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.
+298 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)
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.
.
+87 -----
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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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.
+87 -----
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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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.
+87 -----
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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