Question

solve for x.

Answer 1

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

\end{align}}\)

Answer 2

what is this for?

Answer 3

algebra absolute value

Answer 4

x∈R i have never seen this before

Answer 5

that means x is an element of set of real numbers

Answer 6

Break into cases.
we have x-3=0 if x=3
and we have x+5=0 if x=-5

Answer 7

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

Answer 8

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

Answer 9

I will let you type out case 3
but anyways we need to solve these equations in the two cases I have already mentioned

Answer 10

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

Answer 11

\(|x|=\pm x\) 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

Answer 12

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

Answer 13

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
\[\large \color{black}{\begin{align} &|x-3|+|x+5|=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\
&\implies x-3+x+5=7,\hspace{.33em}\\~\\
&\cancel{\implies x-3-(x+5)=7,\hspace{.33em}}\\~\\
&\implies -(x-3)+x+5=7,\ \ x\in \mathbb{R}\hspace{.33em}\\~\\
&\implies -(x-3)-(x+5)=7,\hspace{.33em}\\~\\
\end{align}}\]

Answer 14

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

Answer 15

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

Answer 16

got -9/2 ,5/2

for case 1 and 3

Answer 17

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

Answer 18

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

Answer 19

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

second that

Answer 20

thnx!