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

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 2)

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MitchÂ´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 MitchÂ´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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