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If c is a constant such that the system of equations given abov

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If c is a constant such that the system of equations given abov  [#permalink]

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New post 12 Mar 2019, 13:35
1
4
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

28% (02:15) correct 72% (02:08) wrong based on 65 sessions

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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?

\(\eqalign{
& \left( A \right)\,\,\,c > 0 \cr
& \left( B \right)\,\,\left| c \right| > 1 \cr
& \left( C \right)\,\,\left| c \right| < 1 \cr
& \left( D \right)\,\,\left| c \right| = 2 \cr
& \left( E \right)\,\,\left| c \right| \ne 2 \cr}\)

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New post 12 Mar 2019, 14:53
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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:


\(\left\{ \matrix{
\,x - cy = 1 \hfill \cr
\,cx - 4y = c \hfill \cr} \right.\,\,\,\,\,\, \cong \,\,\,\,\,\,\left\{ \matrix{
\,x - 1 = cy \hfill \cr
\,cx - c = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{
\,x - 1 = cy \hfill \cr
\,c\left( {x - 1} \right) = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{
\,x - 1 = cy \hfill \cr
\,c \cdot cy = 4y \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \cong \,\,\,\,\,\,\,\left\{ \matrix{
\,x - 1 = cy \hfill \cr
\,y\left( {{c^2} - 4} \right) = 0 \hfill \cr} \right.\)

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

\(\left| c \right| \ne 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{c^2} - 4 \ne 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,\,y = 0\,\,\,\,\,\,\,\left[ {y\left( {{c^2} - 4} \right) = 0} \right] \hfill \cr
\,\,x = 1\,\,\,\,\,\,\,\left[ {x - 1 = cy} \right] \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {x,y} \right){\rm{ = }}\left( {1,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{unique}}\,\,{\rm{solution!}}\,\,\)
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Re: If c is a constant such that the system of equations given abov  [#permalink]

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New post 23 Mar 2019, 23:44
1
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
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New post 24 Mar 2019, 07:21
hellosanthosh2k2 wrote:
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
Hi hellosanthosh2k2,

Thanks for your beautiful contribution (and nice wording).

[To our students: this is an EXCELLENT approach (although at our Quant Class 10 we have still not dealt with line equations properties).]

Just two very small details:

1. You have to deal with the case \(c=0\) apart, of course.
2. You certainly mean parallel distinct lines can never have a solution.
(Another possibility: Parallel lines can never have a SINGLE (x,y) pair as solution.)

Regards,
Fabio.
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Re: If c is a constant such that the system of equations given abov  [#permalink]

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New post 24 Mar 2019, 10:49
x = 1+cy
cx=c+4y
solving for x
(c+4y)/c = 1+cy
y(c^2-4)=0
if c = +/- 2 then any y value will satisfy the equation, which we do not want. so |c|=!2
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If c is a constant such that the system of equations given abov  [#permalink]

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New post 26 Mar 2019, 04:26
fskilnik wrote:
hellosanthosh2k2 wrote:
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
Hi hellosanthosh2k2,

Thanks for your beautiful contribution (and nice wording).

[To our students: this is an EXCELLENT approach (although at our Quant Class 10 we have still not dealt with line equations properties).]

Just two very small details:

1. You have to deal with the case \(c=0\) apart, of course.
2. You certainly mean parallel distinct lines can never have a solution.
(Another possibility: Parallel lines can never have a SINGLE (x,y) pair as solution.)

Regards,
Fabio.


Thank you, sorry missed out two points you brought out.
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Re: If c is a constant such that the system of equations given abov  [#permalink]

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New post 26 Mar 2019, 06:55
hellosanthosh2k2 wrote:
Thank you, sorry missed out two points you brought out.
Hi again, hellosanthosh2k2 !

I am glad you liked the details I mentioned.

I have realized I forgot to give you kudos for your excellent approach. Not anymore... ;)

Regards and success in your studies/activities,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net
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Re: If c is a constant such that the system of equations given abov   [#permalink] 26 Mar 2019, 06:55
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