Answer
$(a)$
The solution set is $[\frac{1}{2}, +\infty)$
$(b)$
The solution set is $(1,4)$
$(c)$
We have the next solutions: $x=6$ and $x=-1$
$(d)$
Solution set is $[-1, 6]$
$*$(For the detailed explanation, see the solving steps below)$*$
Work Step by Step
$(a)$ $2x\geq 1$
At first, we have to divide both sides of the inequality sign by $2$, so we will have only a variable on the left-hand side.
$x\geq\frac{1}{2}$
And we got the solution set, that is $[\frac{1}{2}, +\infty)$
$(b)$ $(x-1)(x-4)\lt 0$
We have to solve this using intervals method. We first have to find critical points (zeros) of the equation and then determine signs of the intervals between these points. Since we need values less than $0$, the negative interval will be our solution set.
$x_1=1$, $x_2=4$
We have the following intervals:
$(-\infty, 1)$ - Positive value
$(1,4)$ - Negative value
$(4, +\infty)$ - Positive value
So the solution set is $(1,4)$
$(c)$ $|2x-5|=7$
In the case of absolute value equation, we have to consider the fact, that the absolute value can be either negative or positive. So, we have to solve two equations:
$2x-5=7$
$2x=12$
$x=6$
AND
$-(2x-5)=7$
$-2x+5=7$
$-2x=2$
$x=-1$
$(d)$ $|2x-5|\leq7$
The absolute value inequality must be solved following the same idea used in part $(c)$.
$2x-5\leq7$
AND
$-(2x-5)\leq7$ => $2x-5\geq-7$
They can be written in a one, equivalent inequality:
$-7\leq2x+-5\leq7$ //Add $5$
$-2\leq2x\leq12$ //Divide by $2$
$-1\leq x\leq6$
So, our solution set is $[-1, 6]$
