fskilnik wrote:

GMATH practice exercise (Quant Class 10)

\(\left\{ \matrix{

\,x - cy = 1 \hfill \cr

\,cx - 4y = c \hfill \cr} \right.\)

If \(c\) is a given constant such that the system of equations given above has exactly one ordered pair (x,y) as its solution, which of the following must be true?

\(\eqalign{

& \left( A \right)\,\,\,c > 0 \cr

& \left( B \right)\,\,\left| c \right| > 1 \cr

& \left( C \right)\,\,\left| c \right| < 1 \cr

& \left( D \right)\,\,\left| c \right| = 2 \cr

& \left( E \right)\,\,\left| c \right| \ne 2 \cr}\)

\(?\,\,\,:\,\,\,c\,\,\,{\rm{for}}\,\,{\rm{unique}}\,\,{\rm{solution}}\,\,\left( {x,y} \right)\,\,\)

\(c = 0\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{

\,x = 1 \hfill \cr

\, - 4y = 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refutes}}\,\,\left( A \right),\left( B \right),\left( D \right)\)

\(c = 3\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{

\,x - 3y = 1 \hfill \cr

\,3x - 4y = 3 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{

\,x - 3y = 1 \hfill \cr

\,x - {4 \over 3}y = 1 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{

\,x - 3y = 1 \hfill \cr

\,\left( {1 + 3y} \right) - {4 \over 3}y = 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{unique}}\,\,{\rm{solution!}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refutes}}\,\,\left( C \right)\)

The answer is (E) by exclusion!

We follow the notations and rationale taught in the

GMATH method.

Regards,

Fabio.

POST-MORTEM:\(\left\{ \matrix{

\,x - cy = 1 \hfill \cr

\,cx - 4y = c \hfill \cr} \right.\,\,\,\,\,\, \cong \,\,\,\,\,\,\left\{ \matrix{

\,x - 1 = cy \hfill \cr

\,cx - c = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{

\,x - 1 = cy \hfill \cr

\,c\left( {x - 1} \right) = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{

\,x - 1 = cy \hfill \cr

\,c \cdot cy = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{

\,x - 1 = cy \hfill \cr

\,y\left( {{c^2} - 4} \right) = 0 \hfill \cr} \right.\)

\(\left| c \right| = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{c^2} - 4 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{infinite}}\,\,{\rm{solutions}}\,\,\,\left( {x,y} \right) = \left( {1 + cy,y} \right)\,\,\,\,\,\,\,\left[ {y\,\,{\rm{free}}} \right]\)

\(\left| c \right| \ne 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{c^2} - 4 \ne 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{

\,\,y = 0\,\,\,\,\,\,\,\left[ {y\left( {{c^2} - 4} \right) = 0} \right] \hfill \cr

\,\,x = 1\,\,\,\,\,\,\,\left[ {x - 1 = cy} \right] \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {x,y} \right){\rm{ = }}\left( {1,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{unique}}\,\,{\rm{solution!}}\,\,\)

_________________

Fabio Skilnik ::

GMATH method creator (Math for the GMAT)

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