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

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How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:00
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 19 Jul 2019, 02:08
Bunuel wrote:
How many different values of \(y\) satisfy \(|y + 3| = |8 + y| + |4 - y|\) ?

A. 0
B. 1
C. 2
D. 3
E. 4

 

This question was provided by Crack Verbal
for the Game of Timers Competition

 



OFFICIAL EXPLANATION: FROM CRACK VERBAL:



|y + 3| is less than |y + 8| and |4 - y| is never negative then the equation will be inconsistent.
When |y + 3| is greater than |y + 8| then |4 - y| will be greater than |y + 3| which will again make the equation inconsistent.

Answer: A
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How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post Updated on: 03 Jul 2019, 08:38
21
5
Concept: Critical Point Approach

Critical points are:
y + 3 = 0; 8 + y = 0 & 4 - y = 0
--> y = -8, - 3, 4

3 critical points --> 4 cases as shown
Image

Case 1: y < -8
--> l8 + yl = -(8 + y); ly + 3l = -(y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> - (y + 3) = - (8 + y) + (4 - y)
--> - y - 3 = - 8 - y + 4 - y
--> y = -4 + 3 = - 1
--> y = -1; Not Possible

Case 2: -8 ≤ y < -3
--> l8 + yl = (8 + y); ly + 3l = -(y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> - (y + 3) = (8 + y) + (4 - y)
--> - y - 3 = 8 + y + 4 - y
--> - y = 12 + 3 = 15
--> y = -15; Not Possible

Case 3: -3 ≤ y ≤ 4
--> l8 + yl = (8 + y); ly + 3l = (y + 3) & l4 - yl = (4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> (y + 3) = (8 + y) + (4 - y)
--> y + 3 = 8 + y + 4 - y
--> y = 12 - 3 = 9
--> y = 9; Not Possible

Case 4: y > 4
--> l8 + yl = (8 + y); ly + 3l = (y + 3) & l4 - yl = -(4 - y)
--> |y + 3| = |8 + y| + |4 - y|
--> (y + 3) = (8 + y) - (4 - y)
--> y + 3 = 8 + y - 4 + y
--> - y = 4 - 3 = 1
--> y = -1; Not Possible

IMO Option A

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Originally posted by Dillesh4096 on 03 Jul 2019, 08:35.
Last edited by Dillesh4096 on 03 Jul 2019, 08:38, edited 1 time in total.
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How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post Updated on: 03 Jul 2019, 09:58
3
Ugh, what an ugly question....
Anyway, back to it. Let's first find intervals for each modulus.
|y + 3|
If positive - |y + 3|, key point is y+3>=0 or y>=-3, so we get |y + 3|=y+3
If negative - |y + 3|, key point is y+3<0 or y<-3, so we get |y + 3|=-(y+3)=-y-3
|8 + y|
If positive - |8 + y|, key point is 8+y>=0 or y>=-8, so we get |8 + y|=8+y
If negative - |8 + y|, key point is 8+y<0 or y<-8, so we get |8 + y|=-(8+y)=-y-8
|4 - y|
If positive - |4 - y|, key point is 4-y>=0 or y<=4, so we get |4 - y|=4-y
If negative - |4 - y|, key point is 4-y<0 or y>4, so we get |4 - y|=-(4-y)=4+y
So, we get
\(1.\) \(y<-8\)
\(2.\) \(-8\leq{y}<-3\)
\(3.\) \(-3\leq{y}\leq4\)
\(4.\) \(y>4\)

As we see no number falls between the above intervals, hence A

Originally posted by mira93 on 03 Jul 2019, 08:09.
Last edited by mira93 on 03 Jul 2019, 09:58, edited 2 times in total.
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How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post Updated on: 03 Jul 2019, 09:42
|y+3| = |8+y| + |4-y|
|y+3| = |8+y+4-y|
|y+3| = |12|
We have four cases
(1) LHS +ve ,RHS -ve
(2) LHS -ve, RHS +ve
(3) LHS +ve, RHS +ve
(4) LHS -ve ,RHS -ve
Case 3&4 and case 1&2 will give same value
So take 1 &3
y+3= -12 —> y= -15
or y+3 =12 —> y=9. y=9/-15

Answer C



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Originally posted by Staphyk on 03 Jul 2019, 08:09.
Last edited by Staphyk on 03 Jul 2019, 09:42, edited 2 times in total.
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:13
Start the solving by opening modulus one at a time.

1 Case : all positive

y = 9

2. first mod negative (LHS Neg)
-y-3 = 8+y + 4-y
y = 15

3. rhs - one neg and one pos ; lhs pos

y+3 = -8-y + 4-y
y = -3

4. rhs- last neg

y+3 = 8+y - 4+y
y = -1

hence 4 different values
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:18
On the number line you have to check place three numbers to check conditions of these box values.
numbers: -8,-3,4

4 cases will be there:
case 1: if x<-8
all -ve
-y-3=-8-y-4+y
Solve for y .
y=9;

Simirly for -8<y<-3 y=-7/3
-3<y<4 y=9
and y>4.. y=-1

therefore, three different values of y exist.
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:25
1
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 + y-4 --> \(y= -1\) --> invalid because not within stated range
\(4>x ≥-3\) --> y+3 = 8+y + 4-y --> \(y= 9\) --> invalid because not within stated range
\(-3>x ≥-8\) --> -y-3 = 8+y + 4-y --> \(y= -15\) --> invalid because not within stated range
\(-8 ≥x\) --> -y-3 = -8-y + 4-y --> \(y= -1\) --> invalid because not within stated range

so there is acceptable value upon validating, so A
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:26
Only one solution when y=9 (y+3=8+y+4-y gives y=12-3=9) satisfies the equation. When all modulus is negative still y=9 (-y-3=-8-y-4+y gives -y=-9; y=9). IMO B
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:27
2
2
How many different values of y satisfy |y+3|=|8+y|+|4−y| ?

A. 0
B. 1
C. 2
D. 3
E. 4

Answer: A

|y+3| = |8+y|+|4-y|
=> |y+8| -|y+3| + |y-4| = 0

range of y: |y+8| |y+3| |y-4|
y >= 4 : + + + => y+8 -y-3+y-4 = 0 => y = 1, but y can't be 1 because y has to be >=4 --> no solution
-3 <=y<4 : + + - => y+8 -y -3 -y +4 =0 => y = 9 , but y can't be 9 because y has to be < 4 --> no solution
-8 <=y<-3: + - - => y+8 +y +3 -y +4 =0 => y = -15 , but y can't be -15 because y has to be >= -8 --> no solution
y < -8 : - - - => y+8 -y-3+y-4 = 0 => y = 1, but y can't be 1 because y has to be < -8 --> no solution
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:29
1
The intervals should be divided on the number line for -8, -3, 4.

In none of the intervals, the equation holds for a particular value of y.

So answer should be A

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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:33
1
solving the expression does not given any value to satisfy the given relation IMO A; 0


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

A. 0
B. 1
C. 2
D. 3
E. 4
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:34
3
1
How many different values of yy satisfy |y+3|=|8+y|+|4−y|

A. 0
B. 1
C. 2
D. 3
E. 4

if y is positive then |y+3| will always be less than |8+y| i.e. |y+3| can never be equal to |8+y|+|4−y|

If y is Negative then |y+3| will always be less than |4−y| i.e. |y+3| can never be equal to |8+y|+|4−y|

i.e. NO SOLUTION

Answer: Option A
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:35
1
Zeros of the equation:
y+3=0
8+y=0
4-y=0

y=-8; -3; 4

1st case: y<=-8

-y-3=-8-y+4-y >>
y=-1 (doesn't meet the criteria y<=-8)

2nd case: -8<=y<=-3
-y-3=8+y+4-y
-y-3=12
-y=15
y=-15 (doesn't meet the criteria -8<=y<=-3)

3rd case: -3<=y<=4
y+3=8+y+4-y
y+3=12
y=9(doesn't meet the criteria -8<=y<=-3)

4th case: y>=4
y+3=8+y-4+y
y=-1 (doesn't meet the criteria y>=4)

No solution
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:35
1
Quote:
|y+3|=|8+y|+|4−y||y+3|=|8+y|+|4−y|

By substituting values in the equation, we realise in all cases LHS < RHS.
for -ve values |4-y| will always be more than |y+3|
for +ve value |8+y| will always be more than |y+3|
Hence 0,
Option A
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:36
1
|y + 3| = |8 + y| + |4 - y|

We will check for mod points at -3,-8,4
if y>4
Then: \(y+3=8+y-4+y\)
y=1. Not possible since Y>4
-3<y<4
Then: \(y+3=8+y+4-y\)
y=11. Not possible since -3<y<4
-8<y<-3
Then: \(-y-3=8+y+4-y\),
y=-15.Not possible since -8<y<-3
-8>y
\(-y-3=-8-y+4-y\)
y=-1. Not possible since -8>y

zero solution. A
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:41
1
1
How many different values of y satisfy |y+3|=|8+y|+|4−y|?
A. 0
B. 1
C. 2
D. 3
E. 4

Let us divide the number line into 4 different regions: -
Region 1 where y>4
Region 2 where -3<y<=4
Region 3 where -8<y<=-3
Region 4 where y<=-8

Region 1 where y>4
expression becomes
y+3 = y+8 + y-4
=> y = -1
Not possible in the region since y>4

Region 2 where -3<y<=4
y+3= y+8+4-y
=> y=9
Not possible in the region since -3<y<=4

Region 3 where -8<y<=-3
-y-3 = y+8 +4-y
=> y = -15
Not possible in the region since -8<y<=-3

Region 4 where y<=-8
-y-3 = -y-8 +4-y
=> y = -1
Not possible in the region since y<=-8

Therefore, it may be concluded that no value of y satisfies the above equation.
=> Zero solutions (0)

IMO A
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How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post Updated on: 03 Jul 2019, 08:44
1
Let us draw a number line:

-8________-3_________0_________4
Clearly the distance of any possible value >4 is farthest from -8,
So it's not possible that

distance between that number and 3 = distance between that number & 4 + distance between that number & 8
Same concept applies for all the other zones, <-8, -8<y<-4.

So zero solutions.
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?  [#permalink]

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New post 03 Jul 2019, 08:44
1
Step-1
Identify the the positive and negative intervals against each absolute expression.

Step-2
Use the property of absolute function, |x|=x when x>0 and |x|=-x when x<0

Now the intervals are :
1)y<-8 2) -8<=y<-3 3) -3<=y<=4 4) y>4

Now testing the expression at the intervals , we have No solution for y.

Ans. (A)
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Re: How many different values of y satisfy |y + 3| = |8 + y| + |4 - y| ?   [#permalink] 03 Jul 2019, 08:44

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