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

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 2)

Click on the image to open it in a much larger version!



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

[GMATGuruNY]

GMATH practice exercise (Quant Class 2)

Click on the image to open it in a much larger version!

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.

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 2)

Click on the image to open it in a much larger version!

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.