## mathmath333 one year ago solve for x.

1. mathmath333

\large \color{black}{\begin{align} |x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}

2. anonymous

what is this for?

3. mathmath333

algebra absolute value

4. anonymous

x∈R i have never seen this before

5. mathmath333

that means x is an element of set of real numbers

6. freckles

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

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

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

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

look and I made a type-o everywhere I have |x+3| I meant to write |x-3|

11. 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

We have to break in 3 cases. 1) x<-5 2)-5<x<3 <,> with equal to sign 3)x>3

13. 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 -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}}

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

We have to break in 3 cases. 1) x<-5 2)-5<x<3 <,> with equal to sign 3)x>3

16. mathmath333

got -9/2 ,5/2 for case 1 and 3

17. 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 -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

also if you try to graph y=|x-3|+|x+5| and y=7 you will see no intersection

19. IrishBoy123

"so there are no solutions to |x-3|+|x+5|=7" second that

20. mathmath333

thnx!