System of equations in two variables

[jodiwv]

Need help solving using method 2
x+1=2/3x
I do not understand the steps. :shock:

Jodiwv

 

[mmm4444bot]

What is method 2 ?

 

[jodiwv]

example: -2x+13=4x-17
using method one you graph two functions
using method two you graph one function
does this help?

 

[mmm4444bot]

Okay, I think they want you to subtract all of the terms to one side, so that the equation equals zero.

In this case, y = 0, so you would graph the line and look for the x-intercept.

Can you subtract (2/3)x from both sides ?

You'll get a result that looks like 0 = m*x + b.

 

[BigGlenntheHeavy]

$\displaystyle x+1 \ = \ \frac{2}{3x}, \ cross-multiplimg \ gives \ 3x^2+3x \ = \ 2, \ \implies \ 3x^2+3x-2 \ = \ 0,$

$\displaystyle now \ solve \ for \ x \ using \ the \ quadratic \ formula.$

 

[jodiwv]

The sample shows f(x)=-6x+30 and g(x)=0
my original problem is x+1=2/3x
my solution was -1 2/3 x +1
but the graph does not work

 

[mmm4444bot]

I interpret the expression 2/3x to mean 2/3 * x.

In other words, x is in the numerator: 2x/3.

Glenn interpreted 2/3x to mean 2/(3x),

In other words, x is in the denominator.

Which is it ?

Also, your sample has two different functions: f and g. I don't understand what two functions have to do with the "original problem".

Finally, you titled this post "System of equations in two variables".

You posted only one equation, and it containes only one variable.

I'm sorry. I simply do not understand what you've been asked to do, and I do not understand your typing.

The solution to the equation x + 1 = 2/3 x is x = -3.

If that is not what you want, then I suggest that you post the entire exercise and type your expressions using proper notation.

 

[Denis]

x+1=2/3x

That means x + 1 = 2x / 3 : whassamattayou BigG?!
SO:
3x + 3 = 2x
x = -3

Jodi, next time you post a problem, post it CLEARLY and IN FULL.
Nobody here is clairvoyant (except Subhotosh!)

 

[BigGlenntheHeavy]

$\displaystyle Denis, \ asleep \ at \ the \ wheel, \ perhaps?$

 

[jodiwv]

Thanks to all three of you. I turned the paper in and got 100%. Thanks again!

 

[lookagain]

From the order of operations, $\displaystyle 2/3x = \frac{2}{3}x = \frac{2x}{3}.$

At the risk of redundancy, the original poster should have either posted one of the two forms
(if in horizontal form), depending on which was meant:

$\displaystyle (2/3)x$ or $\displaystyle 2/(3x)$, using appropriate grouping symbols.

 

[mmm4444bot]

It would be very nice, were I to get $1 for every post on these boards lacking grouping symbols.