Use the distributive property to simplify the binomials.

$\displaystyle 2(3)-2(2x)\leq 9(2)-9(3x)$

Simplify both sides.

$\displaystyle 6-4x\leq 18-27x$

Add $\displaystyle 27x$ on both sides.

$\displaystyle 6-4x+27x\leq 18-27x+27x$

$\displaystyle 6+23x\leq 18$

Subtract six from both sides.

$\displaystyle 6+23x-6\leq 18-6$

$\displaystyle 23x\leq 12$

Divide by 23 on both sides.

$\displaystyle \frac{23x}{23}\leq \frac{12}{23}$

The answer is:  $\displaystyle x\leq \frac{12}{23}$

Use distribution to simplify the right side.

$\displaystyle 3x>-5(2x)-5(-3)$

Simplify the right side.

$\displaystyle 3x>-10x+15$

Add $\displaystyle 10x$ on both sides of the equation.

$\displaystyle 3x+10x>-10x+15+10x$

$\displaystyle 13x>15$

Divide by 13 on both sides.

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

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

Add $\displaystyle 4x$ on both sides.

$\displaystyle 2-4x+4x>6x-8+4x$

$\displaystyle 2>10x-8$

Add 8 on both sides.

$\displaystyle 2+8>10x-8+8$

$\displaystyle 10>10x$

Divide by 10 on both sides.

$\displaystyle \frac{10}{10}>\frac{10x}{10}$

$\displaystyle 1>x$

The answer is:  $\displaystyle x< 1$

Add $\displaystyle 9x$ on both sides to avoid dividing by a negative later in the calculation.  Dividing by a negative value will require switching the sign.

$\displaystyle -9x+7+(9x)>8+9x+(9x)$

The inequality becomes:  

$\displaystyle 7> 18x+8$

Subtract 8 from both sides.

$\displaystyle 7-8> 18x+8-8$

$\displaystyle -1>18x$

Divide by 18 on both sides.

$\displaystyle \frac{-1}{18}>\frac{18x}{18}$

The answer is:  $\displaystyle x< -\frac{1}{18}$

There is a domain restriction with the square root functions.

Both the functions must be zero or greater.

Square both sides.

$\displaystyle (\sqrt{5x-7}) ^2< (\sqrt{2x})^2$

$\displaystyle 5x-7< 2x$

Subtract $\displaystyle 2x$ from both sides, and add 7 on both sides.

$\displaystyle 5x-7+(7)-[2x]< 2x-[2x]+(7)$

$\displaystyle 3x< 7$

Divide by three on both sides to isolate x.

$\displaystyle x< \frac{7}{3}$

Notice that this value will also include all the values that are negative, which cannot satisfy the original equation.  We will need to identify the largest possible x-value of both radicals to determine a viable domain.

Set the two radicals greater or equal to zero and solve for $\displaystyle x$.

$\displaystyle \sqrt{2x}\geq0 \rightarrow 2x\geq 0 \rightarrow x\geq0$

$\displaystyle \sqrt{5x-7}\geq0 \rightarrow5x-7\geq0 \rightarrow 5x\geq7\rightarrow x\geq \frac{7}{5}$

The value of $\displaystyle x$ must also be greater than $\displaystyle \frac{7}{5}$.

Rewrite $\displaystyle \frac{7}{5}\leq x< \frac{7}{3}$ in interval notation.

The answer is:  $\displaystyle [\frac{7}{5},\frac{7}{3})$

Divide by two on both sides.

$\displaystyle \frac{2x^2}{2}>\frac{18}{2}$

The inequality becomes:

$\displaystyle x^2>9$

Square root both sides.

$\displaystyle \sqrt{x^2}>\sqrt{9}$

$\displaystyle \pm x >3$

Split the inequalities.

$\displaystyle x>3$ is one solution.

Simplify $\displaystyle -x>3$ by dividing by negative one on both sides.  This will change the sign.

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

$\displaystyle x< -3$ is the second solution.

Write the terms in interval notation.

The answer is:  $\displaystyle (-\infty, -3) \bigcup(3,\infty)$

Add $\displaystyle 3x$ on both sides.

$\displaystyle -3x+7+(3x)\leq 8x-4+(3x)$

$\displaystyle 7\leq 11x-4$

Add four on both sides.

$\displaystyle 7+4\leq 11x-4+4$

$\displaystyle 11\leq 11x$

Divide by eleven on both sides.

$\displaystyle \frac{11}{11}\leq \frac{11x}{11}$

$\displaystyle 1\leq x$

The answer is:  $\displaystyle x\geq1$

Add $\displaystyle 4x$ on both sides.

$\displaystyle -3x-8+4x< -4x-9+4x$

$\displaystyle x-8< -9$

Add eight on both sides.

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

Simplify both sides.

The answer is:  $\displaystyle x< -1$

Distribute the negative eight into the binomial.

$\displaystyle -16x+24< 17$

Subtract 24 on both sides.

$\displaystyle -16x+24-24< 17-24$

$\displaystyle -16x< -7$

Divide by negative 16 on both sides.  The sign will change direction since we are dividing by a negative value.

$\displaystyle \frac{-16x}{-16}< \frac{-7}{-16}$

The answer is:  $\displaystyle x< \frac{7}{16}$

Distribute the negative three through both terms of the binomial.

$\displaystyle -9x+24>9x+7$

Add $\displaystyle 9x$ on both sides.

$\displaystyle -9x+24+9x>9x+7+9x$

$\displaystyle 24>18x+7$

Subtract seven on both sides.

$\displaystyle 24-7>18x+7-7$

$\displaystyle 17>18x$

Divide by 18 on both sides.

$\displaystyle \frac{17}{18}>\frac{18x}{18}$

The answer is:  $\displaystyle x< \frac{17}{18}$