All Answers 358090 Answers

Thanks Alan,

I understand what you have presented now - I don't remember seeing it presented like that ever before :)

What I don't get is how you can use just the algebra and extract that information, for this exact question, easily.

Let me see how I would extract that:

$$\\x+a-1-xa^2=0 \qquad a\ne0,\;\;x\ne0,\;\;a+x\ne1\\\\
xa^2-1a+(1-x)=0\qquad -1=-x-(1-x)\\\\
xa^2-xa-(1-x)a+(1-x)=0\\\\
xa^2-xa+(1-x)(-a)+(1-x)=0\\\\
xa(a-1)-(1-x)(a-1)=0\\\\
(xa-1+x)(a-1)=0\\\\
$So we have the graphs of$\\\\
xa+x-1=0\qquad and \qquad a-1=0\\\\
-1+xa+x=0\qquad and \qquad a=1\\\\
a=\frac{1-x}{x}\qquad and \qquad a=1\\\\
a=\frac{1}{x}-1 \qquad and \qquad a=1\\\\$$

Now Alan, did you do all that with no working?

Maybe you use mathcad, or a scrap of paper or maybe you are just clever.  

you're all ive ever wanted

i can see it in your smile

i can see it in your eyes 

It plots separate solutions from a single equation in a single colour. e.g. 

.

Thanks Alan,

The bit about the plotting oddity makes sense but the bit about it being 2 graphs really does not.

I mean, I can see how it could be separated so that it is presented and plotted as two graphs.

but it is presented as only one graph so how can you say that it is two  

That is an unusual graph :)

That's because it is actually two graphs:   y = -1 + 1/x  and y = 1

The question that I am now asking myself is how do I determine the ends of the hole around a=1 the graphs indicate that it is more than just one point ????

It's just a plotting oddity (possibly arising because of plotting two functions as though they are 1?).  Plot y=-1+1/x on the same graph and you will see it overwrites the curve without a large hole around x = 1.

.

Hi Rosala,

I have never had pepper and butter popcorn  but it does sound interesting :/    

(I think I would like it)

In Australia hot popcorn is always with salt and butter.  

Fanta would definitely be an improvement on coke.  :)

@@ End of Day Wrap    Thurs 16/7/15     Sydney, Australia Time   7:05pm   ♪ ♫

Hi everyone,

Today our esteemed answerers were Heureka, Alan, CPhill, Geno3141, Radix, Bertie and Dabae. Thank you.  

Nothing,

you need an equal sign with something on both sides if you want to determine the value of x.

You can simplify it

9-6x+1=-6x+10   or    10-6x

step by step working out (1/125)^(1/3) pls

$$\small{\text{$
\begin{array}{rcl}
&&\left( \dfrac{1} {125} \right)^{ \dfrac{1} {3} } \\\\
&=& \dfrac{1} {\left( 125\right)^{ \frac{1} {3} }} \\\\
&=& \dfrac{1} {\left( 5\cdot 5 \cdot 5\right)^{ \frac{1} {3} }}\\\\
&=& \dfrac{1} {\left( 5^3 \right)^{ \frac{1} {3} }}\\\\
&=& \dfrac{1} { 5^{ \frac{3} {3} }}\\\\
&=& \dfrac{1} { 5^1 }\\\\
&\mathbf{=}& \mathbf{ \dfrac{1} {5}}\\\\
\end{array} $}}$$

If it is a calculator output it probably means

$$2e-8=2*10^{-8}$$

otherwise maybe it means

$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{-{\mathtt{8}}} = {\mathtt{0.000\: \!670\: \!925\: \!255\: \!805}}$$

IS THAT RIGHT THAT: X≠0;-a+1

                                  A≠0;-x+1

What is the bar over the top for sabi?    I will ignore it     

$$a \ne 0\;\;\;\;x\ne 0\;\;\;\;x+a\ne 1$$

$$\\\frac{1}{ax}-\frac{a}{x+a-1}=0\\\\
(x+a-1)(ax)[\frac{1}{ax}-\frac{a}{x+a-1}]=0*(x+a-1)(ax)\\\\
(x+a-1)-a(ax)=0\\\\
x+a-1-a^2x=0\\\\
x-a^2x=-a+1\\\\
x(1-a^2)=1-a\\\\
x(1-a)(1+a)=1-a\\\\
1-a=0\quad or \qquad x(1+a)=1\\\\
a=1\;\;\;x\; $can be number except 0 $\quad \quad or \quad x=\frac{1}{1+a}\;\;\;\;where\;\; a\ne 1\\\\$$ 

remembering that

$$a \ne 0\;\;\;\;x\ne 0\;\;\;\;x+a\ne 1$$

The question that I am now asking myself is how do I determine the ends of the hole around a=1 the graphs indicate that it is more than just one point ????

---------------------------------

I will graph it but I will replace 'a' with 'y'

That is an unusual graph :)

His health score will reduce by a factor of 4, so it will be 25% of what it was before. i.e. it will reduce by 75% of its original value.

. 

This question is too general.  Provide some example equations of the type you want to simplify.

.

If you are solving for x, then x = 1/(a+1)    a ≠ -1,  a ≠ 0

if you are solving for a, then there are two solutions:

a = 1      x ≠ 0

and

a = -1 + 1/x    x ≠ 1,  x ≠ 0

.

.

it equals to 4

$$\boxed{
\small{\text{
~~$\frac{1}{2} - \frac{1}{3} = \frac{3}{x}$. What is the value of $x$~?
}}}\\\\\\
\small{\text{$
\begin{array}{rcl}
\dfrac{1}{2} - \dfrac{1}{3} &=& \dfrac{3}{x}\\\\
\dfrac{1}{2}\cdot\dfrac33 - \dfrac{1}{3}\cdot\dfrac22 &=& \dfrac{3}{x}\\\\
\dfrac{3-2}{6} &=& \dfrac{3}{x}\\\\
\dfrac{1}{6} &=& \dfrac{3}{x}\\\\
\dfrac{6}{1} &=& \dfrac{x}{3}\\\\
6 &=& \dfrac{x}{3}\\\\
\dfrac{x}{3} &=& 6 \quad | \quad \cdot 3\\\\
x &=& 6 \cdot 3\\\\
\mathbf{x} & \mathbf{=} & \mathbf{18}\\\\
\end{array}
$}}$$

That is excellent Heather , all except you forgot that the units are  $$m^3$$