b) \( \left.\begin{array}{rl}4+6 x-3 & \leq x+7(x-2) \\ 8 x-2(3 x+4) & \leq 10(x+1)\end{array}\right\} \)

Let’s solve these inequalities step-by-step for some fun with algebra! Starting with the first inequality: \[ 4 + 6x - 3 \leq x + 7(x - 2) \] Simplifying the left side gives: \[ 1 + 6x \leq x + 7x - 14 \] Combine like terms: \[ 1 + 6x \leq 8x - 14 \] Bringing all terms involving \(x\) to one side and constants to the other, you can rearrange to: \[ 1 + 14 \leq 8x - 6x \] Hence, you get: \[ 15 \leq 2x \] Dividing by 2 results in: \[ x \geq 7.5 \] Now let’s tackle the second inequality: \[ 8x - 2(3x + 4) \leq 10(x + 1) \] Distributing gives: \[ 8x - 6x - 8 \leq 10x + 10 \] Combining terms, we have: \[ 2x - 8 \leq 10x + 10 \] Isolating \(x\) leads to: \[ -8 - 10 \leq 10x - 2x \] Which simplifies to: \[ -18 \leq 8x \] Finally, dividing both sides gives: \[ x \geq -\frac{9}{4} \] or \(-2.25\) Tying these solutions together, you have: 1. \( x \geq 7.5 \) 2. \( x \geq -2.25 \) The effective solution for the compound inequalities in interval notation is \( [7.5, \infty) \)!