Question

1. |3x - 2|+x = 14

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.

Explanation:

Step1: Isolate the absolute - value term

Subtract \(x\) from both sides of the equation \(|3x - 2|+x = 14\) to get \(|3x - 2|=14 - x\).

Step2: Set up two separate equations

We know that if \(|a| = b\), then \(a = b\) or \(a=-b\). So we have \(3x - 2=14 - x\) or \(3x - 2=-(14 - x)\).

Solve the first equation \(3x - 2=14 - x\)

Add \(x\) to both sides: \(3x+x - 2=14 - x+x\), which simplifies to \(4x - 2=14\). Then add 2 to both sides: \(4x-2 + 2=14 + 2\), so \(4x=16\). Divide both sides by 4: \(x = 4\).

Solve the second equation \(3x - 2=-(14 - x)\)

Expand the right - hand side: \(3x - 2=-14 + x\). Subtract \(x\) from both sides: \(3x-x - 2=-14 + x-x\), which gives \(2x - 2=-14\). Add 2 to both sides: \(2x-2 + 2=-14 + 2\), so \(2x=-12\). Divide both sides by 2: \(x=-6\).

Step3: Check for extraneous solutions

For \(x = 4\), substitute into the original equation: \(|3\times4 - 2|+4=|12 - 2|+4=|10|+4=10 + 4=14\), so \(x = 4\) is a valid solution.
For \(x=-6\), substitute into the original equation: \(|3\times(-6)-2|+(-6)=|-18 - 2|-6=|-20|-6=20 - 6 = 14\), so \(x=-6\) is also a valid solution.

Answer:

\(x = 4\) or \(x=-6\)