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. 11. |3x + 7|+2x = 20 (dont fear fractions)

Explanation:

Step1: Isolate the absolute - value term

Subtract \(2x\) from both sides of the equation \(|3x + 7|+2x = 20\) to get \(|3x + 7|=20 - 2x\).

Step2: Set up two separate equations

Case 1: \(3x+7 = 20 - 2x\).
Add \(2x\) to both sides: \(3x+2x + 7=20-2x+2x\), which simplifies to \(5x+7 = 20\).
Subtract 7 from both sides: \(5x+7 - 7=20 - 7\), so \(5x=13\).
Divide both sides by 5: \(x=\frac{13}{5}\).
Case 2: \(3x + 7=-(20 - 2x)\).
Expand the right - hand side: \(3x + 7=-20 + 2x\).
Subtract \(2x\) from both sides: \(3x-2x+7=-20 + 2x-2x\), getting \(x+7=-20\).
Subtract 7 from both sides: \(x=-20 - 7=-27\).

Step3: Check for extraneous solutions

For \(x = \frac{13}{5}\):
Left - hand side: \(|3\times\frac{13}{5}+7|+2\times\frac{13}{5}=|\frac{39}{5}+7|+\frac{26}{5}=|\frac{39 + 35}{5}|+\frac{26}{5}=|\frac{74}{5}|+\frac{26}{5}=\frac{74}{5}+\frac{26}{5}=\frac{74 + 26}{5}=20\).
Right - hand side: 20. So \(x=\frac{13}{5}\) is a valid solution.
For \(x=-27\):
Left - hand side: \(|3\times(-27)+7|+2\times(-27)=|-81 + 7|-54=|-74|-54=74-54 = 20\).
Right - hand side: 20. So \(x=-27\) is a valid solution.

Answer:

\(x=\frac{13}{5},x=-27\)