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Sub 505 (Easy)|   Algebra|      
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Bunuel
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 31 Aug 2015, 10:42
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D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x?

(A) 1
(B) 1/2
(C) 1/3
(D) -1/3
(E) -1/2


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D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 31 Aug 2015, 10:48
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Bunuel
If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x?

(A) 1
(B) 1/2
(C) 1/3
(D) -1/3
(E) -1/2


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3x^2+x+12x+4=3x^2+x
12x+4=0
x=-1/3
Answer D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 31 Aug 2015, 14:10
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(x+4)(3x+1) = 3x^2+x

=> 3x^2 + x + 12x +4 = 3x^2 + x
=> 12x+4 = 0
=> 12x = -4
=> x=-1/3

Answer:- D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 31 Aug 2015, 14:18
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Bunuel
If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x?

(A) 1
(B) 1/2
(C) 1/3
(D) -1/3
(E) -1/2


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Expand the expression: (x + 4)(3x + 1) = 3x^2 + x to:

3x^2+x+12x+4 = 3x^2 +x , subtract 3x^2 from both sides
13x + 4 = x
12x = -4
x = -1/3

Answer D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 01 Sep 2015, 03:33
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Bunuel
If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x?

(A) 1
(B) 1/2
(C) 1/3
(D) -1/3
(E) -1/2


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Rearranging the given expression:
(x + 4)(3x + 1) = 3x^2 + x
i.e. 3x^2 + 13x + 4 = 3x^2 + x
i.e. 12x = -4
i.e. x = -4/12 = -1/3

Answer: Option D

Had it been a difficult expression to solve further, then we could have used options to check which on esatisfies the expression after solving it as further as possible
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 01 Sep 2015, 06:24
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Simplifying this equation is my approach here

The equation is

(x + 4)(3x + 1) = 3x^2 + x

or,3x^2 +x +12x +4=3x^2 +x
or,12x=-4
or,x=-4/12
or,x=-1/3

So the Correct answer is D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 17 Jan 2016, 16:23
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Hi, Bunuel!
I solved this problem differently, but not sure if it's mathematically right:
(x + 4)(3x + 1) = 3x^2+ x
(x+4)(3x + 1)=x(3x+1)
(x+4)(3x+1)-x(3x+1)=0
(x+4-x)(3x+1)=0
so, first bracket is invalid, but second 3x=-1 gives the right answer x=-1/3.
What do you think :?:
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 17 Jan 2016, 17:06
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I5O
Hi, Bunuel!
I solved this problem differently, but not sure if it's mathematically right:
(x + 4)(3x + 1) = 3x^2+ x
(x+4)(3x + 1)=x(3x+1)
(x+4)(3x+1)-x(3x+1)=0
(x+4-x)(3x+1)=0
so, first bracket is invalid, but second 3x=-1 gives the right answer x=-1/3.
What do you think :?:
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This is correct. There is no such thing as "so, first bracket is invalid".

Once you get (x+4-x)(3x+1)=0 ---> (4)(3x+1)=0 ---> means that the product of 4 and 3x+1 is =0 ---> This can only happen if either one of them or both of them are zero.

Both can not be zero as one of it is already 4. Thus, it must be then that (3x+1)=0 ---> x=-1/3

Hope this helps.
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 25 Apr 2016, 12:56
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\((x+4)(3x+1)=3x^2+x\)
simplifying the given expression
\(3x^2+13x+4=3x^2+x\)
12x=-4
x=\(\frac{-1}{3}\)
correct answer option D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 16 Apr 2020, 10:42
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x?

(x + 4)(3x + 1) = 3x^2 + x
or,3x^2 +x +12x +4=3x^2 +x
or,12x=-4
or,x=-4/12
or,x=-1/3

correct answer D
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If (x + 4)(3x + 1) = 3x^2 + x, what is a possible value of x? 18 Aug 2025, 05:10
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