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What range of values of x will satisfy the inequality 2x +
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What range of values of x will satisfy the inequality 2x + 3 > 7x  2? A. x < 1/9 or x > 5 B. 1 < x < 1/9 C. 1/9 < x < 1 D. 1/9 < x < 5 E. x < 1/9 or x > 1
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Originally posted by ramzin on 25 May 2011, 07:04.
Last edited by Bunuel on 28 Apr 2014, 02:06, edited 2 times in total.
Renamed the topic, edited the question and the OA.




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Re: What range of values of x will satisfy the inequality 2x +
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28 Apr 2014, 02:06
ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C.
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Re: What range of values of x will satisfy the inequality contd.
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25 May 2011, 07:26
two methods here.
1. using the solutions here. x= 3/2 and x= 2/7 are solutions. check for the regions
a. x<3/2, b. 3/2<x<2/7 and c. x>2/7
values are a (2x+3) > (7x2) giving x <1/9 hence not a solution.
b (2x+3) > (7x2) giving x < 1 a solution.
c (2x+3) > (7x2) giving x>1/9 hence the solution is 1/9 < x < 1
or
2. squaring both sides
(2x+3) ^2 > (7x2)^2
gives 9x^2 8x  1 > 0 giving solution 1/9 < x < 1.
Hence C you can use whichever you are comfortable with.




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Re: What range of values of x will satisfy the inequality contd.
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25 May 2011, 15:42
Amit,
Thank you, but could you please elaborate why in the option A
"a. x<3/2,
values are a (2x+3) > (7x2) giving x <1/9 hence not a solution."
the second part  (7x2) is positive? If we get a number less than 1/9, 10 for instance would not the second part be negative as well?
Regards, Steve.



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Re: What range of values of x will satisfy the inequality 2x +
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28 Apr 2014, 04:27
Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. Hi Bunuel , Thanks for the answer. I did not understand how are the roots 1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.



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Re: What range of values of x will satisfy the inequality 2x +
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28 Apr 2014, 04:38
gauravsoni wrote: Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. Hi Bunuel , Thanks for the answer. I did not understand how are the roots 1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate. \((x+\frac{1}{9})(x1)=0\) > \(x+\frac{1}{9}=0\) or \(x1=0\) > \(x=\frac{1}{9}\) or \(x=1\).
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Re: What range of values of x will satisfy the inequality 2x +
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28 Apr 2014, 05:45
Bunuel wrote: Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. Hi Bunuel , Thanks for the answer. I did not understand how are the roots 1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate. \((x+\frac{1}{9})(x1)=0\) > \(x+\frac{1}{9}=0\) or \(x1=0\) > \(x=\frac{1}{9}\) or \(x=1\).[/quote] That part is fine , i'm solving the quadratic equation 9x^28x1 < 0 as (x9)(x+1) then getting x = 9 , x = 1. I also check out your link for solving quadratic equations in equalities but could relate it.



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Re: What range of values of x will satisfy the inequality 2x +
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28 Apr 2014, 05:52
gauravsoni wrote: Bunuel wrote: Bunuel wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2? A. x < 1/9 or x > 5 B. 1 < x < 1/9 C. 1/9 < x < 1 D. 1/9 < x < 5 E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. Hi Bunuel , Thanks for the answer. I did not understand how are the roots 1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate. \((x+\frac{1}{9})(x1)=0\) > \(x+\frac{1}{9}=0\) or \(x1=0\) > \(x=\frac{1}{9}\) or \(x=1\). That part is fine , i'm solving the quadratic equation 9x^28x1 < 0 as (x9)(x+1) then getting x = 9 , x = 1. I also check out your link for solving quadratic equations in equalities but could relate it.[/quote] (x9)(x+1) is not a correct factoring of 9x^28x1, it should be \((x+\frac{1}{9})(x1)\) (\((9x+1)(x1)\)). Factoring Quadratics: http://www.purplemath.com/modules/factquad.htmSolving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htmHope this helps.
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Re: What range of values of x will satisfy the inequality 2x +
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29 Apr 2014, 01:49
amit2k9 wrote: two methods here.
1. using the solutions here. x= 3/2 and x= 2/7 are solutions. check for the regions
a. x<3/2, b. 3/2<x<2/7 and c. x>2/7
values are a (2x+3) > (7x2) giving x <1/9 hence not a solution.
b (2x+3) > (7x2) giving x < 1 a solution.
c (2x+3) > (7x2) giving x>1/9 hence the solution is 1/9 < x < 1
or
2. squaring both sides
(2x+3) ^2 > (7x2)^2
gives 9x^2 8x  1 > 0 giving solution 1/9 < x < 1.
Hence C you can use whichever you are comfortable with. Hi Bunuel, Thanks for the solution! I get the squaring both sides approach, but it takes over 3 minutes for me to do it that way. Is there a faster way to solve this? I noticed Amit provided another method, but I'm not sure I understand the first approach completely. Could you explain why he used the "" sign in the first and third parts (highlighted in blue above)?



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Re: What range of values of x will satisfy the inequality 2x +
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29 Apr 2014, 04:46
gauravsoni wrote: Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. Hi Bunuel , Thanks for the answer. I did not understand how are the roots 1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate. Hi Gauravsoni, \(9x^28x1<0\) Can be factored 9x^2 9x+x1 <0 9x(x1)+1 (x1) <0 or (9x+1)(x1)<0 or x=1/9 or x=1
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Re: What range of values of x will satisfy the inequality 2x +
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07 May 2014, 10:41
hi bunnel what do u mean when u say the following: Quote: Since both sides of the inequality are nonnegative we can safely square:
4x^2+12x+9>49x^228x+4 > 45x^240x5<0 > 9x^28x1<0 > (x+\frac{1}{9})(x1)<0.
how how we solve inequality which is negative?? whenever we see absolute values on each side of the sign... should we use the squaring approach??
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Re: What range of values of x will satisfy the inequality 2x +
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08 May 2014, 03:24
nandinigaur wrote: hi bunnel what do u mean when u say the following: Quote: Since both sides of the inequality are nonnegative we can safely square:
4x^2+12x+9>49x^228x+4 > 45x^240x5<0 > 9x^28x1<0 > (x+\frac{1}{9})(x1)<0.
how how we solve inequality which is negative?? whenever we see absolute values on each side of the sign... should we use the squaring approach?? We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality) Adding/subtracting/multiplying/dividing inequalities: helpwithaddsubtractmultdividmultipleinequalities155290.htmlAs for your other questions: it depends on a question.
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Re: What range of values of x will satisfy the inequality 2x +
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12 May 2014, 00:13
Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?.. I tried to deduce the answers by substituting possible values from the various option ranges.



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Re: What range of values of x will satisfy the inequality 2x +
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12 May 2014, 00:18
himanshujovi wrote: Bunuel wrote: ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Since both sides of the inequality are nonnegative we can safely square: \(4x^2+12x+9>49x^228x+4\) > \(45x^240x5<0\) > \(9x^28x1<0\) > \((x+\frac{1}{9})(x1)<0\). The "roots" are 1/9 and 1 (Solving Quadratic Inequalities: solvingquadraticinequalitiesgraphicapproach170528.html). "<" sign indicates that the solution must be between the roots: \(\frac{1}{9}<x<1\). Answer: C. If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?.. I tried to deduce the answers by substituting possible values from the various option ranges. If x = 0, then 2x + 3 = 3 and 7x  2 = 2 > 3 > 2. So, 0 is a possible value of x. Hope it helps.
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Re: What range of values of x will satisfy the inequality contd.
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18 Jun 2014, 06:15
amit2k9 wrote: two methods here.
1. using the solutions here. x= 3/2 and x= 2/7 are solutions. check for the regions
a. x<3/2, b. 3/2<x<2/7 and c. x>2/7
values are a (2x+3) > (7x2) giving x <1/9 hence not a solution.
b (2x+3) > (7x2) giving x < 1 a solution.
c (2x+3) > (7x2) giving x>1/9 hence the solution is 1/9 < x < 1
or
2. squaring both sides
(2x+3) ^2 > (7x2)^2
gives 9x^2 8x  1 > 0 giving solution 1/9 < x < 1.
Hence C you can use whichever you are comfortable with. Hi Amit, can you explain how to get the range 1/9 < x < 1 in the second method the solutions are x< 1 and x < 1/9 (how does the sign reverse to the other side ?)



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Re: What range of values of x will satisfy the inequality 2x +
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28 Mar 2016, 08:03
What range of values of x will satisfy the inequality 2x + 3 > 7x  2? Plotting the The halfs of the equation we get the diagram as shown...shaded portion shows the area of interest satisfyin the above inequality. In upper half x=1 is the point of intersection i.e the maximum value of x. Only C satisfies. A. x < 1/9 or x > 5 B. 1 < x < 1/9 C. 1/9 < x < 1 D. 1/9 < x < 5 E. x < 1/9 or x > 1
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Re: What range of values of x will satisfy the inequality 2x +
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10 Apr 2016, 19:27
ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 I took the safest approach, plug in numbers :D A. x=2. 4+3 = 1, 1=1. 7*2 = 14 2 = 16. 16=16 2 is not > than 16, so can't be A. B. x= 1/8. 2/8+3 = 2.75. 7*1/8 = 7/8 2 = 2.875 not true, so can't be B. D. x=4. 8+3=11. 282=26. so not true, D is out. E. same as in A. x=2 doesn't work. only C remains...



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Re: What range of values of x will satisfy the inequality 2x +
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24 Apr 2016, 21:38
Hi, I have some doubts.
In these kind of questions we generally get 4 results for x. In this question for instance, we get the following 
1. x<1 2. x>1 3. x<1/9 4. x>1/9
So should now the range be the inequality that restricts the the values within two limits? for example 1<x<1/9 and cannot be the range right?
Thank you.



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Re: What range of values of x will satisfy the inequality 2x +
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06 Oct 2019, 07:16
ramzin wrote: What range of values of x will satisfy the inequality 2x + 3 > 7x  2?
A. x < 1/9 or x > 5
B. 1 < x < 1/9
C. 1/9 < x < 1
D. 1/9 < x < 5
E. x < 1/9 or x > 1 Plug values x=4( For D and E, it is valid for both) 11>26( not possible) D and E are out. x=0 3>2 A is out. We are left with B and C. Plug x=1/2 C is the winner:)



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Re: What range of values of x will satisfy the inequality 2x +
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08 Oct 2019, 16:44
In the essence of GMAT here is a quick smart solution. A simple trick to find a range is to maximise the equation and minimise it to get the largest and the smallest value. So to maximize let’s take 2x+3>7x2 gives us x=1 And take lhs as 2x3>7x2 we take this since if we have Lhs as negative the +7 will go and make the number smaller. Now we have 9x=1 so x=1/9 Answer C Hit the like button to encourage active discussions
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Re: What range of values of x will satisfy the inequality 2x +
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