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A number 84 is divided into two parts. if the difference between half of the first part and one-third of the second part is 12, find two parts of number.

 Feb 2, 2020
 #1
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1/2a - 1/2b = 12,

a +b =84, solve for a,b

a =54  and  b=30

 Feb 2, 2020
 #3
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+1

Remember that the question said "half of the first part and one-third of the second part".smiley

tomsun  Feb 2, 2020
edited by tomsun  Feb 2, 2020
 #4
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Sorry! I took the 2nd part as 1/2 instead of 1/3.

Guest Feb 2, 2020
 #2
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1st part= x     2nd part=y

we just need to set a system of equations:

x+y=84,

1/2x-1/3y=12.

y=84-x

we now know what y equal to, so we put "84-x" into the second equation.

1/2x-1/3(84-x)=12

1/2x-28+1/3x=12

1/2x+1/3x=12+28

3/6x+2/6x=40

5/6x=40

x=48

y=84-48

y=36

we will put the answer in (x,y):

(48,36)

 Feb 2, 2020
 #5
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+1

 

A number 84 is divided into two parts. if the difference between half of the first part and one-third of the second part is 12, find two parts of number.      

 

Call the first part       A

Call the second part  B

 

We know that                                                       A + B = 84              

and that                                                               A/2 – B/3 = 12       

 

From there it's just substitution                            A = 84 – B         above, where it says A we're going to plug this in   

 

                                                                            (84 – B)/2 – B/3 = 12           

Let's multiply everything by 6 and       

get rid of those pesky denominators                    3 • (84 – B) – 2 • B = 6 • 12             

 

                                                                            252 – 3B – 2B = 72                 

Combine like terms                                              252 – 5B = 72                

Subtract 252 from both sides                               –5B = –180            

Divide both sides by –5                                            B = 36              

 

Plug this 36 back into the original equation          A + 36 = 84                      

                                                                                 A = 84 – 36                   

                                                                                 A = 48                        

 

We could have solved for A first; it doesn't matter.                        

Just remember which one you designated as "the first part"

so you know which one to divide by 2 and which one to divide by 3.

.

 Feb 2, 2020
edited by Guest  Feb 2, 2020
 #6
avatar+272 
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nice

tomsun  Feb 2, 2020
 #7
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Hi there...

 

If you know that "(a+b)=84".

And you know that "((a/2)-(b/3))" equals a difference of "12".

You could then say that "(a/2)=((b/3)+12)".
Which equals "24" on both sides.
That in turn means that​ if "(a/2)" equals "24" then "a" must equal "48".

 

And if "(a+b)=84", that means that "b=(84-48)" equals "36".
 

Just to do a quick check...

"((a/2)-(b/3))" = "((48/2)-(36/3))"
That should equal "12", if not check your math.

Because if "((a/2)-((b/3)+12))" is correct, that in turn means that "((48/2)-((36/3)+12))" should equal "0".

 

I hope that helped a bit?


Kind regards
BizzyX


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 Feb 2, 2020
edited by BizzyX  Feb 2, 2020
edited by BizzyX  Feb 3, 2020
edited by BizzyX  Feb 5, 2020
 #8
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All those damned unnecessary brackets again !

Is this some sort of crusade ?

You can't have too many brackets ?

You can, they're a distraction.

Guest Feb 3, 2020
 #9
avatar+78 
0

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Hi there...


- Parentheses are very useful in math.

If you find them so distracting, maybe math is not for you.

 

Personally I find them extremely helpful.

It's way more confusing without them.

 

-----

 

You do you, and I'll do me...

Let's agree to disagree.

 

 

Kind regards

BizzyX

 

 

PS...: 

I reduced the number of Parentheses a bit.
I hope that will keep you from getting so very distracted?

 

 

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BizzyX  Feb 3, 2020
edited by BizzyX  Feb 3, 2020
 #10
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+1

 

Hi, Bizzy.  I have an answer on this thread, too.  I'm not the guest who remarked about the parentheses, but I agree with that guest.  Parentheses are helpful when they're needed, but you're using them unnecessarily.  They make your calculations look busy.  Is that where you got your screen name?  I recognize you're trying to be helpful - we all are - but it's not a lot of help when we have to plow through a million parentheses. 

Guest Feb 3, 2020
 #11
avatar+78 
0

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Hi there...

 

I'll cut back a little.

It's just a habbit of mine.

 

Having mild dyslexia, I find it helpful to compartmentalize each part of an equation.
But maybe doing that to single elements is a bit of an overkill.

 

I'll keep that in mind from now on, when posting.

 

Kind regards

BizzyX


PS...: As a side note...
My nickname "BizzyX" goes way back into the 1980's
Two of my favorite game-series were the "Bubble Bobble"-series and the "Dizzy"-series.
When a few of my friends and I joined the local "DEMO-Scene" on Commodore-computers(Mainly on the "C=64" and/or "C=128".), my nick was actually "Bizzy Bub".
But I later shortend it to what it is today, "BizzyX".
Mainly because I wasn't part of that "DEMO-Scene" anymore, but it's also easier to remember.


So no...
My nickname has got nothing to do with my, for some, a bit excessive use of "Parentheses".

 


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BizzyX  Feb 5, 2020
edited by BizzyX  Feb 5, 2020

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