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09 Nov 2015, 04:21
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Guys, can you help me with this simple question about factoring an equation ?
A question on quantitative part in OG 2016 ask to solve Y in terms of X. The equation is X^2.Y^2 - X.Y = 6
I already know how to find de answer. What I would like to know is why the first step that I did wasn't correct. This was my first step to find the solution:
X^2.Y^2 - X.Y = 6
X.Y(X.Y - 1) -6 = 0
X.Y - 6 =0
X.Y = 6
Y = 6/X
X.Y(X.Y - 1) -6 = 0
X.Y - 1 = 0
X.Y = 1
Y = 1/X
I know that the factorization in the equation ( X^2.Y^2 - X.Y = 6) is wrong. But can someone help me and explain how to factor this equation in the right way (I'm curious for know it).
Tks everybody
A question on quantitative part in OG 2016 ask to solve Y in terms of X. The equation is X^2.Y^2 - X.Y = 6
I already know how to find de answer. What I would like to know is why the first step that I did wasn't correct. This was my first step to find the solution:
X^2.Y^2 - X.Y = 6
X.Y(X.Y - 1) -6 = 0
X.Y - 6 =0
X.Y = 6
Y = 6/X
X.Y(X.Y - 1) -6 = 0
X.Y - 1 = 0
X.Y = 1
Y = 1/X
I know that the factorization in the equation ( X^2.Y^2 - X.Y = 6) is wrong. But can someone help me and explain how to factor this equation in the right way (I'm curious for know it).
Tks everybody
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
09 Nov 2015, 04:28
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phelipemariano
This question is discussed here: if-xy-0-and-x-2-y-2-xy-6-which-of-the-following-could-be-y-126166.html
Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm
Hope it helps.
09 Nov 2015, 20:34
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phelipemariano
To point out the error you have made here, I ask you to think about this:
How will you solve: \(a(a - 1) - 6 = 0\) ?
Is it the same as: \((a - 1)(a - 6) = 0\) ?
Note that the two are not same.
\(a(a - 1) - 6 = 0\) actually is \(a^2 - a - 6 = 0\)
\((a - 1)(a - 6) = 0\) actually is \(a^2 - 7a + 6 = 0\)
What is the same as \((a - 1)(a - 6) = 0\) ?
When you have \(a(a - 1) - 6(a - 1) = 0\), then you can take (a - 1) common and you are left with \((a - 1)(a - 6) = 0\)
Hope this helps.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
