Solve the equation check for extraneous solution 6|6-4 x|=8 x+4answers are x=1, x=5/2.or x=2, x=5/2, x=5/2or x=1
6|6-4x| = 8 x + 4
+ sign
6 (6x-4) = 8 x+4
36 x - 24 = 8 x + 4
28 x = 28
x = 1
negative sign
6(4x-6) = 8 x + 4
24x - 36 = 8 x + 4
16 x = -40
x = 5/2
so
5/2 or 1
To solve the equation `|6-4x| = 8x+4`, first we can break it down into two separate equations:
1. `6-4x = 8x+4`, when the expression inside the absolute value is positive.
2. `-(6-4x) = 8x+4`, when the expression inside the absolute value is negative.
Let's solve these equations separately and check for extraneous solutions:
Equation 1:
`6-4x = 8x+4`
To solve for x, start by isolating the x terms on one side:
-4x - 8x = 4 - 6
Simplify:
-12x = -2
Divide both sides by -12:
x = -2/(-12)
x = 1/6
So, we have found one potential solution: x = 1/6.
Now let's move on to Equation 2:
`-(6-4x) = 8x+4`
Distribute the negative sign:
-6 + 4x = 8x + 4
Again, isolate the x terms:
-4x - 8x = 4 + 6
Simplify:
-12x = 10
Divide both sides by -12:
x = 10/(-12)
x = -5/6
We have found another potential solution: x = -5/6.
Now, let's check if any of these solutions are extraneous by substituting them back into the original equation:
For x = 1/6:
|6-4(1/6)| = 8(1/6) + 4
|6 - 2/3| = 4/3 + 4
|18/3 - 2/3| = 16/3
|16/3| = 16/3
16/3 = 16/3
For x = -5/6:
|6-4(-5/6)| = 8(-5/6) + 4
|6 + 10/3| = -40/6 + 4
|18/3 + 10/3| = -20/3
|28/3| = -20/3
28/3 ≠ -20/3
From the checks, we can conclude that x = 1/6 is a valid solution, while x = -5/6 is an extraneous solution.
Therefore, the valid solutions for the equation |6 - 4x| = 8x + 4, after checking for extraneous solutions, are x = 1/6.