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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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18 Mar 2015, 05:07
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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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18 Mar 2015, 17:15
I don't know if this is a proper way to approach this problem but I'll give a try. The main equation could be rewritten this way: wy*(2x+3y)=(12x)(12+x). So we have two cases where (2x+3y) is equal either to (12x) or (12+x). If 2x+3y=12x 3x+3y=12 x+y = 4, knowing the value of either x or y would help to check this case. If 2x+3y=12+x x+3y=12, this could be a solution, but before concluding the previous case must be confirmed.
(1) Sufficient. This helps us check the case of 2x+3y=12x >x+y=4. As we know that x+2y=11  >x=112y, we can plug it into x+y=4. So, 112y+y=4 > y=7 and x=3, which is not possible, because in the question it's told that x has a positive value.
(2) Not sufficient. Just know the value of w value does not help us in any way to find the answer to the question.
Answer A



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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18 Mar 2015, 19:30
viktorija wrote: I don't know if this is a proper way to approach this problem but I'll give a try. The main equation could be rewritten this way: wy*(2x+3y)=(12x)(12+x). So we have two cases where (2x+3y) is equal either to (12x) or (12+x). If 2x+3y=12x 3x+3y=12 x+y = 4, knowing the value of either x or y would help to check this case. If 2x+3y=12+x x+3y=12, this could be a solution, but before concluding the previous case must be confirmed.
(1) Sufficient. This helps us check the case of 2x+3y=12x >x+y=4. As we know that x+2y=11  >x=112y, we can plug it into x+y=4. So, 112y+y=4 > y=7 and x=3, which is not possible, because in the question it's told that x has a positive value.
(2) Not sufficient. Just know the value of w value does not help us in any way to find the answer to the question.
Answer A hi viktorija, i will not touch on statement l, as you have written it as sufficient.. the method i would go about is.. the value of x as 9 and y as 1 and w as 3 will only be possible satisfying condition of integers... the two eq can be solved for this.. now statement ll..(2) w = 3... take the Q stem equation..x^2 + 2xwy + 3wy^2 = 144.... put w=3...x^2 + 6xy + 9y^2 = 144... (x+3y)^2=144.. so x+3y=12 or 12.. but since all numbers are positive and we are adding two positive value x+3y=12..... so sufficient.. never discard a choice before trying that value in main eq.. ans D...
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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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18 Mar 2015, 19:47
(2) w = 3 => X2+6XY+9Y2 = 144 => (X+3Y)2=144 => SUFFICIENT
(1) X+2Y = 11 => I cannot find the answer: x+3y=?, but I am not sure that I may fall into the TRAP "Impossible for you is not impossible for everyone".
Anw, for me, I will choose B



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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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Bunuel wrote: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) x + 2y = 11 (2) w = 3
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:We have one equation and three variables. Normally, we would need three equations to solve for the values of the individual variables. Here, though, we are asked to find the value of an expression, and in questions of this sort, often we can find the value of the expression without finding the individual variables. We need (x + 3y) Statement #1: x + 2y = 11 Well, this gives us the value of another expression, and if we knew the value of y, we could simply add that to this expression to get the value of (x + 3y). The problem is: this would require us to solve for the individual variables, and we have only two equations for three variables. For example, we could solve this for x, x = 11 – 2y, and plug this into the prompt equation to eliminate x, but that would still leave us with one equation with both y and w as unknowns. One equation, two unknowns. We cannot solve for anything with this information. This statement, alone and by itself, is not sufficient. Statement #2: w = 3 If we plug this into the prompt equation, we get: \(x^2 + 2x(3)y + 3(3)y^2 = 144\) \(x^2 + 6xy + 9y^2 = 144\) This is the square of a sum, because it fits the pattern: \(a^2 + 2ab + b^2 = (a + b)^2\) where a = x, and b = 3y. This means: \((x + 3y)^2 = 144\). Normally, when we take a square root, we would have to consider the ± sign. Here, though, we are guaranteed that all three numbers, x & y & w, are positive, so the sum (x + 3y) would have to be positive. Thus, we can take a square root and be sure the answer will be positive. \(x + 3y = 12\) This statement lead directly to a numerical answer to the prompt question. This statement, alone and by itself, is sufficient. Answer = (B)
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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 04:54
Now I have a question.. How did MAgoosh arrive from \(x^2+6xy+9y^2\) to \((x +3y)^2\)? Ok, they remembered that \((a + b)^2 = a^2 + b^ 2 + 2ab\), right? Now how would I arrive from the right hand side (RHS) to the LHS? I'm asking because I wouldnt be able to cramup all the infinite possible variations in LHS. It's pretty easy to solve from \((a + b)^2\) to get \(a^2 + b^2 + 2ab\) but i dont know how to solve from the later to the former, if I hadn't earlier known that the former equals the later. I guess there should be a way of telling. If my querry made sense please give an educative and honest reply. Thanks.



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 05:16
Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. I think Magoosh official explanation is incorrect. Bunuel and chetan2u please correct me if I am wrong. As per the question stem, \(x^2 + 2xwy + 3wy^2 = 144\) The RHS of this equation is a perfect square (i.e. \(12^2\)). Hence, LHS must also be a perfect square. \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \((\sqrt{3w}y)^2\)} squaring both sides \(w^2 = 3w\) \(w^2  3w = 0\) \(w(w  3) = 0\) this implies, either \(w = 0\) or \(w = 3\). However, \(w\) cannot be equal to \(0\) as we know that \(x\), \(y\) and \(w\) are all positive (given) therefore, \(w = 3\). if \(w = 3\), we can rewrite the given equation as: \((x)^2 + 2(x)(3y) + (3y)^2 = 144\) \((x + 3y)^2 = 12^2\) hence, \(x + 3y = 12\). therefore, we can find solution to the question irrespective of what is given in the statements. Hence, both the statements are sufficient. the correct answer, therefore, should be D



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 05:39
nalinnair wrote: Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. I think Magoosh official explanation is incorrect. Bunuel and chetan2u please correct me if I am wrong. As per the question stem, \(x^2 + 2xwy + 3wy^2 = 144\) The RHS of this equation is a perfect square (i.e. \(12^2\)). Hence, LHS must also be a perfect square. \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \((\sqrt{3w}y)^2\)} squaring both sides \(w^2 = 3w\) \(w^2  3w = 0\) \(w(w  3) = 0\) this implies, either \(w = 0\) or \(w = 3\). However, \(w\) cannot be equal to \(0\) as we know that \(x\), \(y\) and \(w\) are all positive (given) therefore, \(w = 3\). if \(w = 3\), we can rewrite the given equation as: \((x)^2 + 2(x)(3y) + (3y)^2 = 144\) \((x + 3y)^2 = 12^2\) hence, \(x + 3y = 12\). therefore, we can find solution to the question irrespective of what is given in the statements. Hence, both the statements are sufficient. the correct answer, therefore, should be DIt's not clear how did you come up that \(w = \sqrt{3w}\)?
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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 05:50
nalinnair wrote: Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. I think Magoosh official explanation is incorrect. Bunuel and chetan2u please correct me if I am wrong. As per the question stem, \(x^2 + 2xwy + 3wy^2 = 144\) The RHS of this equation is a perfect square (i.e. \(12^2\)). Hence, LHS must also be a perfect square. \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\)From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \((\sqrt{3w}y)^2\)} squaring both sides \(w^2 = 3w\) \(w^2  3w = 0\) \(w(w  3) = 0\) this implies, either \(w = 0\) or \(w = 3\). However, \(w\) cannot be equal to \(0\) as we know that \(x\), \(y\) and \(w\) are all positive (given) therefore, \(w = 3\). if \(w = 3\), we can rewrite the given equation as: \((x)^2 + 2(x)(3y) + (3y)^2 = 144\) \((x + 3y)^2 = 12^2\) hence, \(x + 3y = 12\). therefore, we can find solution to the question irrespective of what is given in the statements. Hence, both the statements are sufficient. the correct answer, therefore, should be DText in red above is incorrect. If you are trying to make a complete square of the expression \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\), you will not end up getting anywhere. \((x+y*\sqrt{3w})^2 = x^2+3w*y^2+2*x*y*\sqrt{3w} \neq x^2 + 2xwy + 3wy^2\) Hope this helps.



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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 06:30
Quote: \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \( Hi nalinnair, coefficient of y is not w but [m]\sqrt{3w}\) your steps hereafter will change.. since 2ab does not have \(\sqrt{3w}\), you cannot change this into the (a+b)^2 ... you may be correct in your observation to some extent.. And i get your point for considering that, but say w is some other value, Y can change to adjust to that value of w.. I have not tried to find answer for that, but there may be other combinations too... you get\(w= \frac{(−x^2+144)}{(2xy+3y^2)}\)..
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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 13:24
nalinnair wrote: Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. I think Magoosh official explanation is incorrect. Bunuel and chetan2u please correct me if I am wrong. As per the question stem, \(x^2 + 2xwy + 3wy^2 = 144\) The RHS of this equation is a perfect square (i.e. \(12^2\)). Hence, LHS must also be a perfect square. \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \((\sqrt{3w}y)^2\)} squaring both sides \(w^2 = 3w\) \(w^2  3w = 0\) \(w(w  3) = 0\) this implies, either \(w = 0\) or \(w = 3\). However, \(w\) cannot be equal to \(0\) as we know that \(x\), \(y\) and \(w\) are all positive (given) therefore, \(w = 3\). if \(w = 3\), we can rewrite the given equation as: \((x)^2 + 2(x)(3y) + (3y)^2 = 144\) \((x + 3y)^2 = 12^2\) hence, \(x + 3y = 12\). therefore, we can find solution to the question irrespective of what is given in the statements. Hence, both the statements are sufficient. the correct answer, therefore, should be DIt is not clear to me either. But Engr2012, if we can get solutions and even the (1) and (2) then the question is faulty. If u are right above, then it's slightsome from Magoosh. But Mahoosh is no fluke at all. Please Brent of GMATPrepNow, I call on you here. You'll be of an exceptional importance in this quant piece. A kudos for your thought... Posted from my mobile device



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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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Nez wrote: nalinnair wrote: Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. I think Magoosh official explanation is incorrect. Bunuel and chetan2u please correct me if I am wrong. As per the question stem, \(x^2 + 2xwy + 3wy^2 = 144\) The RHS of this equation is a perfect square (i.e. \(12^2\)). Hence, LHS must also be a perfect square. \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \((\sqrt{3w}y)^2\)} squaring both sides \(w^2 = 3w\) \(w^2  3w = 0\) \(w(w  3) = 0\) this implies, either \(w = 0\) or \(w = 3\). However, \(w\) cannot be equal to \(0\) as we know that \(x\), \(y\) and \(w\) are all positive (given) therefore, \(w = 3\). if \(w = 3\), we can rewrite the given equation as: \((x)^2 + 2(x)(3y) + (3y)^2 = 144\) \((x + 3y)^2 = 12^2\) hence, \(x + 3y = 12\). therefore, we can find solution to the question irrespective of what is given in the statements. Hence, both the statements are sufficient. the correct answer, therefore, should be DIt is not clear to me either. But Engr2012, if we can get solutions and even the (1) and (2) then the question is faulty. If u are right above, then it's slightsome from Magoosh. But Mahoosh is no fluke at all. Please Brent of GMATPrepNow, I call on you here. You'll be of an exceptional importance in this quant piece. A kudos for your thought... Posted from my mobile device You will not get a unique value for x+3y from statement 1. This is the crux of the question at hand and is the very reason why this statement is NOT sufficient. There is no 1 unique value of w that will make the LHS a perfect square. Can you write what exactly is the issue here? Clearly, statement 1 will not be sufficient on its own to answer the question asked, what is the value of x+3y? FYI, nalinnair 's solution above is not correct while creating perfect squares from the LHS for statement 1.



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 18:41
chetan2u wrote: Quote: \((x)^2 + 2(x)(wy) + (\sqrt{3w}y)^2\) \([(a + b)^2 = a^2 + 2ab + b^2]\) From this, we can infer that \(w = \sqrt{3w}\) {\(w\) being the coefficient of \(y\) in \(2(x)(wy)\) and \(\sqrt{3w}\) being coefficient of \( Hi nalinnair, coefficient of y is not w but [m]\sqrt{3w}\) your steps hereafter will change.. since 2ab does not have \(\sqrt{3w}\), you cannot change this into the (a+b)^2 ... you may be correct in your observation to some extent.. And i get your point for considering that, but say w is some other value, Y can change to adjust to that value of w.. I have not tried to find answer for that, but there may be other combinations too... you get\(w= \frac{(−x^2+144)}{(2xy+3y^2)}\)..Hi Nez and nalinnair, Further to my above post.. Nalin you are correct to an extent that this is the possiblity of finding the solution in one way.. But I have written an Equation to find teh value of w from the original equation.. \(w= \frac{(−x^2+144)}{(2xy+3y^2)}\).. Now you can see w is dependent on x and y.. there can be thousands of answer for this.. we are just given that x and y are positive, they can be fractions, irrational number etc..
1) One solution you have found is correct, where w=\(\sqrt{3}\), and x+3y=12..here too x and y can have various solutions..
2) say x=10 and y=1.. then \(w=\frac{(−x^2+144)}{(2xy+3y^2)}\).. \(\frac{(−10^2+144)}{(2*10*1+3*1^2)}\).. or w=44/23.. substitute all three you will get 144.. 3) say x=1, y=1..
then \(w=\frac{(−x^2+144)}{(2xy+3y^2)}\).. \(\frac{(−1^2+144)}{(2*1*1+3*1^2)}\).. w=143/5..
So you see there are various combinations possible..
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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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01 Mar 2016, 21:19
chetan2u and Engr2012Thank you for the inputs. I can now see my mistake. Regards, Nalin



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If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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02 Mar 2016, 06:04
Nez wrote: Bunuel wrote: Nez wrote: Can someone put some of these solutions in proper encoding? It's really kinda jagged here and crooked. Looks the scene of a fatal bottle fight.. Edited. Thank you for the suggestion. Thanks for that. I hope others take a cue for better learning... Now I have a question.. How did MAgoosh arrive from \(x^2+6xy+9y^2\) to \((x +3y)^2\)? Ok, they remembered that \((a + b)^2 = a^2 + b^ 2 + 2ab\), right? Now how would I arrive from the right hand side (RHS) to the LHS? I'm asking because I wouldnt be able to cramup all the infinite possible variations in LHS. It's pretty easy to solve from \((a + b)^2\) to get \(a^2 + b^2 + 2ab\) but i dont know how to solve from the later to the former, if I hadn't earlier known that the former equals the later. I guess there should be a way of telling. If my querry made sense please give an educative and honest reply. Thanks. Missed this question of yours. When you have any quadratic equation, check whether the discriminant = 0 > if it is, then yes, the given quadratic equation is a perfect square. Refer to my post ( whatisthevalueof9x230xy25y214094.html#p1652927) for a detailed explanation of the formulae involved. If a given quadratic equation is a perfect square, you need to know how to factorize any given quadratic equation in order to save time during GMAT. You were given \(x^2+6xy+9y^2\) . Now the trick here is to find 2 numbers (a and b) that are such that \(a+b = 6xy\) and \(a*b = 9*y^2*x^2\) > \(a=3xy, b = 3xy\) Thus, \(x^2+6xy+9y^2 = x^2+3xy+3xy+9y^2 = x(x+3y)+3y(x+3y)=(x+3y)(x+3y)=(x+3y)^2\) This method can also be applied to simpler perfect squares > \(x^2+24x+144\) > a,b such that \(a+b=24x, a*b = 144x^2\)>\(a=b=12x\) > \(x^2+12x+12x+144 = x(x+12)+12(x+12)=(x+12)(x+12)=(x+12)^2\) Hope this helps.



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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08 Mar 2016, 19:29
Bunuel wrote: If \(x^2 + 2xwy + 3wy^2 = 144\), and if x & y & w are all positive, then what is the value (x + 3y) ?
(1) \(x + 2y = 11\) (2) \(w = 3\)
Kudos for a correct solution. 1. we know nothing about w, so 1 alone is insufficient. 2. w=3 x^2 +6xy+9^2=144 now... (x+3y)^2 = 144 since we are told that x and y are positive then x+3y=12 sufficient



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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w [#permalink]
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21 Oct 2017, 13:32
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Re: If x^2 + 2xwy + 3wy^2 = 144, and if x & y & w are all positive, then w
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21 Oct 2017, 13:32






