there are four cases
Case 1)
2x+3 > 7x -2
5> 5x
x< 1
Case 2
2x+3 > -(7x-2)
2x+3>-7x +2
9x> -1
x> -1/9
Case 3
-(2x+3) > (7x-2)
-2x-3 >7x-2
-1> 9x
x< -1/9
Case 4
-(2x+3) > -(7x-2)
x> 1
When we substitute the values for solutions of all the above cases in |2x+3|>|7x-2|
X> -1/9 & X < 1 satisfies the above eqn.
So -1/9 < X < 1 is the range of X
Inequality
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Solve the following two equations:Anahatha wrote:what is the range of x?
|2x+3|>|7x-2| Can someone please explain step by step?
2x+3 = 7x-2 and 2x+3 = -(7x-2).
2x+3 = 7x-2
5 = 5x
x = 1.
2x+3 = -(7x-2)
2x+3 = -7x+2
9x = -1
x = -1/9.
x = 1 and x = -1/9 are the critical points: the points at which the graphs of |2x+3| and |7x-2| intersect. To the right and to the left of these critical points, one of the two graphs must yield higher values than the other.
Thus, to determine the range of |2x+3|>|7x-2| -- to determine where |2x+3| yields higher values than |7x-2| -- we need to test a value less than -1/9, a value between -1/9 and 1, and a value greater than 1.
Let x = -1.
|2*(-1)+ 3| > |7*(-1)- 2|
|1| > |-9|
1 > 9.
Doesn't work. We know that x cannot be less than - 1/9.
Let x = 0.
|2*(0)+ 3| > |7*(0)- 2|
|3| > |2|
3 > 2.
This works. We know that x can be between -1/9 and 1.
Let x = 2.
|2*(2)+ 3| > |7*(2)- 2|
|7| > |12|
7 > 12.
Doesn't work. We know that x cannot be greater than 1.
Thus, the range of |2x+3|>|7x-2| is -1/9 < x < 1.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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