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.

Guest Feb 2, 2020

#2**+2 **

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)**

tomsun Feb 2, 2020

#5**+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.

_{.}

Guest Feb 2, 2020

edited by
Guest
Feb 2, 2020

#7**0 **

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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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BizzyX Feb 2, 2020

#8**0 **

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**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.

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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

#10**+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**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