Question

8. Solve the equation. Check for extraneous solutions.

6|6 – 4x| = 8x + 4 (1 point)
x = 1
x = five over two or x = 2
x = five over two
x = five over two or x = 1

Answer 1

Can I please have help on this

Answer 2

Factor 2 out of the absolute bars:
6 * 2 * |3 - 2x| = 4(2x + 1)
divide both sides by 12
|3 - 2x| = 4/12 * (2x + 1)
|3 - 2x| = 1/3 * (2x+1)
square both sides and solve for x.
You will get a quadratic equation with two values for x.
Test each x value with the original equation and discard any extraneous solution introduced by squaring. Keep the solution that satisfies the original equation.

Answer 3

Instead of dividing both sides by 12 it may be easier to divide both sides by 4:
3 * |3-2x| = (2x+1)
square both sides and solve for x.

Answer 4

A second method to solve the problem WITHOUT squaring:
|6 – 4x| is zero when 4x = 6 or x = 1.5
When x < 1.5, 6-4x is positive and therefore, |6-4x| = 6 - 4x
When x > 1.5, 6-4x is negative and therefore, |6-4x| = -(6-4x) = 4x - 6

For x < 1.5: 6(6 - 4x) = 8x + 4
36 - 24x = 8x + 4
32x = 32
x = 1.
This solution falls under the case of x < 1.5 and therefore x = 1 is a valid solution.

For x > 1.5: 6(4x - 6) = 8x + 4
24x - 36 = 8x + 4
16x = 40
x = 40/16 = 5/2 = 2.5.
This solution falls under the case of x > 1.5 and therefore x = 2.5 is a valid solution.

x = 1 and x = 2.5 are the solutions.

You can put both values into the original equation to make sure it satisfies.

Answer 5

Okay, thank you so much for explaining in detail, this really helped

Answer 6

You are welcome.