MaxWong

\(\dfrac{1}{3}+\dfrac{1}{5}\\ =\dfrac{5}{15}+\dfrac{3}{15}\\ =\dfrac{8}{15}\)

I have a quicker way to do fraction addition actually, you can try to learn that or you can just ignore that if you want.

Diagonally multiply and add them up, put it at the numerator's place, multiply the denominators, simplify.

Don't understand? Never mind, Look at the example:

Example:

\(\dfrac{2}{3}+\dfrac{4}{5}\)

Diagonally multiply and add them up: 2 x 5 + 3 x 4 = 10 + 12 = 22

Put it at the numerator's place: we have \(\dfrac{22}{something\; here}\) now.

Multiply the denominators: 3 x 5 = 15. We now have \(\dfrac{22}{15}\), it is improper!!

Therefore we need to Simplify: Final answer = 1 7/15 :D You can check the answer by using the old way :)

\(\dfrac{21}{57}\)

This fraction is not beautiful, why? Because it is not the simplest form :(

Then we will try to make both numerator and denominator smaller and at the same time keep its value the same.

But how??

We will divide it by a fraction which denominator and numerator is the same because if the denominator and numerator is the same, the fraction is equal to 1!

\(\dfrac{21}{57}\div \dfrac{3}{3}\)

Then we will distribute the division sign......

\(\dfrac{21\div 3}{57\div 3}\)

Then divide........

\(\dfrac{7}{19}\)

Yay that fraction is beautiful now because that we cannot do the same process to the fraction. It is because 7 and 19 have no common factors except 1.

What if we have this?

\(\dfrac{42}{126}\)

Repeat the progress with 2 over 2 as they are both divisible by 2!

\(\dfrac{42}{126}\div \dfrac{2}{2}\)

Distribute the division sign:

\(\dfrac{42\div 2}{126\div 2}\)

Divide them:

\(\dfrac{21}{63}\)

This is still not beautiful enough! We have to make it more beautiful!! :(

We can see that they are both divisible by 7, therefore we will repeat the process using 7 over 7.

\(\dfrac{21}{63}\div\dfrac{7}{7}\)

Distribute the division sign:

\(\dfrac{21\div 7}{63\div 7}\)

Divide them:

\(\dfrac{3}{9}\)

NO!! That's not beautiful enough :( 

Repeat the process with 3 over 3.

\(\dfrac{3}{9}\div \dfrac{3}{3}\)

Distribute the division sign:
\(\dfrac{3\div 3}{9\div 3}\)

Divide them:

\(\dfrac{1}{3}\)

Yay that's beautified by me again :D 1 and 3 have no common factors except 1!!

Look at the last 3 digits of the bigger number.(Of course 4875837 is bigger......)

Add the last 3 digits of 4875837(i.e. 837) to the smaller number(i.e. 45), which result is 882.

Put the remaining 4 digits of the bigger number to the result.

Answer: 4875882.

That's my way of quick calculations :D

If you have a enormously large number and a 3 digit number adding together then look at the last 4 digits of the bigger number and repeat what I did.

You do operations like you have learnt in your primary school (or you call it elementary school it's the same) addition subtraction multiplication division.......

Imaginary number is just in the form a times i, the imaginary unit. Something called complex numbers are what you will actually duel with. They are in the form a + bi.

Complex number addition example:

\((3+2i)+(1+4i)\\ = 3+1+2i+4i\\ = 4 + 6i\)

Yeah just like how you duel with variables! Grouping terms!!

Complex number subtraction example:

\((52+6i)-(32 + 14i)\\ = 52 - 32 + 6i - 14i\\ = 20 - 8i\)

Yeah it's just grouping terms again.......

Complex number multiplication example:

\((i+2)(5i-1)\\ =(i)(5i)+(2)(5i)-(1)(i)-(2)(1)\\ =5i^2 + 10i - i - 2\\ =5i^2 + 9i - 2\\ = 9i - 2 - 5\leftarrow\text{Note that }i^2 = -1\\ =9i-7\)

Just what you do with polynomials and adding a rule that i^2 = -1...... Nothing special

Up to this point, there are nothing that feels hard. But the difficulties are all in complex number division.......

Complex number division example:

\(\dfrac{5-2i}{3-4i}\\ =\dfrac{5-2i}{3-4i} \times \dfrac{3+4i}{3+4i}\\ =\dfrac{(5-2i)(3-4i)}{(3-4i)(3+4i)}\\ =\dfrac{7-26i}{9+16}\\ \boxed{\text{Numerator:Just as how I did multilpication above.}\\\text{Denominator: }(a+bi)(a-bi)\equiv a^2+b^2}\\ =\dfrac{7-26i}{25}\\ =\dfrac{7}{25}-\dfrac{26i}{25}\)

Q: How do I know what to multiply in the 1st step?

A: You need a idea called 'conjugates'. Just times the original fraction with a fraction of the original denominator's conjugate over itself(because of the rule of x/x = 1)

Q: How do I think of a conjugate to multiply?

A: Just make sure that after some steps you can use the rule (a+bi)(a-bi) = a^2 + b^2 at the denominator.

Let the original fraction be (6+7i)/(2-5i), Then you will multiply (2+5i)/(2+5i) to it, because after further steps you can use that rule I mentioned at the denominator and that will become a real number(non-imaginary number) at the next step because as you can see, there is no 'i's on the right hand side of the rule (a+bi)(a-bi)=a^2 + b^2.

\(18\div 3 \times 5 - 4\\ = 6\times 5 -4\\ =30 - 4\\ = 26\)

Then I will answer 7 and 8 for Guest #1? :D

7. rs + rt = r(s+t)

8. 2x+10 = 2(x+5)

I am going to do it a different way as Guest #1 did. :)

\(2(x+4)-1=0\\ 2(x+4)=1\leftarrow\text{Add 1 each side}\\ x+4=\dfrac{1}{2}\leftarrow\text{Divide by 2}\\ x = \dfrac{1}{2}-4 = -3\dfrac{1}{2}\leftarrow\text{Minus 4 each side.ANSWER!!YAAAAY!!}\)

Group the u terms: 3u + 4u = 7u

Group the v terms: 7v - 2v = 5v

Add them up: 3u+4u+7v-2v = 7u + 5v

Same answer as SupriseMe

\(ax = y^3\\ 6a = 8^3\\ a = \dfrac{8^3}{6}=\dfrac{256}{3}\\ \text{Therefore }\dfrac{256x}{3}=y^3\)