What range of values of x will satisfy the inequality |2x +

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.

\((x+\frac{1}{9})(x-1)=0\) --> \(x+\frac{1}{9}=0\) or \(x-1=0\) --> \(x=-\frac{1}{9}\) or \(x=1\).

\((x+\frac{1}{9})(x-1)=0\) --> \(x+\frac{1}{9}=0\) or \(x-1=0\) --> \(x=-\frac{1}{9}\) or \(x=1\).[/quote]

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(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]

(x-9)(x+1) is not a correct factoring of 9x^2-8x-1, it should be \((x+\frac{1}{9})(x-1)\) (\((9x+1)(x-1)\)).

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

Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm

Hope this helps.

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)?

Hi Gauravsoni,

\(9x^2-8x-1<0\)

Can be factored 9x^2 -9x+x-1 <0

9x(x-1)+1 (x-1) <0 or (9x+1)(x-1)<0 or x=-1/9 or x=1

hi bunnel

what do u mean when u say the following:



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 non-negative (the same for taking an even root of both sides of an inequality)

Adding/subtracting/multiplying/dividing inequalities: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html

As for your other questions: it depends on a question.

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.

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

What is wrong with the following in the negative scenario B below?

|2x+3| > |7x-2|

(A)
2x + 3 > 7x -2
2x - 7x > -5
-5x > -5
x < -5 / -5
x < 1

(B)
-(2x+3) > 7x-2
2x + 3 < -1 (7x-2)
2x + 3 < -7x + 2
2x + 7x < 2 - 3
9x < -1
x < -1/9 ?

This doesn't match ans choice C

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

The critical points (also known as transition points or key points) for |2x + 3| and |7x - 2| are x = -3/2 and x = 2/7, respectively. These critical points are where the expressions inside the modulus equal zero, signifying a transition from negative to positive values (or vice-versa).

Considering the two critical points, we analyze three ranges:

1. If x < -3/2, then 2x + 3 < 0 and 7x - 2 < 0. Thus, |2x + 3| = -(2x + 3) and |7x - 2| = -(7x - 2). Hence, for this range we'd get: -(2x + 3) > -(7x - 2), which gives x > 1. Discard this range because it contradicts with the range we consider: x < -3/2.

2. If -3/2 ≤ x ≤ 2/7, then 2x + 3 ≥ 0 and 7x - 2 ≤ 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = -(7x - 2). Hence, for this range we'd get: 2x + 3 > -(7x - 2), which gives x > -1/9. Combining with the range we consider, we'd get -1/9 < x ≤ 2/7.

3. If x > 2/7, then 2x + 3 > 0 and 7x - 2 > 0. Thus, |2x + 3| = 2x + 3 and |7x - 2| = 7x - 2. Hence, for this range we'd get: 2x + 3 > 7x - 2, which gives x < 1. Combining with the range we consider, we'd get 2/7 ≤ x < 1.

So, we get two ranges: -1/9 < x ≤ 2/7 and 2/7 ≤ x < 1, which give: -1/9 < x < 1.

Answer: C.

10. Absolute Value

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Hope it helps.

Hi Bunuel,

Can you please explain why sometimes we need to identify the critical points and analyse the three different ranges (such as in this question where you solved above, and then other times you can solve absolute value questions by setting two equations, e.g.

1. +ve = +ve,
2. +ve = -ve

For example:
Scenario 1: x + 1 = 2(x - 1)
Scenario 2: (x + 1) = -2(x - 1)

I've found that on some questions, Method 2 works (and is quicker), whilst other times you need to find critical points and solve 3 equations. Thank you!

Here is my explanation to it. Hope you like it.



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