He has to draw arcs through both points and then arcs that stem from those arcs created in the first step.

if you need help with constructions try going here:

https://mathopenref.com/tocs/constructionstoc.html

The line she constructed is not passing through C....

For styling purposes:

\(\pm(1+i), \pm(1-i)\)

Try Drawing a Venn Diagram!

This is just 3 to the 3rd power, or \(3^3\), which is also known as 27.

This is just 3 to the 4th power, or \(3^4\), which is also known as 81.

Whoops! I meant 3^4 power, which is 81.

No, you are not correct. The correct answer is x=6.

This is the answer assuming the first case I stated (2 different terms being multiplied)

\(\frac{\left(x^2+x-12\right)\left(3x^2+11x-4\right)}{\left(x-2\right)\left(x^2-4\right)}\)

Are you saying that (x ^ 2 + x - 12)/(x - 2) is getting multiplied by (3x ^ 2 + 11x - 4)/(x ^ 2 - 4) as 2 seperate terms, like (x ^ 2 + x - 12)/(x - 2) * (3x ^ 2 + 11x - 4)/(x ^ 2 - 4), or is (3x ^ 2 + 11x - 4)/(x ^ 2 - 4) in the denominator of (x ^ 2 + x - 12)/(x - 2)? Try using LaTeX to make the problem clearer.

Hmm. We seem to be disagreeing. I think you forgot that y = 2 + 5/6x is a different term than 4x-3y=3, lol. Guest probably should have put something to tell us that they are 2 different terms, unless I am the mistaken one. Oh I see you didn't see that it was a system of equations so you just solved for x!

As for the 5/6x confusion you are having, just put the 5/6x in the term you are substituting in for "y", like I did when I said \(\mathrm{Subsititute\:}y=2+\frac{5}{6}\cdot \:x\)