In order to solve this inequality, we need to apply each mathematical operation to all three sides of the equation.  Let's start by subtracting $\displaystyle 1$ from all the sides:

$\displaystyle -4 \leq 2x < 3$

Now we divide each side by $\displaystyle 2$.  Remember, because the $\displaystyle 2$ isn't negative, we don't have to flip the sign:

$\displaystyle -2 \leq x < \frac{3}{2}$

Add 26 on both sides.

$\displaystyle 2x-26+26\leq 22+26$

$\displaystyle 2x\leq 48$

Divide by two on both sides.

$\displaystyle \frac{2x}{2}\leq \frac{48}{2}$

The answer is:  $\displaystyle x\leq 24$

Distribute the negative through the terms of the binomial.

$\displaystyle 15x-18< 2x-3$

Subtract $\displaystyle 2x$ on both sides.

$\displaystyle 15x-18-2x< 2x-3-2x$

$\displaystyle 13x-18< -3$

Add 18 on both sides.

$\displaystyle 13x-18+18< -3+18$

$\displaystyle 13x< 15$

Divide by 13 on both sides.

$\displaystyle \frac{13x}{13}< \frac{15}{13}$

The answer is:  $\displaystyle x< \frac{15}{13}$

The first thing we can do is distribute the $\displaystyle 4$ and the $\displaystyle -8$ into their respective terms:

$\displaystyle 32-12x< 32-8x-16$

Now we can start to simplify by gathering like terms.  Remember, if we multiply or divide by a negative number, we change the direction of the inequality:

$\displaystyle -12x< -8x-16$

$\displaystyle 12x>8x+16$

$\displaystyle 4x>16$

$\displaystyle x>4$

It can be tricky because there's a negative sign in the equation, but we never end up multiplying or dividing by a negative, so there's no need to change the direction of the inequality.  We simply divide by $\displaystyle 2$ and multiply by $\displaystyle 3$:

$\displaystyle \frac{2}{3}x>-18$

$\displaystyle \frac{1}{3}x>-9$

$\displaystyle x>-27$

Remember, anything we do to one side of the inequality, we must also do to the other two sides.  We can start by adding one to all three sides:

$\displaystyle -7\leq2x-1< 15$

$\displaystyle -6\leq2x< 16$

And now we divide each side by two:

$\displaystyle -3\leq x< 8$

The first thing we can do is distribute the negative sign in the middle term.  Because we're not multiplying or dividing the entire inequality by a negative, we don't have the change the direction of the inequality signs:

$\displaystyle -9\leq -x+5< 8$

Now we can subtract $\displaystyle 5$ from the inequality:

$\displaystyle -14\leq -x< 3$

And finally, we can multiply by $\displaystyle -1$.  Note, this time we're multiplying the entire inequality by a negative, so we have to change the direction of the inequality signs:

$\displaystyle 14\geq x>3$

We start by distributing the $\displaystyle 2$:

$\displaystyle x+3 \leq 2x-8$

Now we can add $\displaystyle 8$, and subtract an $\displaystyle x$ from each side of the equation:

$\displaystyle 11 \leq x$

We didn't multiply or divide by a negative number, so we don't have to reverse the direction of the inequality sign.

Start by finding the roots of the equation by changing the inequality to an equal sign.

$\displaystyle x^2=5x+6$

$\displaystyle x^2-5x-6=0$

$\displaystyle (x-6)(x+1)=0$

Now, make a number line with the two roots:

Pick a number less than $\displaystyle -1$ and plug it into the inequality to see if it holds.

For $\displaystyle x=-2$,

$\displaystyle (-2)^2< 5(-2)+6$

$\displaystyle 4< -4$ is clearly not true. The solution set cannot be $\displaystyle x< -1$.

Next, pick a number between $\displaystyle -1\text{ and }6$.

For $\displaystyle x=0$,

$\displaystyle 0^2< 5(0)+6$

$\displaystyle 0< 6$ is true so the solution set must include $\displaystyle -1< x< 6$.

Finally, pick a number greater than $\displaystyle 6$.

For $\displaystyle x=7$,

$\displaystyle 7^2< 5(7)+6$

$\displaystyle 49< 41$ is clearly not the so the solution set cannot be $\displaystyle x>6$.

Thus, the solution set for this inequality is $\displaystyle -1< x< 6$.

First, find the zeroes of the numerator and the denominator. This will give the boundary points of the intervals to be tested.

$\displaystyle x^{2}+7x-18 = 0$

$\displaystyle (x+9)(x-2) = 0$

$\displaystyle x= -9$; $\displaystyle x= 2$

$\displaystyle x-6 = 0$

$\displaystyle x= 6$

Since the numerator may be equal to 0, $\displaystyle x= -9$ and $\displaystyle x= 2$ are included as solutions. However, since the denominator may not be equal to 0, $\displaystyle x= 6$ is excluded as a solution. 

Now, test each of four intervals for inclusion in the solution set by substituting one test value from each:

$\displaystyle (-\infty, -9)$

Let's test $\displaystyle x = -10$:

$\displaystyle \frac{x^{2}+7x-18}{x-6} \ge 0$

$\displaystyle \frac{(-10)^{2}+7(-10)-18}{(-10)-6} \ge 0$

$\displaystyle \frac{100-70-18}{-10-6} \ge 0$

$\displaystyle \frac{12}{-16} \ge 0$

$\displaystyle -\frac{3}{4} \ge 0$

This is false, so $\displaystyle (-\infty, -9)$ is excluded from the solution set.

$\displaystyle (-9, 2)$

Let's test $\displaystyle x= 0$:

$\displaystyle \frac{x^{2}+7x-18}{x-6} \ge 0$

$\displaystyle \frac{0^{2}+7 (0)-18}{0-6} \ge 0$

$\displaystyle \frac{-18}{-6} \ge 0$

$\displaystyle 3 \ge 0$

This is true, so $\displaystyle (-9, 2)$ is included in the solution set.

$\displaystyle (2, 6)$

Let's test $\displaystyle x = 3\;$:

$\displaystyle \frac{x^{2}+7x-18}{x-6} \ge 0$

$\displaystyle \frac{3^{2}+7(3)-18}{3-6} \ge 0$

$\displaystyle \frac{9+21-18}{3-6} \ge 0$

$\displaystyle \frac{12}{-3} \ge 0$

$\displaystyle -4 \ge 0$

This is false, so $\displaystyle (2, 6)$ is excluded from the solution set.

$\displaystyle (6, \infty)$

Let's test $\displaystyle x = 7$:

$\displaystyle \frac{7^{2}+7 (7)-18}{7-6} \ge 0$

$\displaystyle \frac{49+49-18}{7-6} \ge 0$

$\displaystyle \frac{ 80}{1} \ge 0$

$\displaystyle 80 \ge 1$

This is true, so $\displaystyle (6, \infty)$ is included in the solution set.

The solution set is therefore $\displaystyle \left [-9, 2 \right ] \cup (6, \infty)$.