If y=|x+5|−|x−5|

[ErikM411]

Couldn't find any subforum that this question fit into.

A question found on gmatclub that I have a hard time grasping.

I tried applying the "complex absolute value" thingy in the algebra strategy guide, but its still kinda unclear. Anyways, here goes

If y=|x+5|−|x−5|, then y can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21


I get that the midpoints are -5 and 5, but how do I decide on the actual range? I mean, how do I know how far to the right/left the values stretch?

[vintin]

Hello,

Let me try out the solution for you.

Consider the below scenarios for the equation y = |x+5|-|x-5|

Case 1:
when x more than or equal to 5
then y=(x+5)-(x-5) = 10
hence y=10

Case 2:
when -5<x<5
y=(x+5)-(-(x-5)) = 2x
y=2x so y can take 9 values corresponding to x={-4,-3,-2,-1,0,1,2,3,4}

Case 3:
when x less than or equal to -5
y= -(x+5)-(-(x-5))
y=-10


Hence if we combine all 3 cases we get that y can take total of 11 values.

Hope I am correct

Cheers ,
Vinay

[ErikM411]

Hello,

Let me try out the solution for you.

Consider the below scenarios for the equation y = |x+5|-|x-5|

Case 1:
when x more than or equal to 5
then y=(x+5)-(x-5) = 10
hence y=10

Case 2:
when -5<x<5
y=(x+5)-(-(x-5)) = 2x
y=2x so y can take 9 values corresponding to x={-4,-3,-2,-1,0,1,2,3,4}

Case 3:
when x less than or equal to -5
y= -(x+5)-(-(x-5))
y=-10


Hence if we combine all 3 cases we get that y can take total of 11 values.

Hope I am correct

Cheers ,
Vinay

Thanks, the answer is actually 21. But you seem to be on the right track.

I get the "y=10" and y=2x" part, but from there - I'm a bit lost.

"y=2x so y can take 9 values corresponding to x={-4,-3,-2,-1,0,1,2,3,4}" - Here is what I don't understand. How can we know what the stretch, or range, for x is?

[vintin]

Hello,

I defined and divided the complete range of x into 3 cases.

Case 1:
when x more than or equal to 5
then y=(x+5)-(x-5) = 10
hence y=10

Case 2:
when -5<x<5
y=(x+5)-(-(x-5)) = 2x
y=2x

Case 3:
when x less than or equal to -5
y= -(x+5)-(-(x-5))
y=-10


So i hope all the 3 cases which i have taken into consideration is okay with you.

so
for any value of x more or equal to 5 , y will always take the value of 10
for any value of x less than or equal to -5 , y will always take the value of -10.

hence the only range of values left to consider is -5<x<5

Now , since y can take only integer values and x can be only between -5<x<5
and y = 2x
So , the below combinations are available

x=4.5 , then y = 9
x=4 , then y = 8
x=3.5 , then y = 7
x=3 , then y = 6
x=2.5 , then y = 5
x=2 , then y = 4
x=1.5 , then y = 3
x=1 , then y = 2
x=0.5 , then y = 1
x=0 , then y = 0
x=-0.5 , then y = -1
x=-1 , then y = -2
x= -1.5 , then y = -3
x= -2 , then y = -4
x= -2.5 , then y = -5
x= -3 , then y = -6
x= -3.5 , then y = -7
x=-4 , then y = -8
x=-4.5 , then y = -9

so y can take 19 values from this case.

apart from this from case 1 and 3 , we get y =10 and -10

so in total , y has 21 values ..

Hope , i make this clear for you

PS: in my earlier solution i assumed X is also a integer

Cheers,
Vinay

[RonPurewal]

^^ that's a good summary.

in any case, if you don't see any other key to a problem like this one, just start experimenting with numbers. if you just toss enough numbers into this expression, you'll see how it behaves.
it will be particularly clear that it stays "frozen" at 10 for large enough positive values, and at -10 for large enough negative values.

[vintin]

Thanks a lot Ron

[RonPurewal]

you're welcome.