The solution set of the inequation $\frac{7+5x}{4}≥\frac{x}{3}-10, x \in R$ is :

$[-\frac{141}{11}, ∞)$

$\frac{7+5x}{4}\ge \frac{x}{3}-10.$

$12\left(\frac{7+5x}{4}\right)\ge 12\left(\frac{x}{3}-10\right).$

$3(7+5x)\ge4x-120.$

$21+15x\ge4x-120.$

$11x\ge-141.$

$x\ge-\frac{141}{11}.$

$\text{Solution set }=\left[-\frac{141}{11},\infty\right).$