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655-705 (Hard)|   Algebra|      
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fskilnik
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If a and b are constants such that are constants such that ax/(x^2 - 1 04 Feb 2019, 11:47
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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)!

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65% (hard)

Question Stats:

61% (02:13) correct 39% (02:49) wrong based on 188 sessions

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If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

(A) -2
(B) -1
(C) 0
(D) 2
(E) 4
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E
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KanishkM
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If a and b are constants such that are constants such that ax/(x^2 - 1 04 Feb 2019, 12:02
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fskilnik
GMATH practice question (Quant Class 13)

If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

(A) -2
(B) -1
(C) 0
(D) 2
(E) 4

Source: https://www.gmath.net
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ax + b x + b = 2x -1 , for x > 1

a - b = ?

(a+b)x + b = 2x - 1

b = -1, then a+b = 2
a = 3

Now a - b = 3 +1 = 4

E
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fskilnik
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If a and b are constants such that are constants such that ax/(x^2 - 1 04 Feb 2019, 14:33
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fskilnik
GMATH practice question (Quant Class 13)

If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

(A) -2
(B) -1
(C) 0
(D) 2
(E) 4

Source: https://www.gmath.net
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\(? = a - b\)

\({{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{{ax + b\left( {x + 1} \right)} \over {{x^2} - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,x\left( {a + b - 2} \right) + b + 1 = 0\,\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,x > 1\,\,\,\,\left( * \right)\)

\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,b + 1 = 0\,\,\,\, \Rightarrow \,\,\,\,b = - 1 \hfill \cr \\
\,a + b - 2 = 0 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a = 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 4\)


The correct answer is therefore (E).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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jfranciscocuencag
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If a and b are constants such that are constants such that ax/(x^2 - 1 22 Feb 2019, 21:25
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KanishkM
fskilnik
GMATH practice question (Quant Class 13)

If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

(A) -2
(B) -1
(C) 0
(D) 2
(E) 4

Source: https://www.gmath.net

ax + b x + b = 2x -1 , for x > 1

a - b = ?

(a+b)x + b = 2x - 1

b = -1, then a+b = 2
a = 3

Now a - b = 3 +1 = 4

E
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Hi KanishkM :)

How did you end with (a+b)x + b = 2x - 1?
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KanishkM
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If a and b are constants such that are constants such that ax/(x^2 - 1 22 Feb 2019, 21:39
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jfranciscocuencag
KanishkM
fskilnik
GMATH practice question (Quant Class 13)

If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

(A) -2
(B) -1
(C) 0
(D) 2
(E) 4

Source: https://www.gmath.net

ax + b x + b = 2x -1 , for x > 1

a - b = ?

(a+b)x + b = 2x - 1

b = -1, then a+b = 2
a = 3

Now a - b = 3 +1 = 4

E


Hi KanishkM :)

How did you end with (a+b)x + b = 2x - 1?
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Hey jfranciscocuencag

Just expand x^2-1 to x+1 x-1

Cross out the common terms from both LHS and RHS

After that you can take LCM and easily solve the question.

Posted from my mobile device
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If a and b are constants such that are constants such that ax/(x^2 - 1 18 Nov 2024, 00:28
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If \(a\) and \(b\) are constants such that \(\,\,{{ax} \over {{x^2} - 1}} + {b \over {x - 1}} = {{2x - 1} \over {{x^2} - 1}}\,\,\) for every \(x > 1\), what is the value of \(a - b\) ?

\(\frac{ax}{x^2 - 1} + \frac{b}{x-1} = \frac{2x-1}{x^2-1}\)
\(\frac{ax}{x^2 - 1} + \frac{b(x+1)}{x^2-1} = \frac{2x-1}{x^2-1}\)
\(ax + bx + b = 2x - 1\)

a + b = 2
b = -1
a = 3
a - b = 3 - (-1) = 4

IMO E
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