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OpenStudy (anonymous):

What is the solution to the following equation? 1/3 (p+9)=3+2(p-6)

OpenStudy (anonymous):

@iGreen

OpenStudy (vishweshshrimali5):

\[\large{\cfrac{1}{3} (p+9) = 3 + 2(p-6)}\]

OpenStudy (vishweshshrimali5):

First thing first... I don't like that 1/3 stuff, so lets multiply by 3 on both sides of `=` sign.

OpenStudy (vishweshshrimali5):

What do we get? We get : \[\large{p+9 = 3(3+2(p-6))}\]

OpenStudy (vishweshshrimali5):

Any doubt ?

OpenStudy (anonymous):

No

OpenStudy (vishweshshrimali5):

Good... Now, lets simplify that a little bit more... \[\large{p + 9 = 3 * 3 + 3 * 2 * (p-6)}\] \[\large{\implies p + 9 = 9 + 6 * (p-6)}\]

OpenStudy (vishweshshrimali5):

Clear ?

OpenStudy (igreen):

@vishweshshrimali5 That's one way to do it.. We could just distribute 1/3 into the parenthesis: \(\dfrac{1}{3} (p + 9) \rightarrow \dfrac{1}{3}p + 3\) Plug that back in: \(\dfrac{1}{3}p + 3 = 3 + 2(p - 6)\) Now distribute 2 into the parenthesis, do you know how to do that? @-EtherealEbullience

OpenStudy (vishweshshrimali5):

Yep... its just that... writing fraction in latex takes more time ;)

OpenStudy (igreen):

Lol

OpenStudy (anonymous):

1/3p+3=3+2(12p)?

OpenStudy (vishweshshrimali5):

Did you forget something ??

OpenStudy (vishweshshrimali5):

\[\large{a * (\color{red}{x} + \color{green}{y}) = a\color{red}{x} + a\color{green}{y}}\]

OpenStudy (igreen):

No..distributing 2 will give us: \(2(p - 6) \rightarrow (2 \times p) + (2 \times -6)\)

OpenStudy (anonymous):

Oh

OpenStudy (vishweshshrimali5):

Yep... so what would you get ?

OpenStudy (anonymous):

9?

OpenStudy (igreen):

No..

OpenStudy (igreen):

How did you get that?

OpenStudy (igreen):

Can you multiply these? \(2 \times p\) \(2 \times -6\)

OpenStudy (anonymous):

-12

OpenStudy (vishweshshrimali5):

Good...

OpenStudy (vishweshshrimali5):

Now : \[\dfrac{1}{3}p + 3 = 3 + 2*p - 12\]

OpenStudy (vishweshshrimali5):

Can you solve this ?

OpenStudy (anonymous):

1/3p+3=27p?

OpenStudy (vishweshshrimali5):

27 ? How ?

OpenStudy (vishweshshrimali5):

\[\dfrac{1}{3}p + 3 = 3 + 2*p - 12\] Lets see... I am going to leave all the `p's` alone. So: \[\dfrac{1}{3}p + \color{red}3 = \color{red}3 + 2*p - \color{red}12\] I am going to simplify only the red colored part

OpenStudy (vishweshshrimali5):

Sorry, its this: \[\dfrac{1}{3}p + \color{red}3 = \color{red}3 + 2*p - \color{red}{12}\] \[\implies \cfrac{1}{3}p = (\color{red}{3-12-3}) + 2*p\]

OpenStudy (vishweshshrimali5):

What is the value of the red coloured part ?

OpenStudy (anonymous):

-12

OpenStudy (vishweshshrimali5):

Great !!

OpenStudy (vishweshshrimali5):

So, we have: \[\implies \cfrac{1}{3}p = (\color{red}{3-12-3}) + 2*p\] \[\implies \cfrac{1}{3}p = -12 + 2p\] Now lets multiply by 3 both sides

OpenStudy (vishweshshrimali5):

We get: \[\large{3*\cfrac{1}{3} * p = 3 * (-12 + 2*p)}\] \[\large{\implies p = 3*(-12) + 3 * 2*p}\] \[\large{\implies p = -36 + 6p}\] \[\large{\implies p + 36 = 6p}\] Any doubt in any step ?

OpenStudy (anonymous):

No

OpenStudy (vishweshshrimali5):

Good. So, we have: \[\large{p + 36 = 6p}\] \[\large{36 = 6p - p}\] \[\large{5p = 36}\]

OpenStudy (vishweshshrimali5):

Now lets divide by 5 on both sides

OpenStudy (anonymous):

5/36

OpenStudy (vishweshshrimali5):

Why ?

OpenStudy (anonymous):

. . .Because that's an answer choice. And you can't simplify it anymore. . .

OpenStudy (vishweshshrimali5):

See: \[\large{\color{red}{\cfrac{1}{5}} * 5p = \color{red}{\cfrac{1}{5}} * 36}\] \[\large{\implies p = \cfrac{36}{5}}\]

OpenStudy (igreen):

^^

OpenStudy (anonymous):

. . . : I

OpenStudy (igreen):

\(\dfrac{36}{5} = 7.2\)

OpenStudy (vishweshshrimali5):

Well, there we go :)

OpenStudy (anonymous):

. . . x3 Thank you.

OpenStudy (vishweshshrimali5):

Your welcome :)

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