Answer
$(a)$ $[-4, 3)$
$(b)$ $(-2, 0) ∪ (1, +\infty)$
$(c)$ $(1,7)$
$(d)$ $(-1, 4]$
Work Step by Step
$(a)$
$-4\lt 5-3x\leq 17$ //Subtract $5$
$-9\lt -3x \leq 12$ //Divide by $-3$
$3\gt x \geq -4$ //Arrange the order
$-4\leq x \lt 3$
$[-4, 3)$
$(b)$
$x(x-1)(x+2)\gt 0$
We can solve the inequality using intervals. Intervals will be between the points where the expression gets $0$.
$x_1=0$
$x_2=1$
$x_3=-2$
So we have four intervals:
$(-\infty, -2)$ - Negative
$(-2,0)$, - Positive
$(0,1)$ - Negative
$(1, +\infty)$ - Positive
$-2\lt x \lt 0$ and $x \gt 1$
$(-2, 0) ∪ (1, +\infty)$
$(c)$
$|x-4|\lt3$
An inequality involving an absolute value requires two possible solution:
$±(x-4)\lt 3$
$(x-4) \lt 3$
$(x-4) \gt -3$
$-3 \lt x-4 \lt 3$ //Add $4$
$1 \lt x \lt 7$
$(1,7)$
$(d)$
$\frac{2x-3}{x+1}\leq1$ //Move all terms to the left-hand side and simplify
$\frac{2x-3-x-1}{x+1}\leq 0$
$\frac{x-4}{x+1} \leq 0$
Now we can solve it using intervals method. We have two critical points:
$x-4=0$
$x_1=4$
$x+1=0$
$x_2=-1$ //Note, the denominator cannot be negative, so this value corresponds to an open interval
So, we have three intervals:
$(-\infty, -1)$ - Positive
$(-1,4]$ - Negative
$[4, +\infty)$ - Positive
$-1\lt x \leq 4$
$(-1, 4]$