How do you solve the equation and identify any extraneous solutions for #4 abs(4-3x )= 4x + 6#?

To solve the equation (4 \left| 4 - 3x \right| = 4x + 6), first, isolate the absolute value term:

1. Divide both sides by 4:

(\left| 4 - 3x \right| = x + \frac{3}{2})

2. Break the absolute value into two cases:

a) (4 - 3x = x + \frac{3}{2})

b) (4 - 3x = -\left( x + \frac{3}{2} \right))

3. Solve each case separately:

a) (4 - 3x = x + \frac{3}{2})

(-3x - x = \frac{3}{2} - 4)

(-4x = -\frac{5}{2})

(x = \frac{5}{8})

b) (4 - 3x = -\left( x + \frac{3}{2} \right))

(4 - 3x = -x - \frac{3}{2})

(-3x + x = -\frac{3}{2} - 4)

(-2x = -\frac{11}{2})

(x = \frac{11}{4})

4. Check for extraneous solutions by plugging the values back into the original equation:

For (x = \frac{5}{8}):

(4 \left| 4 - 3\left( \frac{5}{8} \right) \right| = 4\left( \frac{5}{8} \right) + 6)

(4 \left| \frac{7}{8} \right| = \frac{5}{2} + 6)

(4 \cdot \frac{7}{8} = \frac{17}{2})

(7 = \frac{17}{2}) - This is false, so (x = \frac{5}{8}) is extraneous.

For (x = \frac{11}{4}):

(4 \left| 4 - 3\left( \frac{11}{4} \right) \right| = 4\left( \frac{11}{4} \right) + 6)

(4 \left| -\frac{5}{4} \right| = 11 + 6)

(4 \cdot \frac{5}{4} = 17)

(5 = 17) - This is false, so (x = \frac{11}{4}) is also extraneous.

Therefore, there are no valid solutions to the equation.