Posted by caesari on 9/18/2012 10:42:00 PM (view original):
So here is another problem:

4(3y x 6)+11y= 93

Solve for y.

Same as:

4(18y) + 11y = 93
72y + 11y = 93
83y = 93
y = 93/83 ~ 1.12

Posted by caesari on 9/18/2012 10:28:00 PM (view original):

Posted by isack24 on 9/18/2012 10:23:00 PM (view original):

Posted by zbrent716 on 9/18/2012 9:50:00 PM (view original):

Posted by sol_phenom3 on 9/18/2012 7:49:00 PM (view original):
Al, that's wrong.  It's 100 DIVIDED by Y(2+3) = 10.  Not 100 MINUS Y(2+3).  

Regardless, Y is 2.  It's simple math.  100 / 2(2+3) = 10.  Because 2(2+3) is either 2*5 or 2*2 + 2*3, either way, that gives you 10.  100 / 10 = 10.  Problem solved.

OK, so if Y=2 there... solve this:

6 ÷ 2 (1+2) =

9.

Where are people getting confused?

That's what I got too.

Order of Opps is probably throwing people off. I did it wrong the first time too. Parentheses FIRST, then divide 6 by 2, then multiply by 3.

Except I don't think you are correct. At least, not if you agree with the first one that Y = 2.

Plug your answer of Y=2 into the original and solve the same way you do to get 9 as an answer.

100 ÷ 2 (2+3) = 10
100 ÷ 2 (5) = 10
Now, by your use of order of operations (ignoring the distributive property)...
50 (5) = 10
250 = 10

Oops.

Posted by isack24 on 9/18/2012 10:51:00 PM (view original):

Posted by caesari on 9/18/2012 10:42:00 PM (view original):
So here is another problem:

4(3y x 6)+11y= 93

Solve for y.

Same as:

4(18y) + 11y = 93
72y + 11y = 93
83y = 93
y = 93/83 ~ 1.12

Again, gotta distribute...

12y+ 24 + 11y = 93
23 y + 24 = 93
23y = 69
y = 3

Unless you meant:

4(3y + 6) + 11y = 93, in which case...

12y + 24 + 11y = 93
23y + 24 = 93
23y = 69
y = 3

Posted by zbrent716 on 9/18/2012 10:53:00 PM (view original):

Posted by isack24 on 9/18/2012 10:51:00 PM (view original):

Posted by caesari on 9/18/2012 10:42:00 PM (view original):
So here is another problem:

4(3y x 6)+11y= 93

Solve for y.

Same as:

4(18y) + 11y = 93
72y + 11y = 93
83y = 93
y = 93/83 ~ 1.12

Again, gotta distribute...

12y+ 24 + 11y = 93
23 y + 24 = 93
23y = 69
y = 3

Nope, look at the parenthetical.  It's not addition.

Oops, meant addition.

I'll shut up now. LOL

--Caesari

Posted by isack24 on 9/18/2012 10:54:00 PM (view original):

Posted by zbrent716 on 9/18/2012 10:53:00 PM (view original):

Posted by isack24 on 9/18/2012 10:51:00 PM (view original):

Posted by caesari on 9/18/2012 10:42:00 PM (view original):
So here is another problem:

4(3y x 6)+11y= 93

Solve for y.

Same as:

4(18y) + 11y = 93
72y + 11y = 93
83y = 93
y = 93/83 ~ 1.12

Again, gotta distribute...

12y+ 24 + 11y = 93
23 y + 24 = 93
23y = 69
y = 3

Nope, look at the parenthetical.  It's not addition.

Good call; read it too quickly.

What about the original equalling 9 though? If that is true, and y=2 from the first, then it doesn't work, no?

Isn't it sick how we get some sort of disgusting satisfaction out of this?

Posted by isack24 on 9/18/2012 10:23:00 PM (view original):

Posted by zbrent716 on 9/18/2012 9:50:00 PM (view original):

Posted by sol_phenom3 on 9/18/2012 7:49:00 PM (view original):
Al, that's wrong.  It's 100 DIVIDED by Y(2+3) = 10.  Not 100 MINUS Y(2+3).  

Regardless, Y is 2.  It's simple math.  100 / 2(2+3) = 10.  Because 2(2+3) is either 2*5 or 2*2 + 2*3, either way, that gives you 10.  100 / 10 = 10.  Problem solved.

OK, so if Y=2 there... solve this:

6 ÷ 2 (1+2) =

9.

Where are people getting confused?

if 9 is the correct answer to this problem ... and I think it is ... then that means that the (1+2) is on the top and not on the bottom of the equation.

SO ... it is really (6/2) (1+2) = 9

therefore, the first equation is:

(100/y)(3+2)=10
500/y = 10
500 = 10y
50 = y

Checking the problem ...

(100/50) (2+3) = 10

Posted by hughesjr on 9/18/2012 11:00:00 PM (view original):

Posted by isack24 on 9/18/2012 10:23:00 PM (view original):

Posted by zbrent716 on 9/18/2012 9:50:00 PM (view original):

Posted by sol_phenom3 on 9/18/2012 7:49:00 PM (view original):
Al, that's wrong.  It's 100 DIVIDED by Y(2+3) = 10.  Not 100 MINUS Y(2+3).  

Regardless, Y is 2.  It's simple math.  100 / 2(2+3) = 10.  Because 2(2+3) is either 2*5 or 2*2 + 2*3, either way, that gives you 10.  100 / 10 = 10.  Problem solved.

OK, so if Y=2 there... solve this:

6 ÷ 2 (1+2) =

9.

Where are people getting confused?

if 9 is the correct answer to this problem ... and I think it is ... then that means that the (1+2) is on the top and not on the bottom of the equation.

SO ... it is really (6/2) (1+2) = 9

therefore, the first equation is:

(100/y)(3+2)=10
500/y = 10
500 = 10y
50 = y

Checking the problem ...

(100/50) (2+3) = 10

That's what the question is ultimately, and both the original and the follow up were specifically copied as written, complete with possible ambiguity. 

But your analysis is based entirely on trying to get to a particular answer, rather that just doing the math. That's the exact reason I led with the other one, because - in talking to people - everyone who approaches it with a clean slate solves for y=2... even those people who also solve the follow-up to be equal to 9 (but both cannot be true).

Yeah, I  would think 50 is the answer to the first.

I think the natural tendency is to distribute y to the parenthetical and create a 100 / 5y = 10 situation.

But it should read just like hughes wrote it.

Really interesting question, though.  At first blush, I would have said 50 and 9, but I definitely questioned myself when I saw it solved as y=2.  

Of course, I haven't done algebra in about 12 years, so I really don't have any clue other than my gut feeling.  

Posted by isack24 on 9/18/2012 11:03:00 PM (view original):
Yeah, I  would think 50 is the answer to the first.

I think the natural tendency is to distribute y to the parenthetical and create a 100 / 5y = 10 situation.

But it should read just like hughes wrote it.

10 * 5(Y+5) = 125

Y=?

50y + 250 = 125
50y = -125
y = -2.5

Solving 

10 x 5 x (-2.5 + 5) = 125
50 x (2.5) = 125
125 = 125

I think it helps to put in the extra multiplication sign :)

I certainly agree about the ambiguity, but unless there is some special rule of which I'm not aware, I think these are the right answers.