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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Feb 18, 2019 1:35 pm

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GMATH practice exercise (Quant Class 2)

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Answer: [spoiler]____(B)__[/spoiler]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

A*B

by GMATGuruNY » Mon Feb 18, 2019 2:34 pm

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

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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.

The correct answer is B.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Mon Feb 18, 2019 4:48 pm

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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