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Example #1I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem? Q.: \(|x+3| - |4-x| = |8+x|\). How many solutions does the equation have? Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) \(x < -8\). \(-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) \(-8 \leq x < -3\). \(-(x+3) - (4-x) = (8+x)\) --> \(x = -15\). We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) \(-3 \leq x < 4\). \((x+3) - (4-x) = (8+x)\) --> \(x = 9\). We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4-x) = (8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not more than 4)
I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
a) \(x < -8\). \(-(x+3) - (4-x) = -(8+x)\) --> \(x = -1\). We reject the solution because our condition is not satisfied (-1 is not less than -8)
if x < -8, (x + 3) is always negative. So, modulus is non-negative and we need to change a sign: |x+3| = - (x+3) for x<-8 For example, if x = -10, |-10+3| = |-7| = 7 -(-10+3) = -(-7) = 7
In other words, |x| = x if x is positive and |x|=-x if x is negative.
Thanks for the explanation Walker. So would this then be correct:
Example #1 Q.: |x+3| - |4-x| = |8+x|. How many solutions does the equation have? Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x < -8. -(x+3) - (4-x) = -(8+x) --> x = -1. |x+3|is always - because x<-8 but l4-xl will always be positive because 4-x= a +, l8+xl will also be negative as well given x<-8
b) -8 \leq x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. on this one lx+3l again can only be negative and l4-xl will be still always be negative but given the range l8+xl may only be postiive hence l8+xl has to be positive or non-negative (o)?
c) -3 \leq x < 4. (x+3) - (4-x) = (8+x) --> x = 9. so on this term first one is positive second term continues to be positive and 3rd term is positive.
d) x \geq 4. (x+3) + (4-x) = (8+x) --> x = -1. on this one first term obviously positive second term is negative therefore the sign changes from - to + so from -(4-x) to +(4-x) since this will always be negative for given x range? And a negative times a negative gets us to a positive?
I just want to make sure I am thinking about this correctly. Thanks walker!
Sorry walker one more question on the below as I review this Thinking of inequality with modulus as a segment at the number line.
For example, Problem: 1<x<9. What inequality represents this condition? Image A. |x|<3 B. |x+5|<4 C. |x-1|<9 D. |-5+x|<4 E. |3+x|<5 Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.
Why would be looking for 4 on the right side? shouldn't this be 5 as the midpoint? And why do we want the left side to be 0 at x=5? Id like to understand this a little better
II. Converting inequalities with modulus into range expression. In many cases, especially in DS problems, it helps avoid silly mistakes.
For example, |x|<5 is equal to x e (-5,5). |x+3|>3 is equal to x e (-inf,-6)&(0,+inf)
How did you convert lx+3l>3 into (-inf,-6) and (0,+inf)
Sorry these maybe simple questions but I just want to grasp the concept firmly
It is quite convenient to remember the definition of modulus. |x| is the distance of x from 0 on the number line. So if |x| = 1, we are looking for points which are at a distance 1 away from 0. If |x| < 1, we are looking for points which are at a distance less than 1 away from 0. If |x| > 1, we are looking for points which are at a distance more than 1 away from 0.
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If |x - 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Great thanks for the explanation on that Karishma - very very very helpful.
Can you also help me with how we converted lx+3l>3 into (-inf,-6) and (0,+inf)?
Is this because x+3>=0 (non-negative) and negative x+3<0 (negative). So we get to x+3=3 so x=0 which fits condition 1 above and this will go to infinity
-(x+3)=3 so you get -x=6 or x=-6 which fits condition two above so x<-6 so therefore it goes to negative infinity?
I think I am close. appreciate the help.
Yes, lx+3l>3 gives us two cases
Case 1: x + 3 >= 0 or x > = -3 Then (x + 3) > 3 or we can say x > 0
Case 2: x + 3 < 0 or x < -3 Then -(x + 3) > 3 or we can say x < -6
So either x > 0 which translates to (0, inf) or x < -6 which translates to (-inf, -6)
OR consider that lx+3l>3 means distance of x from -3 is more than 3. If you go to 3 steps to right from -3, you reach 0. Anything after than is ok. If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
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