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If y = ax + b and y = cx + d for all values of x, where a, b, c, and d

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If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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If y = ax + b and y = cx + d for all values of x, where a, b, c, and d are constants, then all the following must be true EXCEPT:

A. a = c

B. ac = -1

C. a^2 = c^2

D. \(|a| = \sqrt{c^2}\)

E. ac + 1 > 0

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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New post 08 Jul 2015, 04:14
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Bunuel wrote:
If y = ax + b and y = cx + d for all values of x, where a, b, c, and d are constants, then all the following must be true EXCEPT:

A. a = c

B. ac = -1

C. a^2 = c^2

D. \(|a| = [square_rootc^2[/square_root]\)

E. ac + 1 > 0

Kudos for a correct solution.


Given : y = ax + b and y = cx + d

i.e. ax + b = cx + d

i.e. a = c [Option A ruled out]

therefore, a^2 = c^2 [Option C ruled out]

therefore, \(|a| = \sqrt{c^2}\) [Option D ruled out]

and since both a and c have same value and same sign so their product must be positive
i.e. ac > 0
hence ac> -1 [Option E ruled out]

Answer: Option B
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Kudos [?]: 2348 [1], given: 51

Expert Post
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Kudos [?]: 132978 [0], given: 12391

If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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New post 13 Jul 2015, 02:17
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Bunuel wrote:
If y = ax + b and y = cx + d for all values of x, where a, b, c, and d are constants, then all the following must be true EXCEPT:

A. a = c

B. ac = -1

C. a^2 = c^2

D. \(|a| = \sqrt{c^2}\)

E. ac + 1 > 0

Kudos for a correct solution.


800score Official Solution:

Since y is equal to both ax + b and cx + d, we know that:
ax + b = cx + d.

Since it is true for any value of x, let’s plug in x = 0. It yields b = d.
If we plug in x = 1. It yields a + b = c + d. We already know that b = d, so a = c

You may think of the formulas as the two expressions representing the same line. That results in the fact that these expressions are the same.

Let’s look at the choices one by one, to determine which is NOT NECESSARILY true.

Choice (A): We have established above that this must be true.

Choice (B): This CANNOT be true. Since a and c must be the same number, their product cannot be negative.

Choice (C): This means that |a| = |c|. Since a = c, this must be true.

Choice (D): √(c²) is equal to |c|. So the equation in the answer choice becomes |a| = |c|, which is the same as the equation in choice (C), so it must be true.

Choice (E): Since a and c must be the same number, ac + 1 must be positive.

The correct answer is B.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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New post 03 Oct 2016, 07:49
Bunuel
ax + b = cx + d
x(a-c)=d-b
x=(d-b)/ (a-c)

if a=c then x becomes undefined, thats why I chose A.
whats wrong here?

Kudos [?]: 9 [0], given: 56

Expert Post
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Math Expert
User avatar
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Joined: 02 Sep 2009
Posts: 42286

Kudos [?]: 132978 [1], given: 12391

Re: If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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New post 03 Oct 2016, 08:02
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deepak268 wrote:
Bunuel
ax + b = cx + d
x(a-c)=d-b
x=(d-b)/ (a-c)
if a=c then x becomes undefined, thats why I chose A.
whats wrong here?


If a=c, then we cannot write the highlighted part but x*0 = d-b would still be perfectly valid equation, which would mean that d-b=0 and x can take any value.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 132978 [1], given: 12391

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If y = ax + b and y = cx + d for all values of x, where a, b, c, and d [#permalink]

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New post 15 Oct 2017, 08:35
Bunuel wrote:
If y = ax + b and y = cx + d for all values of x, where a, b, c, and d are constants, then all the following must be true EXCEPT:

A. a = c

B. ac = -1

C. a^2 = c^2

D. \(|a| = \sqrt{c^2}\)

E. ac + 1 > 0

Kudos for a correct solution.


a=b; because both are coefficients of x.
assuming a = 1
Eliminate option a, because it is true. If option a is eliminated, then you must also eliminate option c; AV and Root both suggest +/- relationship; therefore the absolute value of a = c; eliminate d. Option E is eliminate because it can be true.

Option B suggests either a or b is a negative, but not both. Since a = b, therefore, B, ac = -1 cannot be true.
Answer is B

Kudos [?]: 4 [0], given: 2

If y = ax + b and y = cx + d for all values of x, where a, b, c, and d   [#permalink] 15 Oct 2017, 08:35
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