Answer:
Step-by-step explanation:
The first step contains the error.
2x - 10 = 22 Add 10 to both sides
+ 10 +10
2x = 32 Divide by 10
2x/2 = 32/2
x = 16
A+b+c=6, a=2c, b=a+c using a and b in the first equation gives you:
2c+a+c+c=6 now use a=2c again
2c+2c+c+c=6
6c=6
c=1, since a=2c
a=2, since b=a+c
b=3
So the number is 231.
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
He bought 3 cucumbers because 3*5=15, 15-6=9, 9*.7=6.3, .3*3=.9, .9+6.3=7.2