Math Discussion

[-Spector-]

Well since we don't have a different sub forum for math talk, I guess I'll just post this here. I was bored so I was taking a math placement test and I came to this problem:

The solution to 5x - ( x - 3) = 2 is also the solution to:

A. 4 x - 3 = 2
B. 4 x = 1
C. 4 x = -5
D. 4 x + 3 = 2
E. I do not know


Here's how I did the problem:

5x - ( x - 3) = 2

5x -x +3 = 2 (Multiplied the negative sign into the parentheses)

4x +3 = 2 (Subtracted x from 5x)

4x = -1 (Subtracted 3 from the left side and added it to the right)

With the final answer being 4x = -1


Did I do something wrong? The only option closest to that is 4x = 1, but I'm not sure how I would of mixed the signs up.

Any help?


P.S. Let's get a Math Theory/Discussion forum FTW!

[!King_Amazon!]

The answer is D, you found that yourself. You just went to far.

[-Spector-]

Damn I'm a dumbass, thanks lol

[Kazilla]

The solution to 5x - ( x - 3) = 2 is also the solution to:

A. 4 x - 3 = 2
B. 4 x = 1
C. 4 x = -5
D. 4 x + 3 = 2
E. I do not know


Here's how I did this problem:

5x - ( x - 3 ) = 2 || (same as) 5x - 1x - 3 = 2.

4x - 3 = 2 || Add 3 to both sides.

4x = 5 || Divide each side by 4

x = 5/4

I think the answer is A

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

The problem I think also lies in the question itself, not only is it a math placement but comprehension as well. It says that "The solution to 5x - ( x - 3) = 2 is also the solution to:" which means, the solution to this equation is the also the solution to one of these.

So after you solve the initial equation you have to plug it into each of the other equations until you come up with a result that = true.

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

Btw, how did your 3 end up positive?

Unless I'm wrong too, its been a while.

[D3V]

Please excuse my dear aunt sally. K_A had it right.

[!King_Amazon!]

Here's how I did this problem:

5x - ( x - 3 ) = 2 || (same as) 5x - 1x - 3 = 2.

4x - 3 = 2 || Add 3 to both sides.

4x = 5 || Divide each side by 4

x = 5/4

I think the answer is A

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

The problem I think also lies in the question itself, not only is it a math placement but comprehension as well. It says that "The solution to 5x - ( x - 3) = 2 is also the solution to:" which means, the solution to this equation is the also the solution to one of these.

So after you solve the initial equation you have to plug it into each of the other equations until you come up with a result that = true.

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

Btw, how did your 3 end up positive?

Unless I'm wrong too, its been a while.

You distribute the negative to the stuff inside of the parenthesis. There's no need to actually solve the problem, because in the process of solving the problem, the problem becomes identical to one of the answers, thus it has the same solution.

[D3V]

Yeah pud.

/closed.

[Kazilla]

You distribute the negative to the stuff inside of the parenthesis. There's no need to actually solve the problem, because in the process of solving the problem, the problem becomes identical to one of the answers, thus it has the same solution.
Saw that just before you stated it, I forgot about that little rule.

[Chruser]

Arbitrary mimeTeX tests.

http://83.226.212.89/cgi-bin/mimetex.cgi?\white\Large\LARGE%20e^x=\sum_{n=0}^\i nfty\frac{x^n}{n!}


http://83.226.212.89/cgi-bin/mimetex.cgi?\white\Large\normalsize\left(\large\be gin{array}{GC+23}\varepsilon_x\\\varepsilon_y\\\va repsilon_z\\\gamma_{xy}\\\gamma_{xz}\\\gamma_{yz}\ end{array}\right)\ {\Large=}\ \left[\begin{array}{CC} \begin{array}\frac1{E_{\fs{+1}x}} &-\frac{\nu_{xy}}{E_{\fs{+1}x}} &-\frac{\nu_{\fs{+1}xz}}{E_{\fs{+1}x}}\\ -\frac{\nu_{yx}}{E_y}&\frac1{E_{y}}&-\frac{\nu_{yz}}{E_y}\\ -\frac{\nu_{\fs{+1}zx}}{E_{\fs{+1}z}}& -\frac{\nu_{zy}}{E_{\fs{+1}z}} &\frac1{E_{\fs{+1}z}}\end{array} & {\LARGE 0} \\ {\LARGE 0} & \begin{array}\frac1{G_{xy}}&&\\ &\frac1{G_{\fs{+1}xz}}&\\&&\frac1{G_{yz}}\end{array} \end{array}\right] \ \left(\large\begin{array} \sigma_x\\\sigma_y\\\sigma_z\\\tau_{xy}\\\tau_{xz} \\\tau_{yz} \end{array}\right)

[Kazilla]

Chruser, I'm curious, what is your IQ?

[Grav]

What is this, strain distribution class? Get out of here Denton.

[Kazilla]

Dude, i had a teacher named Denton, she pissed me the fuck off.