MathsGod1

Unless there is a value for X.

That equation is correct already.

I'll try.

I just did 20mins worth of LaTeX...It's gone...

$$\\2)\\\\
\frac{15}{10}\\\\
\frac{5*3}{5*2}\\\\
\frac{\not{5}\;*3}{\not{5}\;*2}\\\\
\frac{3}{2}\\\\\\$$

$$\\3)\\\\
\frac{49}{56}\\\\
\frac{7*7}{7*8}\\\\
\frac{\not{7}\;*7}{\not{7}\;*8}\\\\
\frac{7}{8}\\\\
4)\\\\
\frac{8}{12}\\\\
\frac{4*2}{4*3}\\\\
\frac{\not{4}\;*2}{\not{4}\;*3}\\\\
\frac{2}{3}$$

It might be wrong/presentation might be bad, because i rushed it -sorry .

The problem is with Latex is, when you type in a lot.

The Coding only sends the last part.

So, this time i had to send the code into parts.

Ok, Melody. 

    $$\\Q1)\\\\
\frac{2}{3}\times\frac{5}{7}\\\\
Times\;the\;numerator\;by\;the\;numerator\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
2\times5=35\\
3\times7=21\\\\
\frac{2}{3}\times\frac{5}{7}\;=\;\dfrac{35}{21}\\\\
You\;can\;simplify\;\quad\frac{35}{21}\;\div7=\quad\frac{5}{3}\\\\
\frac{5}{7}\;is\;your\;answer.$$

$$\\Q2)\\\\
\frac{16}{12}\times\frac{15}{10}\\\\
Times\;the\;numerator\;by\;the\;numerator.\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
16\times15=(numerator*numerator)\\\\
To\;this\;without\;calculator.\\\\
10\times5=50\\
10\times10=100\\
6\times5=30\\
6\times10=60\\\\
50+100=150\quad150+60=210\quad210+30=240\\\\
12\times10=(denomiator*denominator)\\\\$$

$$\\To\;do\;this\;without\;a\;calculator.\\\\
10\times10=100\\
2\times10=20\\\\
100+20=120\\\\
16\times15=240\\\\
12\times10=120\\\\
Similify\quad\frac{240}{120}\quad\div120\quad=\frac{2}{1}\\\\
Answer=\frac{2}{1}$$

$$\\Q3\\\\
6\frac{2}{5}\times1\frac{2}{6}\\\\
6\times1=6\\\\
Times\;the\;numerator\;by\;the\;numerator.\\
Times\;the\;denominator\;by\;the\;denominator.\\\\
2\times2=4\\\
5\times6=30\\\\
\frac{2}{5}\times\frac{2}{6}=\frac{4}{30}\\\\
Simplify\quad\frac{2}{30}\quad\div2\quad=\frac{1}{15}\\\\
Answer=6\frac{1}{15}\\
(I\;think...)$$

English.

But if another language is preferred, there are some other forums (when scrolled down).

Like , Russian and French... Ect.

I use Brodude'smethod.

It seems to be simple and easy.

$$\begin{array}{rrlll}\\\\
2x+8&=&x-9\qquad&(-x\;on\;each\;side)\\\\
2x+8(-x)&=&x-9(-x)\\\\
x+8&=&-9\qquad&(-8\;on\;each\;side)\\\\
x+8(-8)&=&-9(-8)\\\\
x&=&-17
\end{array}$$

Sorry! 

-1+10 = 9

If it was 0+10=10

But as it has gone one below zero (a negative) it's 9.

-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 

Let's be specific here.

Grand Piano weighs: Aprrox 252KG

Sun's weighs: 1.989E30 kg

Divide the Pianos weight by the sun's to see how many fit in.

$${\frac{{\mathtt{1\,989\,000\,000\,000\,000\,000\,000\,000\,000\,000}}}{{\mathtt{252}}}} = {\mathtt{7\,892\,857\,142\,857\,142\,857\,142\,857\,142.857\: \!142\: \!857\: \!142\: \!857\: \!1}}$$

Ha...This is it (Approx).

I'm not 100% on how to do this. I'd prefer it if you'd show me.  

I like your way of teaching.

Also, I've seen something like this in my book...

Would you have to factorise to cancel out? (Or am I on the wrong track?)

So far I know:

• Fractions
• Squaring
• Square root
• New Lines
• Colours
• Bigger (lines etc...)
• Array

Maybe other small things, but I've covered the basics...I think?