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mathmath333

  • one year ago

solve for x.

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} |x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    what is this for?

  3. mathmath333
    • one year ago
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    algebra absolute value

  4. anonymous
    • one year ago
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    x∈R i have never seen this before

  5. mathmath333
    • one year ago
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    that means x is an element of set of real numbers

  6. freckles
    • one year ago
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    Break into cases. we have x-3=0 if x=3 and we have x+5=0 if x=-5 |dw:1433970138088:dw|

  7. freckles
    • one year ago
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    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

  8. freckles
    • one year ago
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    And case 2 -5<x<3 if we have this case then |x+3|=-(x-3) and |x+5|=x+5 so you look at solving -(x-3)+(x+5)=7

  9. freckles
    • one year ago
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    I will let you type out case 3 but anyways we need to solve these equations in the two cases I have already mentioned

  10. freckles
    • one year ago
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    look and I made a type-o everywhere I have |x+3| I meant to write |x-3|

  11. mathmath333
    • one year ago
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    \(|x|=\pm x\) 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

  12. Pawanyadav
    • one year ago
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    We have to break in 3 cases. 1) x<-5 2)-5<x<3 <,> with equal to sign 3)x>3

  13. freckles
    • one year ago
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    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 -5<x<3 if we have this case then |x-3|=-(x-3) and |x+5|=x+5 so you look at solving -(x-3)+(x+5)=7 And lastly case 3 x>3 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 \[\large \color{black}{\begin{align} &|x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies x-3+x+5=7,\hspace{.33em}\\~\\ &\cancel{\implies x-3-(x+5)=7,\hspace{.33em}}\\~\\ &\implies -(x-3)+x+5=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ &\implies -(x-3)-(x+5)=7,\hspace{.33em}\\~\\ \end{align}}\]

  14. Pawanyadav
    • one year ago
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    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

  15. Pawanyadav
    • one year ago
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    We have to break in 3 cases. 1) x<-5 2)-5<x<3 <,> with equal to sign 3)x>3

  16. mathmath333
    • one year ago
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    got -9/2 ,5/2 for case 1 and 3

  17. freckles
    • one year ago
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    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 -5<x<3 if we have this case then |x-3|=-(x-3) and |x+5|=x+5 so you look at solving -(x-3)+(x+5)=7 And lastly case 3 x>3 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

  18. freckles
    • one year ago
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    also if you try to graph y=|x-3|+|x+5| and y=7 you will see no intersection

  19. IrishBoy123
    • one year ago
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    "so there are no solutions to |x-3|+|x+5|=7" second that

  20. mathmath333
    • one year ago
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    thnx!

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