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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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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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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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