## Precalculus (6th Edition)

The solution is $\Big(-\infty,\dfrac{3}{2}\Big]\cup\Big[\dfrac{7}{2},\infty\Big)$
$|10-4x|+1\ge5$ Take $1$ to the right side of the inequality: $|10-4x|\ge5-1$ $|10-4x|\ge4$ Solving this absolute value inequality is equivalent to solving two separate inequalities, which are: $10-4x\ge4$ and $10-4x\le-4$ $\textbf{Solve the first inequality:}$ $10-4x\ge4$ Take $10$ to the right side: $-4x\ge4-10$ $-4x\ge-6$ Take $-4$ to divide the right side and reverse the direction of the inequality sign: $x\le\dfrac{-6}{-4}$ $x\le\dfrac{3}{2}$ $\textbf{Solve the second inequality:}$ $10-4x\le-4$ Take $10$ to the right side: $-4x\le-4-10$ $-4x\le-14$ Take $-4$ to divide the right side and reverse the direction of the inequality sign: $x\ge\dfrac{-14}{4}$ $x\ge\dfrac{7}{2}$ Expressing the solution in interval notation: $\Big(-\infty,\dfrac{3}{2}\Big]\cup\Big[\dfrac{7}{2},\infty\Big)$
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