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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 10:05
As y can be negative of positive for absolute value to hold, then any 2 values for y.
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 10:33
Answer is 0.
Take four possible cases of y, i.e, y<8, 8<y<3, 3<y<4 and y>4. In each case, solve the equation, y gets a value outside of assumed range of each case.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 10:37
here the solution of the question
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 10:52
Four scenarios
4>y>0
y+3=y+8+4y y=9
Not possible
y>4
y+3=8+y+y4 y=1
Not possible
y<8 y3=y8+4y y=1 Not possible
3>y>8 y3=8+y+4y y=15 Not possible
0>y>3 y+3=8+y+4y y=9
Not possible
Thus, 0 values



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 11:12
It took me way too long to solve this problem (3m and 46s) I followed the 3 step approach explained in the Absolute Value theory here ( https://gmatclub.com/forum/mathabsolut ... 86462.html), and got the answer 0, which is option A. But obviously from the time standpoint, close to 4 mins is terrible for this kind of question. Can one of the experts help me with a quicker/alternate solution that can be done in less than 2 mins? Here are the details of what I did: Step 1: Identified the critical points, 8, 3 and 4 and set conditions, y < 8, 8 <= y < 3, 3 <= y < 4, and y >= 4 Step 2: For each one of the regions listed in step 1, used conditions to open modulus and solve for y Step 3: Checked if the resulting y value falls within the region for each one of the ranges in Step 1. None of the y values did, so the answer is 0



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 11:26
Basically there can be different combinations made
Firstly all r positive
1 y+3=8+y+4−y
y+3=8+y+4y y=9
Putting 9 in equation Both sides are not equal
2. If one is negative
y+3=8y+4y y=7/3
This satisfies the eqaution
3 both negative
y+3=y84+y y=15
Again this doesn't satify
Now y3=y8+4y y=1 again this doesn't satisfy equation ,
Then y3=y8+4y y=1
Then y3=y84+y y=9
This again doesn't satisfy
There is only one solution as 7/3
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 11:51
firstly analyzing the problem i was confused whether to plug in or form equation with cases '
however considering the equation y+3=8+y+4−y let A= y+3 B= 8+y C=4−y
A is always positive : or zero B is always positive or zero C is always positive or zero
lets compare A and B y+3=8+ythey can never be equal distance of one number and 3 on number line cannot be equal to distance of same number and 8 so no value exists
adding C to the equation further complicates the issue as C is positive and distance of B and C together cannot be equal to A
so answer A : no value exist



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 12:09
y+3=8+y+4y The center points can be gotten by equating the absolute values to zero. Hence the key points are 3, 8 and 4. The possible conditions are y<8; 8<=y<3; 3<=y<4; and y>=4. When y<8, (y+3)=(8+y)+(4y) Solving for y yields y=1 Since 1 does not fit into the range of values y<8, we reject this solution. When 8<=y<3, (y+3)=(8+y)+(4y) y=15. 15 does not fall within the above range hence we reject this solution as well. 3<=y<4; (y+3)=(8+y)+(4y) Solving for y, y=9. And nine falls outside the range, hence we reject this solution as well. y>=4 (y+3)=(8+y)(4y) y=1, which is not greater than 4, hence we reject this solution as well. This means there is no solution to this equation. Hence choice A is the right answer.
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 12:30
there is no such y that will match the equation: positive numbers are out because y+3 will be always less that 8+y (keeping in mind that 4y may not be negative. zero is out negative numbers will make 4y big positive number
answer  A (0)



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 12:56
This equation has 3 critical points, 3,8, 4. Hence, 4 conditions to check. 1. y>4 (y+3)=(8+y)(4y) Since, x=x if x>/=0. x=x if x<0 y=1 , but condition is y>4. No solution in this range 2. 3<y</=4 (y+3)=(8+y)+(4y) y=11, outside the range. No solution 3. 8<y</=3 (y+3)=(8+y)+(4y) y=15
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 15:39
Explanation Given; y+3 = 8+y + 4y Solving for y; Condition 1 y+3 = 8+y+4y This gives; y = 9 Testing y = 9 in the equation above We have; 9+3 = 8+9 + 49 Thus, 12 = 17  5 Therefore, 12 = 12 (y=9 satisfies the equation) Condition 2 If we negate the left side of the equation, we have y3 = 8+y+4y This gives; y=15 Testing y=15 in the original equation We have; 15+3 = 815+4+15 12 = 12 (y = 15 does not satisfy the equation) Condition 3 Negating only one side of the equation on the right side, we have y+3 = 8+y  (4y) y+3 = 8+y4+y y+3 = 4+2y Thus, y=1 Testing y=1 in the original equation We have; 1+3 = 81 + 4+1 2 = 12 (y=1 does not satisfy the equation) Therefore, the equation has only one solution and it's answer choice B. Posted from my mobile device
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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 17:53
There are 4 ranges: Y≤ 8, Y =1 8≤Y≤3, Y= 15 3≤Y≤4, Y = 15 Y≥4, Y = 1
So, none of them comply with the ranges, so (A)
None of them satisfy the equation, so (A)



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 18:08
C  2 values possible



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 20:00
tried plugging in several numbers and realized it never matches so A)



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 21:49
The three critical values are 8, 3 and 4
We can have 4 conditions:
1. y < 8: (y + 3) =  (8 + y) + (4 – y) => y = 1 This does not fall in the range of y < 3. Therefore, we can reject the solution.
2. 8 <= y < 3: (y + 3) = (8 + y) + (4 – y) => y = 15 This does not fall in the range of 8 <= y < 3. Therefore, we can reject the solution.
3. 3 <= y <4 (y + 3) = (8 + y) + (4 – y) => y = 9 This does not fall in the range of 3 <= y <4. Therefore, we can reject the solution.
4. y >= 4 (y + 3) = (8 + y) – (4 – y) => y = 1 This does not fall in the range of y >= 4. Therefore, we can reject the solution.
Therefore, no value of y satisfy. Answer A.



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How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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Updated on: 03 Jul 2019, 22:22
QUESTION: How many different values of y satisfy y+3=8+y+4−y ?
Each y+3, 8+y, and 4−y has a value of 0 or positive number.
For y>= 3, y+3 is always less than 8+y > (y+3) < (y+8) > 3 < 8 Hence, y+3 is also always less than 8+y + 4−y
For y< 3, y+3 is always less than 4y > (y+3) < (4y) > 3 < 4 Hence, y+3 is also always less than 8+y + 4−y
As discussed, for any y value, y+3 will never be equal to 8+y+4−y
Answer is (A)
Originally posted by chondro48 on 03 Jul 2019, 22:05.
Last edited by chondro48 on 03 Jul 2019, 22:22, edited 1 time in total.



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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03 Jul 2019, 22:20
This is a tricky question. First of we should understand that (Y+3), (8+Y) & (4Y) will change signs at 3, 8 & +4 respectively. These 3 points are the minimum ( f(y)=0) for given 3 modulus terms independently.
These 3 points will create 4 intervals on a number line as per following:
(1) Y< 8 (2) 8 <= Y < 3 (3) 3 <= Y < 4 (4) Y >= 4
Open all 3 modulus in each interval and solve for Y. Value which we get for Y in each interval doesn't belong to corresponding interval.
So, effectively we will have "0" values of Y which can satisfy given equation.
ANSWER : A



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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04 Jul 2019, 17:37
Mahmoudfawzy83 wrote: to find the possible values of y in \(y+3=8+y+4−y\), there are 4 intervals to examine: \(x≥4\) > y+3 = 8+y + y4 > \(y= 1\) > invalid because not within stated range \(4>x ≥3\) > y+3 = 8+y + 4y > \(y= 9\) > invalid because not within stated range \(3>x ≥8\) > y3 = 8+y + 4y > \(y= 15\) > invalid because not within stated range \(8 ≥x\) > y3 = 8y + 4y > \(y= 1\) > invalid because not within stated range
so there is acceptable value upon validating, so A Hello Mahmoudfawzy83! How do we know which term we have to change to neg depending the range? Kind regards!



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Re: How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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04 Jul 2019, 17:51
hi jfranciscocuencagjfranciscocuencag wrote: How do we know which term we have to change to neg depending the range? By trying any value from the range, and see whether it will stay positive or negative. lets take \(x>4\) choose any number ... 5 or 100 .. the same will do: when substituting y with 5, y+3 will be 8, so keep it> +y+3 when substituting y with 5, 8+y will be 13, so keep it > +8+y when substituting y with 5, 4y will be 1, so invert the signs > 4+y lets take \(8 >x\) choose any number ... 9 or 100 .. the same will do: when substituting y with 9, y+3 will be 6, so invert the signs> y3 when substituting y with 9, 8+y will be 1, so invert the signs > 8y when substituting y with 9, 4y will be 13, so keep it > +4y It is tedious, but by practice, you will used to it. It is a very important step to use with every absolute value question.
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How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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04 Jul 2019, 18:19
choose any number ... 5 or 100 .. the same will do: when substituting y with 5, y+3 will be 8, so keep it> +y+3 when substituting y with 5, 8+y will be 13, so keep it > +8+y when substituting y with 5, 4y will be 1, so invert the signs > 4+y lets take \(8 >x\) choose any number ... 9 or 100 .. the same will do: when substituting y with 9, y+3 will be 6, so invert the signs> y3 when substituting y with 9, 8+y will be 1, so invert the signs > 8y when substituting y with 9, 4y will be 13, so keep it > +4y Mahmoudfawzy83I have some doubt regarding the second example: when substituting y with 9, y+3 will be 6, so invert the signs> y3 when substituting y with 9, 8+y will be 1, so invert the signs > 8y The numbers are 6 and 1 so they didn't change the sign, they are still neg. Why do we have to invert the sign here?




How many different values of y satisfy y + 3 = 8 + y + 4  y ?
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