mathmath333
  • mathmath333
solve for x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} |x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
what is this for?
mathmath333
  • mathmath333
algebra absolute value

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anonymous
  • anonymous
x∈R i have never seen this before
mathmath333
  • mathmath333
that means x is an element of set of real numbers
freckles
  • freckles
Break into cases. we have x-3=0 if x=3 and we have x+5=0 if x=-5 |dw:1433970138088:dw|
freckles
  • freckles
So let's look at case 1 x<-5 if x<-5 then |x+3|=-(x-3) and |x+5|=-(x+5) so you look at solving -(x-3)-(x+5)=7
freckles
  • freckles
And case 2 -5
freckles
  • freckles
I will let you type out case 3 but anyways we need to solve these equations in the two cases I have already mentioned
freckles
  • freckles
look and I made a type-o everywhere I have |x+3| I meant to write |x-3|
mathmath333
  • mathmath333
\(|x|=\pm x\) so we should havr four cases like this \(\large \color{black}{\begin{align} &|x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies x-3+x+5=7,\hspace{.33em}\\~\\ &\implies x-3-(x+5)=7,\hspace{.33em}\\~\\ &\implies -(x-3)+x+5=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies -(x-3)-(x+5)=7,\hspace{.33em}\\~\\ \end{align}}\) can this be solved like this
Pawanyadav
  • Pawanyadav
We have to break in 3 cases. 1) x<-5 2)-5 with equal to sign 3)x>3
freckles
  • freckles
I guess you an do the cases like that just make sure you check the solutions by pluggin into original because you only need to look at really So let's look at case 1 x<-5 if x<-5 then |x-3|=-(x-3) and |x+5|=-(x+5) so you look at solving -(x-3)-(x+5)=7 And case 2 -53 which we have both |x-3|=x-3 and |x+5|=x+5 (x-3)+(x+5)=7 so of the cases you wrote out you only need to look at your case 1,3,4 \[\large \color{black}{\begin{align} &|x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies x-3+x+5=7,\hspace{.33em}\\~\\ &\cancel{\implies x-3-(x+5)=7,\hspace{.33em}}\\~\\ &\implies -(x-3)+x+5=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies -(x-3)-(x+5)=7,\hspace{.33em}\\~\\ \end{align}}\]
Pawanyadav
  • Pawanyadav
In first case both mode . open with minus sign. In second |x-3| only open with minus sign. In third case both mode. Open with plus sign now solve the 3 eq.'s
Pawanyadav
  • Pawanyadav
We have to break in 3 cases. 1) x<-5 2)-5 with equal to sign 3)x>3
mathmath333
  • mathmath333
got -9/2 ,5/2 for case 1 and 3
freckles
  • freckles
So let's look at case 1 x<-5 if x<-5 then |x-3|=-(x-3) and |x+5|=-(x+5) so you look at solving -(x-3)-(x+5)=7 And case 2 -53 which we have both |x-3|=x-3 and |x+5|=x+5 (x-3)+(x+5)=7 Case 1) you got x=-9/2 but -9/2<-5 is false so x=-9/2 is not a solution Case 3) you got x=5/2 but 5/2>3 is not true so x=5/2 is not a solution so there are no solutions to |x-3|+|x+5|=7
freckles
  • freckles
also if you try to graph y=|x-3|+|x+5| and y=7 you will see no intersection
IrishBoy123
  • IrishBoy123
"so there are no solutions to |x-3|+|x+5|=7" second that
mathmath333
  • mathmath333
thnx!

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