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Sub 505 Level|   Algebra|                                    
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Bunuel
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

5 - 6/x = x --> (5x - 6)/x = x --> 5x -6 = x^2 --> (x-3)(x-2)=0 --> x=3 or x=2.

Answer: C.
That's cool Bunuel.
I can approach this kind of equation with Delta method. However I'm very interested in your factoring method.
Would you please share how you quickly factor the equation x^2 - 5x + 6 = 0 into (x-3)(x-2) = 0. Thanks
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akhandamandala
Bunuel
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

5 - 6/x = x --> (5x - 6)/x = x --> 5x -6 = x^2 --> (x-3)(x-2)=0 --> x=3 or x=2.

Answer: C.
That's cool Bunuel.
I can approach this kind of equation with Delta method. However I'm very interested in your factoring method.
Would you please share how you quickly factor the equation x^2 - 5x + 6 = 0 into (x-3)(x-2) = 0. Thanks

Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

Hope it helps.
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that's great, it helps a lot. I may save 30 seconds from this method
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Walkabout
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

We need to simplify the equation 5 – 6/x = x. We start by multiplying the entire equation by x and we obtain:

5x – 6 = x^2

x^2 – 5x + 6 = 0

(x – 3)(x – 2) = 0

x = 3 or x = 2.

We see that x has 2 possible values.

The answer is C.
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As its a quadric equation, there can be 2 values of x. So do you really to calculate or can select answer 2 without calculating what is the value of x?
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sisirkant
As its a quadric equation, there can be 2 values of x. So do you really to calculate or can select answer 2 without calculating what is the value of x?

Quadratic equation can have 0, 1 or two roots, so you should do some additional math here.
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Hi All,

You'll have to do a bit of algebra to deduce the number of solutions to this equation…

5 - (6/x) = x

First, multiply everything by x…

5x - 6 = x^2

Now, move everything "to the right"….

0 = x^2 - 5x + 6

You can now factor this into two terms…

0 = (x -2)(x - 3)

And answer the question… There are 2 solutions

Final Answer:

GMAT assassins aren't born, they're made,
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Bunuel
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

\(5 - \frac{6}{x} = x\);

\(\frac{5x - 6}{x} = x\);

\(5x -6 = x^2\);

\((x-3)(x-2)=0\);

\(x=3\) or \(x=2\).

Answer: C.

Well i wanted to reconfirm , we eliminated the possibility of x=0 , because equating x=0 in the equation doesn't solve it , Right ??
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Siddhuftr
Bunuel
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

\(5 - \frac{6}{x} = x\);

\(\frac{5x - 6}{x} = x\);

\(5x -6 = x^2\);

\((x-3)(x-2)=0\);

\(x=3\) or \(x=2\).

Answer: C.

Well i wanted to reconfirm , we eliminated the possibility of x=0 , because equating x=0 in the equation doesn't solve it , Right ??

x is not 0 because after solving the question we got that x = 3 or x = 2 and did not get that x = 0, which naturally means that x = 0 does not satisfy the given equation.

x cannot be 0 because x is in the denominator of one term, 6/x, and we cannot divide by 0. If x were 6/x would not be defined.
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Walkabout
If 5 - 6/x = x, then x has how many possible values?

(A) None
(B) One
(C) Two
(D) A finite number greater than two
(E) An infinite number

We need to simplify the equation 5 – 6/x = x. We start by multiplying the entire equation by x and we obtain:

5x – 6 = x^2

x^2 – 5x + 6 = 0

(x – 3)(x – 2) = 0

x = 3 or x = 2.

We see that x has 2 possible values.

The answer is C.

JeffTargetTestPrep
Thank you for this helpful reply. To confirm, you cannot just get rid of the x in the denominator of 6 by multiplying it by x on the right side of the equation, right?

For example:
5 - (6/x)=x
5-6=x^2
This is incorrect because you would either 1.) need to find like terms on the left side or move the entire (6/x) to the right side by adding it to the right side or moving the 5 to the right side by subtracting it from the right side?

I realize this is basic algebra, but that's where my simple mistakes tend to happen :) Thank you again.
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Hi woohoo921,

You CAN multiply both sides of the equation by "X" to remove that fraction - but that means you have to multiply EVERYTHING by X (and in your example, you're only multiplying two of the three pieces of the equation).

With 5 - (6/X) = X

if you multiply both sides by X, you actually have:

5(X) - (6/X)(X) = X(X)
5X - 6 = X^2

You can then move the pieces around and form a quadratic (which you can solve):

X^2 - 5X + 6 = 0
(X - 6)(X + 1) = 0

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]
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hi, was solving this Q on OG website and I was super confused because the variables looked different, I am the only one who saw this weird glitch? (find attached)
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Screenshot 2025-06-07 125534.png
Screenshot 2025-06-07 125534.png [ 4.45 KiB | Viewed 973 times ]

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AnuK2222
hi, was solving this Q on OG website and I was super confused because the variables looked different, I am the only one who saw this weird glitch? (find attached)

Probably just a typo in your version. Here's how it appears in other versions of the OG.


Attachment:
GMAT-Club-Forum-g4udq3vw.png
GMAT-Club-Forum-g4udq3vw.png [ 86.88 KiB | Viewed 967 times ]
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