
#1
Apr2512, 10:36 AM

P: 333

Is this a correct a way of thinking for solving absolute value equations? Say I have 2x+6x+3=x and want to solve for x, then I have:
For 2x+6 2x + 6 if x ≥ 3 2x  6 if x < 3 For x+3 x+3 if x ≥ 3 x3 if x<3 For x x if x ≥ 0 x if x < 0 Am I supposed to look at the cases where x is in a valid interval? e.g. I can't have 2x+6(x+3) = x because x can't be equal to or greater than both 0 and 3. If this is the case, then why can't I have (2x6)  (x+3) = x? This is where x<3 and x<0, isn't this valid? If x is less than 3 then it's also less than 0 



#2
Apr2512, 11:31 AM

P: 876

Yes, that is also a valid equation for x. You just have to look at each case, evaluating one absolute value at a time.
If 2x+6x+3=x, then 2x + 6  x + 3 = x when x >= 3 and 2x  6  x + 3 = x when x < 3. Now we evaluate x + 3 in this pair of equations to get four possibilities: 2x + 6  x  3 = x when x >=3 and x >=  3 2x + 6 + x + 3 = x when x >=3 and x <  3 2x  6  x  3 = x when x < 3 and x >=  3 2x  6 + x + 3 = x when x < 3 and x < 3 Out of the four, only the first and last equations correspond to real values of x. Now we evaluate the x in each of those two equations to get four possibilities: 2x + 6  x  3 = x when x >= 3 and x >= 0 2x + 6  x  3 = x when x >= 3 and x < 0 2x  6 + x + 3 = x when x < 3 and x >= 0 2x  6 + x + 3 = x when x < 3 and x < 0 The third equation does not correspond to any values of x, so we now have 3 equations without absolute values whose solutions are the same as those of the original equation with absolute values. x + 3 = x when x >= 0 x + 3 = x when x >= 3 and x < 0 x  3 = x when x < 3 Since the first and last equation have no solutions, the middle equation contains the only valid solution for the original equation. 



#3
Apr2512, 03:45 PM

Sci Advisor
P: 5,937

The left side can be simplified. 2x+6  x+3 = x+3.
So you have x+3 = x. You have 3 cases. x < 3, x 3 = x, (no solution) 3 < x < 0, x + 3 = x or x = 3/2 0 < x, x+3 = x, (no solution) 



#4
Apr2512, 06:20 PM

P: 333

Solving absolute value equations
Thank you for the replies, I don't want to sound arrogant or anything but at the moment I'm kind of more interested in whether the way I currently think of it is accepted and if it is, why the last part of my initial post doesn't make sense (aside from the fact that you just can't get a solution when you solve it, I'm more interested in the intervals)




#5
Apr2512, 06:46 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,885

If 3< x< 0, 2x+ 6= 2x+ 6 and x+3= x+ 3 but x is still x. The equation becomes 2x+ 6 (x+3)= x. That gives x+ 3= x which reduces to 2x= 3= or x= 3/2. Since that is between 3 and 0, that is a valid solution: 2x+6x+3= 3+ 63/2= 3 3/2= 3/2= 3/2. If 0< x, 2x+ 6= 2x+ 6, x+3= x+ 3, and x= x. The equation becomes 2x+ 6 (x+3)= x. That gives x+ 3= x or 3= 0. Again, that is not true so there is no x larger than 0 that satisfies the equation. It is not a matter of "x< 3" not being valid it is simply that, in that case, the equation reduces to one that has no solution. 



#6
Apr2612, 07:07 PM

P: 333

^ Ah ok, I was thinking all valid intervals yielded valid equations...thank you



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