48÷2(9+3)
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Young Gotti
Well-known member
288
T-AD
Well-known member
I thought it was...
48/2(9+3) = 48/2(12) = 48/24 = 2
Rodja
Board Sponsor
It is. Order of operations is parenthesis first.
diablosho
Active member
The reason I did it the way I did is because eventually variables will make their way into the equation, and you can't simply add the variables together as you can with constants. Thus, it is important to know how to solve this problem the way I did.
For instance:
48/2(x+y) = 48/(2x + 2y) = (48/2x) + (48/2y) = 24x + 24y. You can't add the 2x + 2y without knowing what x and y equal, thus, it is important to be able to expand the parenthesis by integrating the 2 into each variable.
For instance:
48/2(x+y) = 48/(2x + 2y) = (48/2x) + (48/2y) = 24x + 24y. You can't add the 2x + 2y without knowing what x and y equal, thus, it is important to be able to expand the parenthesis by integrating the 2 into each variable.
Rodja
Board Sponsor
Yes, but they aren't treated the same way to begin with. What you did still breaks a fundamental math operation. It worked it this instance, but it may not always work.
diablosho
Active member
No, I don't think I did! Care to explain? I'm in Calculus, but I still mess up my algebra some times, so I'm not really not trying to be argumentative.
Rodja
Board Sponsor
Order Of Operations:
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction
What you did skips the parenthesis and jumps to the multiplication. On the parenthesis portion of OoO, it is if you can do anything. For example, if they are constants, then you solve (e.g. (9+3)). However, if it is either a mix of variables or constants (e.g. (x+8)), then you cannot do anything with it. Your method worked in this instance, but it may not always work and it is extremely critical to get down fundamentals of math since they never change.
I've done more Algebra than I care to remember. Contrary to what many think, Exercise Physiology is a ton of math. On my comps, there were equations that took over a page after conversions and calculations.
Torobestia
Well-known member
It's 288, where in ****'s name did you all learn your math?
EDIT: LMAO I'm so bad. Sorry. Yes, 2
EDIT: LMAO I'm so bad. Sorry. Yes, 2
rxp1997
Active member
actually what he did will always work
3(12)=36
in other numbers, 3(6+6)=3(6) + 3(6)= 18 + 18 = 36
in other numbers, 3(10+2)=3(10) + 3(2)= 30 + 6= 36
we could go all day...
he just chose to brake out the parantheses first using the applicable rules
diablosho
Active member
While you are correct in theory, the parenthetical equation when multiplied by a constant is nothing more than a factored parenthesis. So, in order to tackle the parenthesis first, you MUST factor in the constant multiple. If you do not, then the variable equation will not work (and there are not two different theorems for solving a parenthesis with variables and parentheses with constants). When you evauluate (5+5), you are assuming 1*(5+5), thus (5*1) + (5*1). That is what the theorem states for evaluating a parenthesis. This is why the theorem applies to variables as well.
Fs-->(5+5)=1(5+5)=((5*1)+(5*1))=((5)+(5))=(5+5)<--Ff equals Fs
Fs-->10(5+5)=((10*5)+(10*5))=((50)+(50))=(50+50)=10(5+5)<--Ff equals Fs
Fs-->(x+y)=1(x+y)=((x*1)+(y*1))=((x)+
)=(x+y)<--Ff equals FsFs-->10(x+y)=((10*x)+(10*y))=((10x)+(10y))=(10x+10y)=10(x+y)<--Ff equals Fs
You follow the same order to solve each type of equation. Each step is simply a way of evaluating a parenthesis, and by following backwards, you will see that while parenthesis first, to solve the parenthesis, you must factor in the constant multiple to each part of the parenthesis.
Young Gotti
Well-known member
i don't do math, i just count money!
T-AD
Well-known member
While you are correct in theory, the parenthetical equation when multiplied by a constant is nothing more than a factored parenthesis. So, in order to tackle the parenthesis first, you MUST factor in the constant multiple. If you do not, then the variable equation will not work (and there are not two different theorems for solving a parenthesis with variables and parentheses with constants). When you evauluate (5+5), you are assuming 1*(5+5), thus (5*1) + (5*1). That is what the theorem states for evaluating a parenthesis. This is why the theorem applies to variables as well.
Fs-->(5+5)=1(5+5)=((5*1)+(5*1))=((5)+(5))=(5+5)<--Ff equals Fs
Fs-->10(5+5)=((10*5)+(10*5))=((50)+(50))=(50+50)=10(5+5)<--Ff equals Fs
Fs-->(x+y)=1(x+y)=((x*1)+(y*1))=((x)+
Fs-->10(x+y)=((10*x)+(10*y))=((10x)+(10y))=(10x+10y)=10(x+y)<--Ff equals Fs
You follow the same order to solve each type of equation. Each step is simply a way of evaluating a parenthesis, and by following backwards, you will see that while parenthesis first, to solve the parenthesis, you must factor in the constant multiple to each part of the parenthesis.
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diablosho
Active member
That may be, but I always feel it is very important to understand WHY you do something so that you can properly apply that knowledge to future problems. This math will ONLY get harder from here, and if he doesn't understand WHY it is that he does what he does, he WILL be screwed later, when he gets to solving pythagorean's theorem/point-slope formula/slope-intercept and trigonometric functions/derivatives/integrals/etc. Learning now is the best way to progress, because ALL of these concepts build on top of the knowledge he is acquiring now. So that being said, I truly hope I didn't confuse the O.P., I really just wanted to help him understand what is happening in the background when we solve these equations. Anywho, I gotta go to school now. I have a Calc. test today (for which I'm REALLY nervous!), so please pray for me LOL! Just kidding, it would be really petty of me to ask that! Anywho, let's everybody have a great day!
--Brian
MidwestBeast
Legend
x
:yup:
:yup:
Rodja
Board Sponsor
Algebra is as hard as it will get for 99% of people. Now, I'll admit that I'm really rusty on my calculus skills as I haven't used those in almost a decade since I graduated HS. That being said, calculus really isn't that hard as long as you take the time to understand the concepts of it and then build off of it. Even to this day, I still hate derivatives, though.
diablosho
Active member
It's funny, I thought I hated limits until derivatives. Then, I thought I hated derivatives until differentials and propogated errors, which led me to hating integrals. Really, the only hard part for me is trig identities. Now I realize I just hate math (although I LOVE math too! LOL)! Calculus is cool, but very difficult.
diablosho
Active member
EXACTLY! And almost 100% of the time, when you get something wrong, it goes back to an algebra or trig mistake. If you don't pay very close attention the whole time, you end up solving for z instead of x, or something dumb like that!
vpop91
New member
It's 288. If you got 2, you assumed an extra parenthesis on the bottom. Typically I write divide by x as multiplying a (1/x) to avoid this sort of thing.
It's ((48)*(9+3))/(2) as written. To do it in my head I'd make it (48/2)*(9+3), then 2*(12*12).
Paste it into wolfram alpha (dot) com if you don't believe me. I can't post links.
It's ((48)*(9+3))/(2) as written. To do it in my head I'd make it (48/2)*(9+3), then 2*(12*12).
Paste it into wolfram alpha (dot) com if you don't believe me. I can't post links.
diablosho
Active member
Try this:
Let x equal (9+3) (or even 12 if you want to) in 48/2(9+3)
We now have 48/2x. Looks clearer now right! This reduces to 24/x, or 24/(9+3), or 24/12, or 2/1, or 2! I think we're right!
Let x equal (9+3) (or even 12 if you want to) in 48/2(9+3)
We now have 48/2x. Looks clearer now right! This reduces to 24/x, or 24/(9+3), or 24/12, or 2/1, or 2! I think we're right!
Red Dog
Well-known member
It's completely dependent on which operation you perform first: multiplication or division. PEMDAS suggests multiplication first (x=2) while SOO weighs multiplication and division equally and moves from left to right (x=288).
Wolfram uses SOO and yields x=288.
It's an intentionally ambiguous equation intended to elicit both answers from a population.
Wolfram uses SOO and yields x=288.
It's an intentionally ambiguous equation intended to elicit both answers from a population.
DAdams91982
Board Sponsor
Someone gets it.
Even with PEMDAS rules it is still 288... why? Because when you were in those weeee little grades we were taught that even though M is written before D they still hold the same weight and should we worked from left to right.
diablosho
Active member
But also that to differentiate two equations from each other when they are being multiplied by each other, you must separate them by parentheses. Since they are not, they must be ONE operation, hence one denominator, not two.
DAdams91982
Board Sponsor
Put this in Google 48/2(9+3). That is exactly how it is written.
it is effectively 24 x 12 = 288.
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Red Dog
Well-known member
It can be written two different ways due to the ambiguity produced by the "divide" symbol.
Depending on how you write it, you'll get two different answers -- which is why nobody uses that symbol.
Strictly based on SOO, and utilizing the ridiculous and ambiguous divide sign exactly as it's written in the equation, it's 288.
Depending on how you write it, you'll get two different answers -- which is why nobody uses that symbol.
Strictly based on SOO, and utilizing the ridiculous and ambiguous divide sign exactly as it's written in the equation, it's 288.
diablosho
Active member
Now I see what you're saying! No matter how you attempt to solve it, you have to assume a pair of parentheses. Either the parentheses go around the "48/2" or the "2(9+3)" to give you "(48/2)*(9+3)" or "48/(2(9+3))". I guess this isn't going anywhere (and fast!Red Dog said:
) but I am curious to see how this turns out! I still believe it is 2, but I'm not really sure how to word my response. I have had to be VERY careful with adding parentheses when using my calculator in calculus, so I have learned not to trust the calculator too much. To ME, it seems obvious the answer is 2. I think it is because in order for the (9+3) to be multiplied by the entire fraction, the fraction must be clearly defined, and it must be bounded by another set of parenthese to indicate multiplication.
biglifter85
Member
Exactly