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x + 9 = 18 + -2x

$$\begin{array}{rcl}
x + 9 &=& 18 + -2x \quad | \quad +2x \\
2x+x+9 &=& 18 \\
3x+9 &=& 18 \quad | \quad -9\\
3x &=& 18-9 \\
3x &=& 9 \quad | \quad :3\\
x &=& \frac{9}{3} \\
x &=& 3
\end{Array}

\text{Proof:} \\
3+9=18-2*3 \\
12 = 18-6 \\
12 = 12 \text{okay!}$$

Why don't you people write continuously ......it would be more fun......looks like someone's serious about conserving energy Hun?😏

1/2-what=-1 3/4

$$\begin{array}{rcl}
\frac{1}{2} -x &=& -1 \frac{3}{4} \quad | \quad *(-1)\\\\
-\frac{1}{2}+x &=& 1 \frac{3}{4} \quad | \quad +\frac{1}{2}\\\\
x &=& 1 \frac{3}{4}+\frac{1}{2}\\\\
x &=& \frac{4*1+3}{4}+\frac{1*2}{2*2}\\\\
x &=& \frac{7}{4}+\frac{2}{4}\\\\
x &=& \frac{9}{4}\\\\
x &=& \frac{8+1}{4}\\\\
x &=& \frac{8}{4}+\frac{1}{4}\\\\
x &=& 2+\frac{1}{4}\\\\
x &=& 2\frac{1}{4}
\end{array}$$

Let's see...working with the LHS, we have

- sin(theta)/cot(theta)-sec(theta) = -cos(theta)

-sin(x) / [cos(x)/sin(x)] - 1/cos(x)

-sin^2(x) / cos (x) - 1/cos(x)

- [ 1 + sin^2(x) ] / cos(x) =  -cos(x)   ???.....nope....no way this is true....!!!

And

cos(theta)+tan(theta)xsin(theta)= sec(theta)

Working on the LHS, we have....

cos(x) + [ sin^2(x) / cos(x) ]

[cos^2(x) + sin^2(x) ] / cos(x) =

1 / cos(x)  =  sec(x).......  and this one is true....!!!!

Yes it is an entertaining thread - I will give you that     

There is nothing to excuse Heureka, you beat me fair and square.  LOL  

1/z + x/y = 5/y - 3x / 2y     rearrange

3x / 2y  + x / y  = 5/y - 1/z    simplify

5x / 2y  = [5z - y ] / yz

5x = 2[5z - y ] / z

x  = (2/5) [ 5z - y ] / z = [ 10z - 2y ] /  [5z] = 2 - 2y/5z = 2 -  (2/5) (y/z)....but (y/z)  = 5/3..so we have

x =2 - (2/5)(5/3)  =  2 - 2/3  = 6/3 - 2/3 = 4/3

Hi,  Melody

excuse please

Hi Alan,

thank you very much.

I have replaced 024 with 045

You beat me by a hair heureka     

 next,if x:y:z=2:5:3 , find the value(s) of x satisfyinh the formular in (a)

$$\\y=\frac{x}{2}*5=\frac{5x}{2}, \qquad z=\frac{x}{2}*3=\frac{3x}{2}\\\\\\
x=\frac{10z-2y}{5z}\\\\
x=\frac{10*\frac{3x}{2}-2*\frac{5x}{2}}{5*\frac{3x}{2}}\\\\
x=\frac{15x-5x}{\frac{15x}{2}}\\\\
x=10x*\frac{2}{15x}\qquad \mbox{Note x cannot equal 0}\\\\
x=\frac{4}{3}$$

I found the answer in (a) is $${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}{{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{zy}}\right)}}$$ , how to solve (b)?

$$x=\dfrac{10zy-2y^2}{5zy} = 2-\dfrac{2}{5}*\dfrac{y}{z} \quad | \quad \dfrac{y}{z} = \dfrac{5}{3}\\\\
x= 2 - \frac{2}{3} = 1\frac{1}{3}$$

I think that your part a question can be simplified further to

$$x=\dfrac{10z-2y}{5z}$$

You've had a mental blackout there somewhere heureka!! (Don't worry, I have them all the time!):

(In your step II you should probably replace 024 with 045)

.

This link must go in my next laugh! 

I think today CPhill is saving words Hun?💯% saved

 And that's the truth....!!!!!

Yeah, we probably are! And we should save " energy" shouldn't we ...........Melody?

😏♻

Rosala and I aren't lazy......we just live by this motto.....

Well, you are welcome to join Melody!😎

Hey, how are those comments on your laziness worth 3 points ??????        LOL

Oh you are a lazy pair LOL

Yeah! You know I'm on holidays and tomorrows school so anyways the teachers will squeeze me in the firsts day so it's really boring working out a 100 word question! 

Me too, rosala.....I don't want to have to work too hard.....!!!!

@@ End of Day Wrap     Wed  14/1/15     Sydney, Australia       Time 8:30pm       ♪ ♫

Good evening everyone  

Lots of questions today.   Some on earlier pages have not yet been answered, if you want to repost please follow the repost instructions laid out in the Lantern post, and scattered through the forum. I must put a copy in the Sticky Topics.

Anyway, great answers were delivered by Heureka, YehChi. Geno3141, CPhill, Rampager1998, Sangyun, MathWarrier, GenericUserName, Tetration, KingPiFirePlasma, Rosala, and Sasini.   A big thankyou to each or you.  

Interest posts:

1) End your day with a Laugh.   Thanks Rosala  

Oh melody but they are all high class ones I'm looking for some easy ones!!!ya know like ,,,,,,,,,,, 2+2 😜😊😂