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my fault. I should have said ,I know. I know the result, the doudt is how to get there? I'm suposed to help my 10 years old daughter resolving it for school. coulkd you show a way to help her understand it?

Thank you CPhill

1.833333333333333

4x - 3y = 11    ...(1)

10x + 2y = -1 ...(2)

Multiply the first equation by 2 and the second by 3 so that there are the same number of y's in each:

8x - 6y = 22    ...(3)

30x + 6y = -3  ...(4)

Now add (3) and (4) term by term:

38x + 0 = 19

or

38x = 19

or x = 1/2

Put this back into equation (1)

4*(1/2) - 3y = 11

or

2 - 3y = 11

Subtract 2 from both sides:

-3y = 9

Divide both sides by -3

y = -3

.

Yeah. Anonymous is correct. 

All you got to do is type in the decimal and hit "enter". Tada! 

It's a definite integral so:

.

Holy h**l, that was hard to solve haha!

i manage to multiply 0.47x^2-2000x+1000000 with 100

then divide 47x^2-200000x+100 000 000 with 47

then i just solved it with pq- formula which gave me x = 3675 & x= 579 

Thanks alot for your help, couldnt solve it without you <3 

you can change it by multoplying everything by 100 if you want to.  :))

$$\\0=0.47x^2-2000x+1000000\\\\
0=47x^2-200000x+100000000\\\\$$

Thank you so much !

it did help alot and i got the correct answer  with 579.

But i struggle myself when trying to solve the equation with pq-formula since it has to be x^2+px-q ( cant be 0.47 x^2) could i possibly multiply so x^2 becomes integer ?

Thanks anon :)

I think it has been solved here: http://www.enotes.com/homework-help/b-c-pi-prove-that-identity-true-sin-sin-b-sin-c-154655

$$\\0.53=\frac{(1000-x)^2}{x^2}\\\\
0.53x^2=(1000-x)^2\\\\
0.53x^2=1x^2-2000x+1000000\\\\
0=1x^2-0.53x^2-2000x+1000000\\\\
0=0.47x^2-2000x+1000000\\\\$$

now you can use the quadratic formula - I will use the site calc.

$${\mathtt{0.47}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000\,000}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{10\,000}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{53}}}}{\mathtt{\,-\,}}{\mathtt{100\,000}}\right)}{{\mathtt{47}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{10\,000}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{53}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{100\,000}}\right)}{{\mathtt{47}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{578.700\: \!023\: \!557\: \!336\: \!538\: \!1}}\\
{\mathtt{x}} = {\mathtt{3\,676.619\: \!125\: \!378\: \!833\: \!674\: \!7}}\\
\end{array} \right\}$$

so

$$x\approx 579\qquad or \qquad x\approx 3677$$

Does that help?

Thank you for your answer, this is actully an chemist question but invloves math with 2nd grade equation..

i want to find out x

0.53 = (1000-x)*(1000-x)    /  x^2

this is the real euqation. But i struggle to find get rid of the x^2 in the denominator

would be so gratefull if you understand this :-)

I think that you misunderstood me Sasini.

Can you show me how you got the 2nd last step? 

plz give me those steps. i need them now

I think that you have lost me on the step before that Sasini.   

(And I am usually pretty good at these)

@@ End of Day Wrap      Sun 14/12/14       Sydney,   Australia      Time 12:45 am   (Really Mon morn)        ♬

Hi all,

There were not that many questions or people about today.  I have still been very busy.  Our little forum, often referred to as Camelot, is growing up.  Even when the number of questions is less than usual there is an ever increasing percentage of difficult questions.  This forum is very new.  It is still evolving rapidly.  Sometimes I think of it as a young river gouging its way through the mountians. It is exciting to be a part of this adventure.

Our esteemed answerers today were CPhill, geno3141 and Brittany.  Thanks  

I think Nauseated had a standard Nauseating whinge as well. 

Now for some interest posts:

1) This is a great puzzle.  Thanks flflvm97

Mon 15/12/14

1) I liked Nauseated's answer to this one - it was funny.  Thanks Nauseated.

Not quite anon  :)

$${\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}^{{\mathtt{2}}} = \left(\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,\times\,}}\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)\right)$$

$${\mathtt{1\,000}}{\mathtt{\,\times\,}}\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right){\mathtt{\,-\,}}{x}{\left({\mathtt{1\,000}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}$$

$$1000000-1000x-1000x+x^2$$

$$x^2-2000x+1000000$$

Now you cannot solve anything because there is no equal sign.

really you cannot even simplify because the numerator and the denominator have no common factors.

You can expand the top if you want but that is all you can do :)

1/4 = 2/8

so

$${\frac{{\mathtt{5}}}{{\mathtt{8}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{8}}}} = {\frac{{\mathtt{7}}}{{\mathtt{8}}}}$$

why is  

 $$2x\times x=2x^2$$

because

$$2x\times x=2\times x\times x$$        there is an invisable times sign between the 2 and the x

and

$$2x^2=2\times x^2 = 2\times x \times x$$                 because  $${{\mathtt{x}}}^{{\mathtt{2}}}$$   means  $${\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{x}}$$

Does that help?

Thanks flflvm97

I have given this the green Puzzle icon. :)

And I have added it to the Sticky Topic "Puzzles"

Maybe when you become comfortable with the forum you might like to organise our puzzles for us.

I have saved most of the good ones but I just 'shove them in'  They are not actually organised at all. :(

If you - or anyone else want a job like this then message me please.  

i assume (1000-x)^2 = 1 000 000 - x^2 ?

the equation says =>   0.53 = (1000-x)^2 / x^2

I have added this to the Sticky Topic - "Great Answers to Learn From"