Problem 66 - College Algebra Essentials

Question

Solve each absolute value inequality. $$|3(x-1)+2| \leq 20$$

Step-by-Step Solution

Answer

The solution to the inequality \(|3(x-1)+2| \leq 20\) is \(-19/3 \leq x \leq 7\).

Step 1: Divide the absolute value expression into two inequalities

The absolute value inequality \(|3(x-1)+2| \leq 20\) can be written as two separate inequalities: \(3(x-1)+2 \leq 20\) and \(-3(x-1)-2 \leq 20\).

Step 2: Solve the first inequality

Start with the inequality \(3(x-1)+2 \leq 20\). First, expand the equation to get \(3x-3+2 \leq 20\), this simplifies to \(3x-1 \leq 20\). Then to isolate x, subtract 1 from both sides, yielding \(3x \leq 21\). Finally, divide both sides by 3 resulting in \(x \leq 7\)

Step 3: Solve the second inequality

Move to the second inequality \(-3(x-1)-2 \leq 20\). Like before, first expand this equation to get \(-3x+3-2 \leq 20\), this simplifies to \(-3x+1 \leq 20\). Then, to isolate x, subtract 1 from both sides, yielding \(-3x \leq 19\). Finally, divide both sides by -3 and remember to flip the inequality sign because of the negative divisor, so this results in \(x \geq -19/3\) or approximately \(x \geq -6.33\) after converting to decimal notation.

Step 4: Find the intersection of the solutions

The solutions to the inequality are the values of x that satisfy both \(x \leq 7\) and \(x \geq -19/3\). So, the final solution to the inequality is \(-19/3 \leq x \leq 7\).