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

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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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.

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.
_________________

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

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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