## Precalculus (6th Edition)

The solution is $\Big(-\infty,\dfrac{1}{2}\Big]$
$\dfrac{2x-5}{-8}\le1-x$ Change the sign of the denominator and the sign of the fraction on the left side: $-\dfrac{2x-5}{8}\le1-x$ Rearrange: $x-1\le\dfrac{2x-5}{8}$ Take $8$ to multiply the left side: $8(x-1)\le2x-5$ $8x-8\le2x-5$ Take $2x$ to the left side and $8$ to the right side: $8x-2x\le-5+8$ $6x\le3$ Take $6$ to divide the right side and simplify: $x\le\dfrac{3}{6}$ $x\le\dfrac{1}{2}$ Expressing the solution in interval notation: $\Big(-\infty,\dfrac{1}{2}\Big]$