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12 Mar 2019, 13:35
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:07) correct
63%
(02:06)
wrong
based on 347
sessions
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Not Attempted Yet
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?
\(\eqalign{\\
& \left( A \right)\,\,\,c > 0 \cr \\
& \left( B \right)\,\,\left| c \right| > 1 \cr \\
& \left( C \right)\,\,\left| c \right| < 1 \cr \\
& \left( D \right)\,\,\left| c \right| = 2 \cr \\
& \left( E \right)\,\,\left| c \right| \ne 2 \cr}\)
\(\left\{ \matrix{\\
\,x - cy = 1 \hfill \cr \\
\,cx - 4y = c \hfill \cr} \right.\)
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?
\(\eqalign{\\
& \left( A \right)\,\,\,c > 0 \cr \\
& \left( B \right)\,\,\left| c \right| > 1 \cr \\
& \left( C \right)\,\,\left| c \right| < 1 \cr \\
& \left( D \right)\,\,\left| c \right| = 2 \cr \\
& \left( E \right)\,\,\left| c \right| \ne 2 \cr}\)
23 Mar 2019, 23:44
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Rearranging equations in \(y = mx + c\) form,
\(y = (x/c) - (1/c)\)
\(y = (cx/4) - (c/4)\)
If both equations were to be parallel, then both equations must have same slope,
so \((1/c) = (c/4)\)
\(c^2 = 4\)
\(|c| = 2\)
But question says, both equations have one solution, we all know parallel lines can never have a solution,
so |c| cannot be 2 => E
\(y = (x/c) - (1/c)\)
\(y = (cx/4) - (c/4)\)
If both equations were to be parallel, then both equations must have same slope,
so \((1/c) = (c/4)\)
\(c^2 = 4\)
\(|c| = 2\)
But question says, both equations have one solution, we all know parallel lines can never have a solution,
so |c| cannot be 2 => E
General Discussion
12 Mar 2019, 14:53
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fskilnik
\(?\,\,\,:\,\,\,c\,\,\,{\rm{for}}\,\,{\rm{unique}}\,\,{\rm{solution}}\,\,\left( {x,y} \right)\,\,\)
\(c = 0\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,x = 1 \hfill \cr \\
\, - 4y = 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {x,y} \right) = \left( {1,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refutes}}\,\,\left( A \right),\left( B \right),\left( D \right)\)
\(c = 3\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,x - 3y = 1 \hfill \cr \\
\,3x - 4y = 3 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,x - 3y = 1 \hfill \cr \\
\,x - {4 \over 3}y = 1 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{\\
\,x - 3y = 1 \hfill \cr \\
\,\left( {1 + 3y} \right) - {4 \over 3}y = 1 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{unique}}\,\,{\rm{solution!}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refutes}}\,\,\left( C \right)\)
The answer is (E) by exclusion!
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
POST-MORTEM:
\(\left\{ \matrix{\\
\,x - cy = 1 \hfill \cr \\
\,cx - 4y = c \hfill \cr} \right.\,\,\,\,\,\, \cong \,\,\,\,\,\,\left\{ \matrix{\\
\,x - 1 = cy \hfill \cr \\
\,cx - c = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{\\
\,x - 1 = cy \hfill \cr \\
\,c\left( {x - 1} \right) = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{\\
\,x - 1 = cy \hfill \cr \\
\,c \cdot cy = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{\\
\,x - 1 = cy \hfill \cr \\
\,y\left( {{c^2} - 4} \right) = 0 \hfill \cr} \right.\)
\(\left| c \right| = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{c^2} - 4 = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{infinite}}\,\,{\rm{solutions}}\,\,\,\left( {x,y} \right) = \left( {1 + cy,y} \right)\,\,\,\,\,\,\,\left[ {y\,\,{\rm{free}}} \right]\)
\(\left| c \right| \ne 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{c^2} - 4 \ne 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{\\
\,\,y = 0\,\,\,\,\,\,\,\left[ {y\left( {{c^2} - 4} \right) = 0} \right] \hfill \cr \\
\,\,x = 1\,\,\,\,\,\,\,\left[ {x - 1 = cy} \right] \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {x,y} \right){\rm{ = }}\left( {1,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{unique}}\,\,{\rm{solution!}}\,\,\)
