What is the solution to the following equation?
1/3 (p+9)=3+2(p-6)
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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 ?
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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 ;)
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No..distributing 2 will give us:
\(2(p - 6) \rightarrow (2 \times p) + (2 \times -6)\)
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OpenStudy (anonymous):
Oh
OpenStudy (vishweshshrimali5):
Yep... so what would you get ?
OpenStudy (anonymous):
9?
OpenStudy (igreen):
No..
OpenStudy (igreen):
How did you get that?
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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 ?
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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
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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
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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. . .
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