Can someone please explain how I find the extraneous solution? I'm stuck and don't know what to do. I tried looking it up, but I don't understand what to do... SS below in comments

Question

luvvdani1516 · Answer

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surjithayer · Answer

you are correct.
x=-7 is an extraneous solution.
|x-5|=2(x+1)
when x=-7
|-7-5|=2(-7+1)
|-12|=2(-6)
12=-12
which is not true ,hence x=-7 is an extraneous solution.
solving
x-5=-2(x+1)
x-5=-2x-2
x+2x=-2+5
3x=3
x=3/3=1
when x=1
|x-5|=2(x+1)
|1-5|=2(1+1)
|-4|=2(2)
4=4
which is true.
so x=1 is a solution.

luvvdani1516 · Answer

@surjithayer wrote:

you are correct. x=-7 is an extraneous solution. |x-5|=2(x+1) when x=-7 |-7-5|=2(-7+1) |-12|=2(-6) 12=-12 which is not true ,hence x=-7 is an extraneous solution. solving x-5=-2(x+1) x-5=-2x-2 x+2x=-2+5 3x=3 x=3/3=1 when x=1 |x-5|=2(x+1) |1-5|=2(1+1) |-4|=2(2) 4=4 which is true. so x=1 is a solution.

Ohhhh thank you!

surjithayer · Answer

yw

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