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Re: Math: Absolute value (Modulus)
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14 Jun 2015, 00:44
pacifist85 wrote: I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:
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. We reject the solution because our condition is not satisfied (1 is not less than 8) Why are the all the signs here negative? b) −8≤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.) Why are the signs in the first part here negative and not the second? c) −3≤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.) Why is this similar to the original equation? d) x≥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) This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?
Thank you! hi pacifist85, 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) Why are the all the signs here negative?we have three mod here .. 1)x+3.. this term will be positive because of mod sign but if x<8, say 9.. x+3=9+3=6... so the value of x+3 is actually negative but l6l=6.. therefore to get the value of x we put a negative sign when we open modulus sign.. 2)4−x.. here 4(9)=13 so the values do not change whether with mod sign or without.. so the existing ive sign remains.. 3)8+x... when x=9, 8=(9)=1.. same as 1 above .. you can similarly check for other three parts.. hope it helped
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Re: Math: Absolute value (Modulus)
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14 Jun 2015, 22:14
pacifist85 wrote: I am not sure I undertand this completely. I have a problem with the signs we use depending on the conditions:
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. We reject the solution because our condition is not satisfied (1 is not less than 8) Why are the all the signs here negative? b) −8≤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.) Why are the signs in the first part here negative and not the second? c) −3≤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.) Why is this similar to the original equation? d) x≥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) This makes more sense as x will be positive. But is this why all the signs outside the brackets are positive?
Thank you! IT is derives from the definition of absolute value: x = x if x is positive and x if x is negative. You have three transition points: 8, 3 and 4. When x < 8, all three expressions (x + 8), (x + 3) and (x  4) will be negative (check for say x = 10 to understand). So when you remove the absolute value from x + 8, you will get (x + 8) (because x + 8 is negative).Similarly, when you remove absolute value from x + 3, you will get  (x + 3) because x+3 is negative and so on... Similarly, when 8 < x < 3, (x is to the right of 8) (x+8) will be positive but other two expressions will be negative (check for say x = 5 to understand). and so on...
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Re: Math: Absolute value (Modulus)
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07 Jan 2017, 04:24
Hi everyone,
I'm having problems understanding the example 1 ( as written below). Why are they key points 8, 3 and 4, instead of 3, 4, 8?
Many thanks
Example #1 Q.: x+3−4−x=8+xx+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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not less than 8)
b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) > x=−15x=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) > x=9x=9. We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not more than 4)



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Re: Math: Absolute value (Modulus)
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07 Jan 2017, 05:08
lauramo wrote: Hi everyone,
I'm having problems understanding the example 1 ( as written below). Why are they key points 8, 3 and 4, instead of 3, 4, 8?
Many thanks
Example #1 Q.: x+3−4−x=8+xx+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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not less than 8)
b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) > x=−15x=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) > x=9x=9. We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not more than 4) x  a = b means x is b units away from a. x + a = b x  (a) = b means x is b units away from a. Hence, if we have x + 3, it means the transition point is 3. For more on this, check: https://www.veritasprep.com/blog/2011/0 ... edoredid/
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Re: Math: Absolute value (Modulus)
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20 Jan 2017, 09:15
If x is an integer, what is the value of x? (1) –2(x + 5) < –1 (2) –3x > 9
from S1 it should be x>4.5...........but in explanation its given x>4.5. pls explain the logic



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Re: Math: Absolute value (Modulus)
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21 Jan 2017, 03:33
gupta87 wrote: If x is an integer, what is the value of x? (1) –2(x + 5) < –1 (2) –3x > 9
from S1 it should be x>4.5...........but in explanation its given x>4.5. pls explain the logic –2(x + 5) < –1 Divide by 2 and flip the sign: x + 5 > 0.5; Deduct 5: x > 0.5  5 x > 4.5 Discussed here: ifxisanintegerwhatisthevalueofx109983.htmlHope it helps.
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Re: Math: Absolute value (Modulus)
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21 Jan 2017, 07:55
Bunuel wrote: gupta87 wrote: If x is an integer, what is the value of x? (1) –2(x + 5) < –1 (2) –3x > 9
from S1 it should be x>4.5...........but in explanation its given x>4.5. pls explain the logic –2(x + 5) < –1 Divide by 2 and flip the sign: x + 5 > 0.5; Deduct 5: x > 0.5  5 x > 4.5 Discussed here: ifxisanintegerwhatisthevalueofx109983.htmlHope it helps. understood.but please correct me where am i going wrong?? 2x10<1 => 2x<9 =>2x>9 => x>4.5



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Re: Math: Absolute value (Modulus)
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22 Jan 2017, 03:41
gupta87 wrote: Bunuel wrote: gupta87 wrote: If x is an integer, what is the value of x? (1) –2(x + 5) < –1 (2) –3x > 9
from S1 it should be x>4.5...........but in explanation its given x>4.5. pls explain the logic –2(x + 5) < –1 Divide by 2 and flip the sign: x + 5 > 0.5; Deduct 5: x > 0.5  5 x > 4.5 Discussed here: ifxisanintegerwhatisthevalueofx109983.htmlHope it helps. understood.but please correct me where am i going wrong?? 2x10<1 => 2x<9 =>2x>9 => x>4.52x < 9 Divide by 2 and flip the sign x > 4.5,
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Re: Math: Absolute value (Modulus)
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09 Feb 2018, 00:30
regarding this x  3 well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? \(x\) could be 5 no ? can you advice in which case can i square both sides od modulus ? can i do it x+4 + x  3 = 10 How about this question \(x2 + x+5  x3 > x+13\) woukd i be able to square both sides ? if not please explain why ? thank you in advance for your kind explanation and have an awesome day



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Re: Math: Absolute value (Modulus)
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09 Feb 2018, 01:04
dave13 wrote: regarding this x  3 well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? \(x\) could be 5 no ? can you advice in which case can i square both sides od modulus ? can i do it x+4 + x  3 = 10 How about this question \(x2 + x+5  x3 > x+13\) woukd i be able to square both sides ? if not please explain why ? thank you in advance for your kind explanation and have an awesome day Hey Dave, I am not sure I understand your question but here is something that might help you: Why do we take the range as 'x  3 > 0' ? or 'x + 5 > 0' ? The definition of absolute value will help you understand this. It is discussed here: https://www.veritasprep.com/blog/2014/0 ... thegmat/Can you square x+4 + x  3 = 10? Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as x+4 + x  3 > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign. \(x+4 + x  3 = 10\) \(x + 4^2 + x  3^2 + 2*x+4*x  3 = 10^2\) The term 2*x+4*x  3 will continue to have absolute value sign. How will you take care of it?
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Re: Math: Absolute value (Modulus)
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09 Feb 2018, 12:32
VeritasPrepKarishma wrote: dave13 wrote: regarding this x  3 well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? \(x\) could be 5 no ? can you advice in which case can i square both sides od modulus ? can i do it x+4 + x  3 = 10 How about this question \(x2 + x+5  x3 > x+13\) woukd i be able to square both sides ? if not please explain why ? thank you in advance for your kind explanation and have an awesome day Hey Dave, I am not sure I understand your question but here is something that might help you: Why do we take the range as 'x  3 > 0' ? or 'x + 5 > 0' ? The definition of absolute value will help you understand this. It is discussed here: https://www.veritasprep.com/blog/2014/0 ... thegmat/Can you square x+4 + x  3 = 10? Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as x+4 + x  3 > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign. \(x+4 + x  3 = 10\) \(x + 4^2 + x  3^2 + 2*x+4*x  3 = 10^2\) The term 2*x+4*x  3 will continue to have absolute value sign. How will you take care of it? Hi VeritasPrepKarishma, many thanks for you reply and the article i wish i could answer your question, i dont know how i would take care off it but there is one math wizzard called pushpitkc (perhaps he learnt from you ) who actually applied techique of squaring modulus in this question https://gmatclub.com/forum/howmanyval ... l#p2011543 i extracted his solution here (in wine red with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened otherwise your laptop risks to be frozen like mine so see below this magician squares modulus so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ? How many values of x satisfy the equation x79=11?
A. 1 B. 2 C. 3 D. 4 E. 5
If x7 = a, the equation will become a  9 = 11 ( how did he break this equation into two btw ?
Squaring on both sides, \(a^2\)−18a+81=121 \(a^2\)−18a+81=121
\(a^2\)−18a−40=0 \(a^2\)−18a−40=0
Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B)



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Re: Math: Absolute value (Modulus)
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12 Feb 2018, 05:48
VeritasPrepKarishma wrote: dave13 wrote: regarding this x  3 well \(x\) could be any value be negative or positive ? why are saying to put value greater than 3? \(x\) could be 5 no ? can you advice in which case can i square both sides od modulus ? can i do it x+4 + x  3 = 10 How about this question \(x2 + x+5  x3 > x+13\) woukd i be able to square both sides ? if not please explain why ? thank you in advance for your kind explanation and have an awesome day Hey Dave, I am not sure I understand your question but here is something that might help you: Why do we take the range as 'x  3 > 0' ? or 'x + 5 > 0' ? The definition of absolute value will help you understand this. It is discussed here: https://www.veritasprep.com/blog/2014/0 ... thegmat/Can you square x+4 + x  3 = 10? Yes you can. It is an equation. You can always square an equation. Even if it were an inequality such as x+4 + x  3 > 10, you can still square it since both sides are definitely non negative. But what will we achieve by squaring it? It doesn't remove the absolute value sign. \(x+4 + x  3 = 10\) \(x + 4^2 + x  3^2 + 2*x+4*x  3 = 10^2\) The term 2*x+4*x  3 will continue to have absolute value sign. How will you take care of it? Hi VeritasPrepKarishma, many thanks for you reply and the article i wish i could answer your question, i dont know how i would take care off it but there is one math wizzard called pushpitkc (perhaps he learnt from you ) who actually applied techique of squaring modulus in this question https://gmatclub.com/forum/howmanyval ... l#p2011543 i extracted his solution here (in wine red with question stem and answer choices ), so you dont overload your laptop with many internet browser tabs opened otherwise your laptop risks to be frozen like mine so see below this magician squares modulus so my question when can i apply squaring to modulus ? in which cases can I apply and which cases it wont work ? How many values of x satisfy the equation x79=11?
A. 1 B. 2 C. 3 D. 4 E. 5
If x7 = a, the equation will become a  9 = 11 ( how did he break this equation into two btw ?
Squaring on both sides, \(a^2\)−18a+81=121 \(a^2\)−18a+81=121
\(a^2\)−18a−40=0 \(a^2\)−18a−40=0
Solving for a, a = 20 or 2
If x7 = 20 x could take the value 27 or 13 However, it is not possible to get a value of x where x7 = 2.
Therefore, there are 2 values of x which satisfy the equation(Option B) Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign. \(x^2 = (x)^2 = (x)^2\) \(x + 2^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left) Look at the example you gave: \(x79=11\) Just denote x  7 as a. You get \(a  9 = 11\) Square it and get 'a'. But you don't need a. You need the value of x. For that, we know \(a = 20 = x  7\) (as assumed before) Squaring again will get the answer. How about an expression such as \(2x + 3 = x  4\) Again squaring will help since absolute value sign from both sides will go away. How about \(x+4 + x  3 = 10\) When you square the left hand side, you use \((a + b)^2 = a^2 + b^2 + 2ab\) where \(a = x + 4\) and \(b = x  3\) So \((x+4 + x  3 )^2 = x + 4^2 + x  3^2 + 2*x + 4*x  3\) \(= (x + 4)^2 + (x  3)^2 + 2*x + 4*x  3\) Here is the problem. 2*x + 4*x  3 still has absolute value signs.
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Re: Math: Absolute value (Modulus)
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18 Feb 2018, 08:26
Hi VeritasPrepKarishmamany thanks for taking to explain ! :) i still have some questions :) you say: " Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign." what does "within the sign" mean ? Another question: ok so we have \(x+4 + x  3 = 10\) tha formula we use: \((a + b)^2 = a^2 + b^2 + 2ab\) Do you mean this formula wont work because \(a = x + 4\) and \(b = x  3\) whereas is in other equations we only one modulus on each side like this one \(2x + 3 = x  4\) or this one \(a  9 = 11\) but on the other hand \(a = x + 4\) and \(b = x  3\) i could write / divide left side into two modulus x + 4 and x  3 so \(x + 4\) > following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get > \(x^2+8x+16 = 0\) \(x  3\)> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get > \(x^2  9x +9 = 0\) Another point: you write this > "So\((x+4 + x  3 )^2\)= \( x + 4^2 + x  3^2 + 2*x + 4*x  3\) \(= (x + 4)^2 + (x  3)^2 + 2*x + 4*x  3\)" i dont understand why after this (x+4 + x  3 )^2 you still keep modulus sign :? we square it right for example here everything is simple and clear > \(x + 2^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left) < you square and modulus sign is gone:) i would appreciate your explanation best, D.



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Re: Math: Absolute value (Modulus)
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19 Feb 2018, 03:29
dave13 wrote: Hi VeritasPrepKarishmamany thanks for taking to explain ! i still have some questions you say: " Note here that in this case squaring will take care of the absolute value sign since the whole expression on the left is within the sign." what does "within the sign" mean ? It means the entire left side is within the absolute value sign. Note that these 2 cases are different: \(x + y^2 = (x + y)^2 = x^2 + y^2 + 2xy\) \((x + y)^2 = x^2 + y^2 + 2*x*y = x^2 + y^2 + 2*x*y\) So if x were say 4 and y were 2, in the first case, we would have \((4)^2 + 2^2 +2*(4)*2 = 4^2 + 2^2  2*4*2\) and in the second case we would have \((4)^2 + 2^2 + 2*4*2 = 4^2 + 2^2 + 2*4*2\) So to solve the second case, we need to know the signs of x and y because even after squaring, we still have absolute value signs. dave13 wrote: Another question: ok so we have \(x+4 + x  3 = 10\)
tha formula we use: \((a + b)^2 = a^2 + b^2 + 2ab\)
Do you mean this formula wont work because
\(a = x + 4\) and \(b = x  3\) whereas is in other equations we only one modulus on each side like this one \(2x + 3 = x  4\) or this one \(a  9 = 11\)
but on the other hand \(a = x + 4\) and \(b = x  3\) i could write / divide left side into two modulus x + 4 and x  3
so \(x + 4\) > following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get > \(x^2+8x+16 = 0\)
\(x  3\)> following this formula \((a + b)^2 = a^2 + b^2 + 2ab\) i get > \(x^2  9x +9 = 0\) So the same case is here. When you have \(a = x + 4\) and \(b = x  3\), the 2ab term retains the absolute signs and hence you cannot solve it. To solve an equation, you need to get rid of all absolute value signs. dave13 wrote: Another point: you write this > "So\((x+4 + x  3 )^2\)= \( x + 4^2 + x  3^2 + 2*x + 4*x  3\) \(= (x + 4)^2 + (x  3)^2 + 2*x + 4*x  3\)" i dont understand why after this (x+4 + x  3 )^2 you still keep modulus sign we square it right for example here everything is simple and clear > \(x + 2^2 = (x + 2)^2 = x^2 + 2^2 + 2*x*2\) (No absolute value sign left) < you square and modulus sign is gone:) Yes, the absolute value sign is gone from \(x + 2^2\) and from \(x  3^2\) but what about from \(2*x + 2*x  3\) ? (this is the 2ab term).
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Re: Math: Absolute value (Modulus)
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04 Jun 2018, 21:35
Can someone please explain how we change signs in the steps below? 3steps approach for complex problems Let’s consider following examples, Example #1 Q.: x+3−4−x=8+xx+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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not less than 8) b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) > x=−15x=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.) c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) > x=9x=9. We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.) d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not more than 4) (Optional) The following illustration may help you understand how to open modulus at different conditions. Image Answer: 0



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Re: Math: Absolute value (Modulus)
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04 Jun 2018, 21:49
chinmay96 wrote: Can someone please explain how we change signs in the steps below? 3steps approach for complex problems Let’s consider following examples, Example #1 Q.: x+3−4−x=8+xx+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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not less than 8) b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) > x=−15x=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.) c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) > x=9x=9. We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.) d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not more than 4) (Optional) The following illustration may help you understand how to open modulus at different conditions. Image Answer: 0 Explained here: https://gmatclub.com/forum/x34x8x ... l#p1241355 Hope it helps.
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Re: Math: Absolute value (Modulus)
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04 Jun 2018, 22:47
Got it! Thanks a ton! Bunuel wrote: chinmay96 wrote: Can someone please explain how we change signs in the steps below? 3steps approach for complex problems Let’s consider following examples, Example #1 Q.: x+3−4−x=8+xx+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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not less than 8) b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) > x=−15x=−15. We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.) c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) > x=9x=9. We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.) d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) > x=−1x=−1. We reject the solution because our condition is not satisfied (1 is not more than 4) (Optional) The following illustration may help you understand how to open modulus at different conditions. Image Answer: 0



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Re: Math: Absolute value (Modulus)
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08 Jun 2018, 08:39
walker wrote: ABSOLUTE VALUE3. Check conditions for each solution:a) \(x=5\) has to satisfy initial condition \(x1>=0\). \(51=4>0\). It satisfies. Otherwise, we would have to reject x=5. b) \(x=3\) has to satisfy initial condition \(x1<0\). \(31=4<0\). It satisfies. Otherwise, we would have to reject x=3. .... Example #1Q.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions: a) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8) b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.) c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.) d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not more than 4) (Optional) The following illustration may help you understand how to open modulus at different conditions. Answer: 0 Hey abhimahna, for step "3. Check conditions for each solution:", shouldn't we always have to plug in the value of x in the original equation with absolute value  or it's fine to directly evaluate if the value of x satisfies by just comparing each value of x to its inequality that we got from the points? I believe it's way faster if we just compare it to the inequality, as the example 1 above is doing. For example: If x2=2x3, what are the possible values for x? If you see the uploaded photo, you'll see that we get 2 same values of x (i.e. x=1) out of which one satisfies its inequality (x<3/2) and the other doesn't satisfy its inequality (x>=2). So we would count that as one solution. What the same value satisfied 2 inequalities, then would be consider that as 2 solutions or just 1? Range 1: \(x<3/2\) (x2) = (2x3) x=1 (Satisfies) Range 2: \(\frac{3}{2}\leq{x} < 2\)  (x2) = (2x3) x=5/3 (Satisfies) Range 3: \(x\geq2\) (x2)=(2x3) x=1 (Doesn't satisfy the range)
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Re: Math: Absolute value (Modulus)
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06 Jul 2018, 09:20
Thanks Sir for this amazing tutorial. Could you please explain how you solved this question because i cant get your solution clearly.
I. 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. x1<9 D. 5+x<4 E. 3+x<5 Solution: 10sec. Traditional 3steps method is too timeconsume technique. First of all we find length (91)=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.



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Re: Math: Absolute value (Modulus)
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25 Jul 2018, 06:15
hi walker thanks for explainig the concept can anyone please tell me the solution of 2x+1=5x2 please?




Re: Math: Absolute value (Modulus)
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25 Jul 2018, 06:15



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