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If x  2 = x + 3, x could equal
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07 Mar 2016, 09:22
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73% (01:04) correct 27% (01:02) wrong based on 575 sessions
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If x  2 = x + 3, x could equal A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution
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Re: If x  2 = x + 3, x could equal
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07 Mar 2016, 20:50
is the answer 'B' I back solved the question by putting answer choices in the question. 1/2 gives 5/2 on both sides. Thanks
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Re: If x  2 = x + 3, x could equal
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07 Mar 2016, 20:58
I started drawing a number line, but then I thought better of it and just plugged in the values form the answers. A) 5: LHS: 52 = 7, RHS: 5+3 = 2, no good B) 1/2: LHS: 1/2  2 = 2.5, RHS: 1/2 + 3 = 2.5, Correct
Thus B



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Re: If x  2 = x + 3, x could equal
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07 Mar 2016, 21:04
Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution By definition, xa = distance of x from a. In other words, x2 = distance of x from 2 and x+3 = distance of x from 3. Thus, if x2 = x+3 > distance of x from 2 = distance of x from 3. Thus, x = halfway between 3 and 2. Distance between 3 and 2 = 5 units. Thus x is 2.5 units away from 2 or 3 , giving you x = 22.5 or = 3+2.5 = 0.5, B is thus the correct answer.



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Re: If x  2 = x + 3, x could equal
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08 Mar 2016, 00:39
Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution I just backsolved and was able to see that B was the answer for me. Both sides end up with the same result of 5/2.



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Re: If x  2 = x + 3, x could equal
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08 Mar 2016, 01:15
solving equation we get 1/2
Ans ; B



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Re: If x  2 = x + 3, x could equal
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11 Jun 2016, 11:18
x  2 = x + 3 We see both the sides are positive, hence we can square both the sides and we get X^2 + 4 4x = X^2 + 9 6x 10x= 5 X= 1/2 Answer is B.
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Re: If x  2 = x + 3, x could equal
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12 Jun 2016, 02:19
Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution Since there are two Equations. Lets solve by Critical Value x  2 = x + 3 Critical Value will be 2, 3 hence we Have 3 Solutions Solution No 1 When X < 3 ( RHS Will be Negative ) There fore x2 =  (x + 3 ) => 2x = 1 => x = 1/2 which is not in line with x< 3 , hence rejected Solution No 2 When 3 < X <= 2 ( LHS will be Negative ) (x2) = x+3 => 2x= 1 => x = 1/2 which fits in the range Hence Solution. We can check by substituting in x  2 = x + 3 Therefore 5/2= 5/2 Solution 3 when X > 2 both sides positive x2 = x+3 No Value hence rejected



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Re: If x  2 = x + 3, x could equal
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12 Jun 2016, 03:59
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side (x2)^2= (x+3)^2 by solving we get x= 1/2 . Please correct me if i missed something.



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Re: If x  2 = x + 3, x could equal
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12 Jun 2016, 04:33
sudhirmadaan wrote: Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side (x2)^2= (x+3)^2 by solving we get x= 1/2 . Please correct me if i missed something. .. Hi sudhir, yes, whenever you have ONLY mods on both sides.. example a+b2>ab4... you can square both sides to fet your answer.. and you have correctly found your answer here
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Re: If x  2 = x + 3, x could equal
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24 Jun 2016, 08:12
Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution Given x  2 = x + 3 The best way to get rid of the modulus is to square on both sides.\((x  2)^2\) = \((x+3)^2\) => \(x^2\) + 4 + 4x = \(x^2\) + 9 + 6x => x = 1/2. Correct answer is B...



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Re: If x  2 = x + 3, x could equal
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24 Jul 2016, 13:39
Razween wrote: Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution Since there are two Equations. Lets solve by Critical Value x  2 = x + 3 Critical Value will be 2, 3 hence we Have 3 Solutions Solution No 1 When X < 3 ( RHS Will be Negative ) There fore x2 =  (x + 3 ) => 2x = 1 => x = 1/2 which is not in line with x< 3 , hence rejected Solution No 2 When 3 < X <= 2 ( LHS will be Negative ) (x2) = x+3 => 2x= 1 => x = 1/2 which fits in the range Hence Solution. We can check by substituting in x  2 = x + 3 Therefore 5/2= 5/2 Solution 3 when X > 2 both sides positive x2 = x+3 No Value hence rejected In the first condition in which X is less than 3, why isn't the (x2) term negative as well? The expression (x2) is negative for all values where X is less than 2.



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Re: If x  2 = x + 3, x could equal
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25 Jul 2016, 01:18
astroshagger wrote: Razween wrote: Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution Since there are two Equations. Lets solve by Critical Value x  2 = x + 3 Critical Value will be 2, 3 hence we Have 3 Solutions Solution No 1 When X < 3 ( RHS Will be Negative ) There fore x2 =  (x + 3 ) => 2x = 1 => x = 1/2 which is not in line with x< 3 , hence rejected Solution No 2 When 3 < X <= 2 ( LHS will be Negative ) (x2) = x+3 => 2x= 1 => x = 1/2 which fits in the range Hence Solution. We can check by substituting in x  2 = x + 3 Therefore 5/2= 5/2 Solution 3 when X > 2 both sides positive x2 = x+3 No Value hence rejected In the first condition in which X is less than 3, why isn't the (x2) term negative as well? The expression (x2) is negative for all values where X is less than 2. Yes you are right. If \(x < 3\), then \(x  2 = (x  2)\) and \(x + 3 = (x + 3)\); If \(3 \leq x \leq 2\), then \(x  2 = (x  2)\) and \(x + 3 = x + 3\); If \(x > 2\), then \(x  2 = x  2\) and \(x + 3 = x + 3\).
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If x  2 = x + 3, x could equal
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27 Jul 2017, 02:59
x  2  x + 3=0 We substitute 2,3 in the above expression accordingly and plot the same in number line. Now minimum distance between the two points is 5 which is less than RHS. hence no real solution. Option E.
Please advise.



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Re: If x  2 = x + 3, x could equal
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29 Jul 2017, 09:02
chetan2u wrote: sudhirmadaan wrote: Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side (x2)^2= (x+3)^2 by solving we get x= 1/2 . Please correct me if i missed something. .. Hi sudhir, yes, whenever you have ONLY mods on both sides.. example a+b2>ab4... you can square both sides to fet your answer.. and you have correctly found your answer here hi so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots... please correct me if I am wrong ... thanks in advance ...



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Re: If x  2 = x + 3, x could equal
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29 Jul 2017, 09:14
ssislam wrote: hi
so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots...
please correct me if I am wrong ...
thanks in advance ...
Hello ssislam, In this case, it will not hide as we are taking the square root (or rather assuming) the square root of x in the entire equation. For some mod questions, you can take do a square on both sides and try to solve the question. Hope this helps
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Re: If x  2 = x + 3, x could equal
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29 Jul 2017, 20:51
B is the right option x2=x3 2x=3+2 2x=1 x=1/2



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If x  2 = x + 3, x could equal
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30 Jul 2017, 10:16
Bunuel wrote: If x  2 = x + 3, x could equal
A. 5 B. 1/2 C. 1/2 D. 5 E. No real solution With absolute value equations this simple, for me, the method below is easy and fast (less than a minute). The only way the two sides will be equal is if quantities inside the absolute value brackets are 1) equal or 2) equal but with opposite signs. So A = B or B* 1. Remove the absolute value bars 2. LHS = RHS or LHS = RHS 3. Set up the two equations Case 1: x  2 = x + 3, OR Case 2: x  2 = (x + 3) 4. Solve Case 1: x  2 = x + 3 2 = 3. Not a solution CASE 2: x  2 = (x + 3) x  2 = x  3 2x = 1 x = \(\frac{1}{2}\) 5. Check x = \(\frac{1}{2}\) in equation. > \(\frac{5}{2}\) = \(\frac{5}{2}\) Correct. x = \(\frac{1}{2}\). Answer BHope it helps * and B = A or A. Technically there are four cases. Case 3 is RHS = LHS. Case 4 is RHS =  LHS. These two cases are identical to Cases 1 and 2. Case 3: +RHS = +LHS, (x + 3) = (x  2) (= Case 1) Case 4: +RHS = LHS, (x + 3) = ( x + 2). Multiply both sides by (1) > (x  3) = (x  2) (= Case 2)
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If x  2 = x + 3, x could equal
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