Do recall that squaring and the *radical sign* cancel each other out... like so:(

)

= a
When you put it that way, it isn't enough :P
(

)

= a
(

)

=?
so you start with
(

)

=

8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
Answer:
the first one
Step-by-step explanation:
The first equation is -5m+15
and the first one is 15-5m
Which is basically the same thing.
Glad to help
Then answer you seek is 133.
Answer:
1/7
Step-by-step explanation:
1 - 0 1
------ = -------
0 - ( - 7) 7