## Precalculus (6th Edition)

The solution is $\Big[-\dfrac{11}{5},\infty\Big)$
$\dfrac{4x+7}{-3}\le2x+5$ Change the sign of the denominator and the sign of the numerator on the left side: $\dfrac{-4x-7}{3}\le2x+5$ Take $3$ to multiply the right side: $-4x-7\le3(2x+5)$ $-4x-7\le6x+15$ Take $6x$ to the left side and $7$ to the right side: $-4x-6x\le15+7$ $-10x\le22$ Rearrange: $10x\ge-22$ Take $10$ to divide the right side and simplify: $x\ge-\dfrac{22}{10}$ $x\ge-\dfrac{11}{5}$ Expressing the solution in interval notation: $\Big[-\dfrac{11}{5},\infty\Big)$