T=5((x−2)/4)+1

- anonymous

T=5((x−2)/4)+1

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

and solve for what?

- anonymous

x

- anonymous

sorry

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

steps please.

- anonymous

I answered that.

- anonymous

i still dont understand it though

- anonymous

Do you know why you can equate two sides of an equation?

- anonymous

\[x=\frac{2}{5} (3+2 T) \]

- anonymous

i know about the order of operations thats not what I need to know.

- anonymous

i need all steps from beginning please!

- anonymous

Robtobey how did you get your solution?

- anonymous

there is no 10 in the equation...

- anonymous

*facepalm*

- anonymous

jwaddell
your equation is little confusing. if solve for x. then x would equal some number with T in them.

- anonymous

yes i know

- anonymous

if you solve it that way then.
\[x=(4(\frac{T}{5})-1)+2\]

- anonymous

WOW the T is throwing me off

- anonymous

first, distribute 5 to (x-2)/2 sop, we get (5x-10)/4 + 1=0, if T=0
then, the LCD is 4, so it will become
(5x-10+4)/4=0
5x-6=0
5x=6
x=1.2

- anonymous

if T=0
then
x=1.1999999999

- anonymous

or, if the value of T is not given, then, x=(4T+6)/5

- anonymous

Pat yours makes sense thanks!

- anonymous

you are welcome.

- anonymous

thanks for the medal.

- anonymous

no prob you wanna help me with one more?

- anonymous

sure.

- anonymous

\[A= (3m+2)/(m-1)\]

- anonymous

solve for m

- anonymous

m=(A(m-1)-2)/3

- anonymous

that is not right pat. since the other m is there also.

- anonymous

yeah, i had a mistake, i did not see it. thanks

- anonymous

m=(A+2)/(A-3)
what did you get?

- anonymous

can u show me the steps too please

- anonymous

m=(-2-A)/3-A

- anonymous

(3m+2)/(m-1)=A, then cross multiply, you will get
3m+2=Am-A
3m-Am=-2-A
m(3-A)=-2-A
m=(-2-A)/(3-A)

- anonymous

would't it be Am-3m=2-A???

- anonymous

or actually Am-3m-2+A

- anonymous

jwaddell06,
Do you still want to see how the first problem was solved, now that the dust has settled?

- anonymous

\[T=\frac{5 (x-2)}{4}+1 \]\[T-1=\frac{5 (x-2)}{4} \]\[4(T-1)=5 (x-2)\]\[\frac{4(T-1)}{5}=(x-2)\]\[\frac{4(T-1)}{5}+2=x \]Simplify the left side to obtain:\[\frac{2}{5} (2 T+3)=x \]

- anonymous

now why wouldnt you multiply the 5 into x-2

- anonymous

The object is to isolate x on the right hand equation side. The easiest path for me was to divide by 5 leaving x - 2. When +2 is added to each side, x is left alone.

- anonymous

That makes sense a little but i thought whenever you were given something like 5(x-2) you were supposed to make it 5x-10 then isolate x from there.

- anonymous

Well the final result should be OK. Give me a minute to work it out.

- anonymous

\[4(T-1)=5 (x-2) \]\[4(T-1)=5 x-10 \]\[4(T-1)+10=5 x \]\[\frac{4(T-1)+10}{5}=x \]\[\frac{2}{5} (2 T+3)=x \]

- anonymous

ok i get it up to the last part how do you get 2/5(2T+3)

- anonymous

\[\frac{4(T-1)+10}{5} \]Simplify the numerator\[\frac{(4 T+6)}{5}\]Factor the numerator\[\frac{(2 (2 T+3))}{5}=x \]Factor the expresssion\[\frac{2}{5} (2 T+3) \]
I actually use Mathematica to do the calculations and had to resolve the problem by hand to answer you follow up questions.

- anonymous

what is mathematica??

- anonymous

I have to break off for about 10 minutes. Will be back then to answer any other questions.

- anonymous

i dont understand how to simplify the numerator like that. sorry its been 10 years since I did algebra if not longer.

- anonymous

what is mathematica?? Here is an introductory video.
http://www.wolfram.com/solutions/education/students/

- anonymous

Simplification is a nicety, but not required to present a valid problem solution.

- anonymous

You're still looking at this .-.

Looking for something else?

Not the answer you are looking for? Search for more explanations.