Create an equation that results in one extraneous solution

⎮x⎮= 8 - 3x


x = 8 - 3x . . . . . . . . (1)
-x = 8 - 3x . . . . . . . .(2)
From (1), 4x = 8
x = 2
From (2), 2x = 8
x = 4

Putting x = 2 in the original equation:
⎮2⎮= 8-3(2)
2 = 8 - 6
2 = 2

Putting x = 4 in the original equation:
⎮4⎮= 8-3(4)
4 = 8 - 12
4 = -4

Therefore, the valid solution is x = 2, and the extraneous solution is x = 4

Hi there!

The answer to this question is: 1.198945

Step-by-step explanation:

Treat eˣ like a "x" if you can temporarily replace eˣ with "x"

so now it's (x²- 11) (x²+6)=0 then replace "x" with eˣ

then you should get e²ˣ = 11 and e²ˣ = -6

next in order to solve for x, you must ln both sides

and you get 2x = ln(11) and you should get 1.198945

keep in mind: e²ˣ = -6 will be undefined when you ln both sides