$\displaystyle \frac{2}{3}\left | 9x+6\right | -2\geq 1$

$\displaystyle \frac{2}{3}\left | 9x+6 \right |\geq 3$

$\displaystyle \left | 9x+6 \right |\geq \frac{9}{2}$

At this point, you've isolated the absolute value and can solve this problems for both cases, $\displaystyle 9x+6\geq \frac{9}{2}$ and $\displaystyle 9x+6\leq -\frac{9}{2}$.  Beginning with the first case:

$\displaystyle 9x+6\geq \frac{9}{2}$

$\displaystyle 9x+\frac{12}{2}\geq\frac{9}{2}$

$\displaystyle 9x\geq -\frac{3}{2}$

$\displaystyle x\geq -\frac{1}{6}$

Then for the second case:

$\displaystyle 9x+6\leq -\frac{9}{2}$

$\displaystyle 9x+\frac{12}{2}\leq -\frac{9}{2}$

$\displaystyle 9x\leq -\frac{21}{2}$

$\displaystyle x\leq -\frac{7}{6}$

Since the absolute value with x in it is alone on one side of the inequality, you set the expression inside the absolute value equal to both the positive and negative value of the other side, 11 and -11 in this case. For the negative value -11, you must also flip the inequality from less than to a greater than. You should have two inequalities looking like this.

$\displaystyle -4x-5< 11$ and $\displaystyle -4x-5>-11$

Add 5 to both sides in each inequality.

$\displaystyle -4x< 16$ and $\displaystyle -4x>-6$

Divide by -4 to both sides of the inequality. Remember, dividing by a negative will flip both inequality symbols and you should have this.

$\displaystyle x>-4$ and $\displaystyle x< \frac{3}{2}$

We can set $\displaystyle A = 2, B = 3i$ in the cube of a binomial pattern:

$\displaystyle \left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$\displaystyle = 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$\displaystyle = 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$\displaystyle = 8 + 36i + 54 i^{2} + 27 i^{3}$

$\displaystyle = 8 + 36i + 54 (-1) + 27 (-i)$

$\displaystyle = 8 + 36i - 54 - 27 i$

$\displaystyle = -46 + 9i$

Multiply these complex numbers out in the typical way:

$\displaystyle {}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that $\displaystyle i^2=-1$ by definition. Then, grouping like terms we get

$\displaystyle {} (-10-3)+(5i-6i) = -13-i$

which is our final answer.

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

$\displaystyle (5+8i)(6-2i)=30-10i+48i-16i^2$

Remember that $\displaystyle i=\sqrt-1$, so $\displaystyle i^2=-1$.

Substitute in $\displaystyle -1$ for $\displaystyle i^2$

$\displaystyle 30-10i+48i-16i^2=30+38i-16(-1)=30+38i+16=46+38i$

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

$\displaystyle (2+5i)(-8-i)=-16-2i-40i-5i^2$

Remember that $\displaystyle i=\sqrt-1$, so $\displaystyle i^2=-1$.

Substitute in $\displaystyle -1$ for $\displaystyle i^2$

$\displaystyle -16-2i-40i-5i^2=-16-42i-5(-1)=-16+5-42i=-11-42i$

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

$\displaystyle (2+6i)(5+3i)=10+6i+30i+18i^2$

Remember that $\displaystyle i=\sqrt-1$, so $\displaystyle i^2=-1$.

Substitute in $\displaystyle -1$ for $\displaystyle i^2$

$\displaystyle 10+6i+30i+18i^2=10+36i+18(-1)=10-18+36i=-8+36i$

Remember that 

$\displaystyle i = \sqrt{-1}\\ i^2 = \sqrt{-1}^2 = -1\\ i^3 = i^2\cdot i = -1 \cdot i = -i\\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1$

So the powers of $\displaystyle i$ are cyclic. This means that when we try to figure out the value of an exponent of $\displaystyle i$, we can ignore all the powers that are multiples of $\displaystyle 4$ because they end up multiplying the end result by $\displaystyle 1$, and therefore do nothing.

This means that 

$\displaystyle ai^{93} + bi^{35} + ai^{24} - bi^{86}\\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2$

Now, remembering the relationships of the exponents of $\displaystyle i$, we can simplify this to:

$\displaystyle ai - bi + a - (-b) = ai-bi+a+b\\ = (a+b) + (a-b)i = 40+20i$

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships: 

$\displaystyle a+b=40$

$\displaystyle a-b=10$

No matter how you solve it, you get the values $\displaystyle a=30$, $\displaystyle b=10$.

Step 1: Split the $\displaystyle \sqrt{-243}$ into $\displaystyle \sqrt{243}\cdot\sqrt{-1}$.

Step 2: Recall that $\displaystyle \sqrt{-1}=i$, so let's replace it.

We now have: $\displaystyle \sqrt{243}\cdot i$.

Step 3: Simplify $\displaystyle \sqrt{243}$. To do this, we look at the number on the inside.

$\displaystyle 243=3\cdot3\cdot3\cdot3\cdot3$.

Step 4: Take the factorization of $\displaystyle 243$ and take out any pairs of numbers. For any pair of numbers that we find, we only take $\displaystyle 1$ of the numbers out.

We have a pair of $\displaystyle 3's$, so a $\displaystyle 3$ is outside the radical.
We have another pair of $\displaystyle 3's$, so one more three is put outside the radical.

We need to multiply everything that we bring outside: $\displaystyle 3*3=9$

Step 5: The $\displaystyle i$ goes with the 9...$\displaystyle 9i$

Step 6: The last $\displaystyle 3$ after taking out pairs gets put back into a square root and is written right after the $\displaystyle i$

It will look something like this: $\displaystyle 9i\sqrt{3}$

There are two ways to simplify this problem: 

Method 1: 

$\displaystyle We\ know\ that\ i^2=-1,\ when\ we\ replace\ that\ we\ get:$

$\displaystyle 5\sqrt{-12}$

$\displaystyle =5i\sqrt{12}(take\ the\ i\ out)$

$\displaystyle =5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$\displaystyle =10i{\sqrt{3}}\ reduce$

Method 2: 

$\displaystyle We\ know\ that\ the\ square\ root\ of\ anything\ squared\ is\ just\ itself$

$\displaystyle \sqrt{i^2}=i$

$\displaystyle 5\sqrt{12i^{2}}=5i\sqrt{12}$

$\displaystyle =5i\sqrt{4}\sqrt{3} (the\ perfect\ square\ 4\ will\ go\ into\ 12)$

$\displaystyle =10i{\sqrt{3}}\ reduce$