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If A and B are two fixed constants such that the system

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If A and B are two fixed constants such that the system  [#permalink]

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New post 18 Feb 2019, 14:42
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A
B
C
D
E

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  25% (medium)

Question Stats:

79% (02:21) correct 21% (02:19) wrong based on 19 sessions

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GMATH practice exercise (Quant Class 2)

\(\left\{ \matrix{
\,2\left( {x + y} \right) - A = 0 \hfill \cr
\,3x + By - 6 = 0 \hfill \cr} \right.\)

If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?

(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

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Re: If A and B are two fixed constants such that the system  [#permalink]

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New post 18 Feb 2019, 15:50
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fskilnik wrote:
GMATH practice exercise (Quant Class 2)

\(\left\{ \matrix{
\,2\left( {x + y} \right) - A = 0 \hfill \cr
\,3x + By - 6 = 0 \hfill \cr} \right.\)

If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?

(A) 10
(B) 12
(C) 14
(D) 16
(E) 18


The system will have an infinite number of solutions for (x, y) if the two equations are THE SAME:
2(x+y) - A = 0 --> 2x + 2y - A = 0 --> 6x + 6y - 3A = 0
3x + By - 6 = 0 --------------------------> 6x +2By - 12 = 0
The equations in blue will be the same if 3A=12 and 2B=6, with the result that A=4, B=3, and A*B=12.

.
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Re: If A and B are two fixed constants such that the system  [#permalink]

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New post 18 Feb 2019, 18:08
fskilnik wrote:
GMATH practice exercise (Quant Class 2)

\(\left\{ \matrix{
\,2\left( {x + y} \right) - A = 0 \hfill \cr
\,3x + By - 6 = 0 \hfill \cr} \right.\)

If \(A\) and \(B\) are two fixed constants such that the system of equations given above has more than one ordered pair \((x,y)\) in its solution set, what is the value of \(AB\) ?

(A) 10
(B) 12
(C) 14
(D) 16
(E) 18

Mitch´s nice solution is based on the fact that (A and B already fixed,) the solution sets of ordered pairs (x,y) that satisfy each equation alone represent (each one) a line in the rectangular coordinate system.

We have two lines in the plane, hence they are concurrent (just one point (x,y) in common), parallel and distinct (no point (x,y) in common) or parallel and coincident (every point (x,y) in common).

From the question stem, we are looking for the last scenario, and Mitch´s solution is validated.

Let me present an alternate solution, using not only algebraic operations, but only algebraic arguments/insights. (We offer this problem in our SECOND class... no Analytic Geometry yet!)


\(? = A \cdot B\,\,\,\,\left( {{\rm{system}}\,\,{\rm{with}}\,\,{\rm{more}}\,\,{\rm{than}}\,\,{\rm{one}}\,\,{\rm{solution}}} \right)\)

\(\left\{ \matrix{
\,2\left( {x + y} \right) - A = 0 \hfill \cr
\,3x + By - 6 = 0 \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\,\left\{ \matrix{
\,2x + 2y = A\,\,\,\left( { \cdot \,3} \right) \hfill \cr
\,3x + By = 6\,\,\,\left( { \cdot \,2} \right) \hfill \cr} \right.\,\,\,\,\, \cong \,\,\,\,\,\left\{ \matrix{
\,6x + 6y = 3A\,\,\left( * \right) \hfill \cr
\,6x + 2By = 12 \hfill \cr} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,y\left( {6 - 2B} \right) = 3\left( {A - 4} \right)\,\,\,\,\left( {**} \right)\)

\(6 - 2B \ne 0\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,y = {{3\left( {A - 4} \right)} \over {6 - 2B}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x = {1 \over 6}\left\{ {3A - 6\left[ {{{3\left( {A - 4} \right)} \over {6 - 2B}}} \right]} \right\}\,\,\,\,\mathop \Rightarrow \limits^{A,B\,\,{\rm{given}}} \,\,\,\,\left( {x,y} \right)\,\,\,{\rm{unique}}\,\,{\rm{solution}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{impossible}}\)

\(6 - 2B = 0\,\,\,\, \Rightarrow \,\,\,\,B = 3\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,A = 4\,\,\,\, \Rightarrow \,\,\,\,\,? = AB = 12\)


The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

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Re: If A and B are two fixed constants such that the system   [#permalink] 18 Feb 2019, 18:08
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If A and B are two fixed constants such that the system

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