Question

your turn: by using two separate equations, solve each of these absolute value equations. be sure to check and to reject extraneous solutions. the first one is done for you again. you can thank me later. 9. |x - 3|+5x = 27

Explanation:

Step1: Consider the two - case of absolute value

Case 1: When \(x - 3\geq0\) (i.e., \(x\geq3\)), the equation \(|x - 3|+5x=27\) becomes \((x - 3)+5x=27\).
Simplify the left - hand side: \(x-3 + 5x=27\), which is \(6x-3 = 27\).
Add 3 to both sides: \(6x=27 + 3=30\).
Divide both sides by 6: \(x = 5\). Since \(5\geq3\), \(x = 5\) is a valid solution for this case.

Step2: Consider the second case

Case 2: When \(x - 3<0\) (i.e., \(x<3\)), the equation \(|x - 3|+5x=27\) becomes \(-(x - 3)+5x=27\).
Expand the left - hand side: \(-x + 3+5x=27\), which simplifies to \(4x+3 = 27\).
Subtract 3 from both sides: \(4x=27 - 3=24\).
Divide both sides by 4: \(x = 6\). But since for this case \(x<3\) and \(6>3\), \(x = 6\) is an extraneous solution and should be rejected.

Answer:

\(x = 5\)