You have |x+3| - |4-x| = |8+x|
First, look at the three values independently of their absolute value sign, in other words:
|x+3| - |4-x| = |8+x|
(x+3) - (4-x) = (8+x)
Now, you're looking at x < - 8, so x is a number less than -8. Let's pretend x = -10 here to make things a bit easier to understand.
when x=-10
I.) (x+3)
(-10+3)
(-7)
II.) (4-x)
(4-[-10]) (double negative, so it becomes positive)
(4+10)
(14)
III.) (8+x)
(8+-10)
(-2)
In other words, when x < -8, (x+3) and (8+x) are NEGATIVE. To solve problems like this, we need to check for the sign change.
Here is how I do it step by step.
I.) |x+3| - |4-x| = |8+x|
II.) IGNORE absolute value signs (for now) and find the values of x which make (x+3), (4-x) and (8+x) = to zero as follows:
(x+3)
x=-3
(-3+3) = 0
(4-x)
x=4
(4-4) = 0
(8+x)
x=-8
(8+-8) = 0
Order them from least to greatest: x=-8, x=-3, x=4 These become our ranges for x as follows:
x<-8
-8≤x<-3
-3≤x<4
x>4
So, we test values less than the smallest number, values of x between the smallest and largest number, and values of x greater than the greatest number.
So, now we test the original (x+3) - (4-x) = (8+x) with x values. This is where the sign changes in the equation become important. We need to find the number of solutions for this problem so we need to see for which values of x the problem is valid or not valid. For example:
When x < -8
(x+3) is a negative number
(4-x) is a positive number
(8+x) is a negative number
So
-(x+3) - (4-x) = -(8+x)
-x-3 -4+x = -8-x
-7=-8-x
1=-x
x=-1
Now, we are looking at values for x < -8, yet the result we got was x = -1. -1 DOES NOT fall in the range or x < -1. If you don't understand why simply draw a number line, mark down x< -8 and x=-1. Is -1 less than -8? Nope! Therefore, -1 is NOT a valid solution.
You can repeat this step for the remaining ranges of x.
I hope this helped you!![:-D]( "Very Happy")
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.
When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4
We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.
Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4
The only point is that we don't include the transition points twice.
Hope the role of '=' sign is clear.
Hi Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?
Thanks in advance,
RRSNATHAN.
First, look at the three values independently of their absolute value sign, in other words:
|x+3| - |4-x| = |8+x|
(x+3) - (4-x) = (8+x)
Now, you're looking at x < - 8, so x is a number less than -8. Let's pretend x = -10 here to make things a bit easier to understand.
when x=-10
I.) (x+3)
(-10+3)
(-7)
II.) (4-x)
(4-[-10]) (double negative, so it becomes positive)
(4+10)
(14)
III.) (8+x)
(8+-10)
(-2)
In other words, when x < -8, (x+3) and (8+x) are NEGATIVE. To solve problems like this, we need to check for the sign change.
Here is how I do it step by step.
I.) |x+3| - |4-x| = |8+x|
II.) IGNORE absolute value signs (for now) and find the values of x which make (x+3), (4-x) and (8+x) = to zero as follows:
(x+3)
x=-3
(-3+3) = 0
(4-x)
x=4
(4-4) = 0
(8+x)
x=-8
(8+-8) = 0
Order them from least to greatest: x=-8, x=-3, x=4 These become our ranges for x as follows:
x<-8
-8≤x<-3
-3≤x<4
x>4
So, we test values less than the smallest number, values of x between the smallest and largest number, and values of x greater than the greatest number.
So, now we test the original (x+3) - (4-x) = (8+x) with x values. This is where the sign changes in the equation become important. We need to find the number of solutions for this problem so we need to see for which values of x the problem is valid or not valid. For example:
When x < -8
(x+3) is a negative number
(4-x) is a positive number
(8+x) is a negative number
So
-(x+3) - (4-x) = -(8+x)
-x-3 -4+x = -8-x
-7=-8-x
1=-x
x=-1
Now, we are looking at values for x < -8, yet the result we got was x = -1. -1 DOES NOT fall in the range or x < -1. If you don't understand why simply draw a number line, mark down x< -8 and x=-1. Is -1 less than -8? Nope! Therefore, -1 is NOT a valid solution.
You can repeat this step for the remaining ranges of x.
I hope this helped you!
rrsnathan wrote:
VeritasPrepKarishma wrote:
guerrero25 wrote:
I am trying to understand the Modules questions - I took this from GMAT club's quant book .
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
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)
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.)
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.)
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)
thanks !
Q. |x+3| - |4-x| = |8+x|. How many solutions does the equation have?
I could not follow why the equal signs are considered ? e.g -8 <= x < -3 , -3 <= x < 4, x >=4 ..Appreciate if someone can explain the logic?
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)
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.)
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.)
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)
thanks !
|x| = x when x >= 0 (x is either positive or 0)
|x| = -x when x < 0 (note here that you can put the equal to sign here as well x <= 0 because if x = 0,
|0| = 0 = -0 (all are the same)
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.
When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4
We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.
Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4
The only point is that we don't include the transition points twice.
Hope the role of '=' sign is clear.
Hi Have a small doubt sounds silly but i need to understand this basic.
I could understand this part "There are 3 key points here: -8, -3, 4".
But why is that for all the cases like a) x < -8. -(x+3) - (4-x) = -(8+x) negative sign is added before the three brackets?
Thanks in advance,
RRSNATHAN.